Part of a light ray striking an interface between air and water is refracted, and part is reflected, as shown. The index of refraction of air is 1.00 and the index of refraction of water is 1.33. The frequency of the light ray is 7.85 x 10^16 Hz.(a) If angle 1 measures 40°, find the value of angle 2.(b) If angle 1 measures 40°, find the value of angle 3.(c) Calculate the speed of the light ray in the water.(d) Calculate the wavelength of the light ray in the water.(e) What is the largest value of angle 1, that will result in a refracted ray?
Answers
We will use Snell's law, which states:
[tex]n_1\sin \theta_1=n_2\sin \theta_2[/tex]Where n1 and n2 are the refraction indexes and their respective angles are "theta1" and "theta2".
For part A we replace the values:
[tex]1\sin 40=1.33\sin \theta_2[/tex]Now we solve for "theta2" first by dividing both sides by 1.33:
[tex]\frac{\sin40}{1.33}=\sin \theta_2[/tex]Now we use the inverse function for sine:
[tex]\arcsin (\frac{\sin 40}{1.33})=\theta_2[/tex]Solving the operation:
[tex]28.9=\theta_2[/tex]For part B, since "theta1" and "theta3" are angles of reflection, according to the reflection law, these angles are equal, therefore:
[tex]\theta_3=\theta_1=40[/tex]For part C. The index of refraction is defined as:
[tex]n=\frac{c}{v}[/tex]Where "c" is the speed of light in a vacuum and "v" is the speed of light in the medium. Replacing the values:
[tex]1.33=\frac{3\times10^8\text{ m/s}}{v}[/tex]Now we solve for "v":
[tex]v=\frac{3\times10^8\text{ m/s}}{1.33}[/tex]Solving the operation:
[tex]v=2.26\times10^8\text{ m/s}[/tex]For part d. We will use the following formula:
[tex]\lambda=\frac{v}{f}[/tex]Where "v" is the speed and "f" is the frequency. Replacing we get:
[tex]\lambda=\frac{2.26\times10^8\text{ m/s}}{7.85\times10^{16}s^{-1}}[/tex]Solving the operations:
[tex]\lambda=0.288\times10^{-8}m[/tex]For part e. The largest value of the angle of incidence that will result in refraction is 90 degrees.
Related Questions
A Tour de France cyclist at a rate of 6.5 m/s and is 333 m
ahead of a crew van with powerdrink refills. The van is
traveling at 15 m/s and is accelerating at a constant rate of
0.4 m/s2
How much time will it take for the crew van to catch up
with the cyclist?
Answers
Given data:
* The speed of the cyclist is 6.5 m/s.
* The initial distance of the cyclist from the van is 333 m.
* The initial velocity of the van is 15 m/s.
* The acceleration of the van is,
[tex]a=0.4ms^{-2}[/tex]Solution:
Let x be the distance from the van initial position at which the cyclist and van meet.
Let the cyclist meet the van at time t.
By the kinematics equation, the position of the cyclist at time t is,
[tex]x-333=u_ct+\frac{1}{2}a_ct^2[/tex]where u_c is the speed of the cyclist, a_c is the acceleration of the cyclist, t is the time taken and 333-x is the distance traveled by the cyclist at time t,
The acceleration of the cyclist is zero.
Substituting the known values,
[tex]\begin{gathered} x-333=6.5t+0 \\ x-333=6.5t \end{gathered}[/tex]By the kinematics equation, the position of the van after time t is,
[tex]x=u_vt+\frac{1}{2}a_vt^2[/tex]where u_v is the velocity of the van, a_v is the acceleration of the van, and t is the time taken,
Substituting the known values,
[tex]\begin{gathered} x=15t+\frac{1}{2}\times0.4\times t^2 \\ x=15t+0.2t^2 \end{gathered}[/tex]Substituting this value of x in the kinematics equation of the cyclist,
[tex]\begin{gathered} (15t+0.2t^2)-333=6.5t \\ 15t+0.2t^2-6.5t-333=0 \\ 0.2t^2+8.5t-333=0 \end{gathered}[/tex]By solving the quadratic equation,
[tex]\begin{gathered} t=\frac{-8.5\pm\sqrt[]{8.5^2-(4\times0.2\times(-333))}}{2\times0.2} \\ t=\frac{-8.5\pm\sqrt[]{^{}8.5^2+(4\times0.2\times333)}}{2\times0.2} \\ t=\frac{-8.5\pm18.4}{0.4} \\ t=24.8\text{ s or-67.25 s} \end{gathered}[/tex]As the value of time cannot be negative.
Thus, the time at which the cyclist and van meet is 24.8 seconds.
A flywheel with a moment of inertia of 3.45 kg·m2is initially rotating. In order to stopits rotation, a braking torque of -9.40 N·m is applied to the flywheel. Calculate the initialangular speed of the flywheel if it makes 1 complete revolution from the time the brake isapplied until it comes to rest
Answers
Given data
*The given moment of inertia is I = 3.45 kg.m^2
*The given braking torque is T = -9.40 N.m
*The angular distance traveled is
[tex]\theta=(1\times2\pi)rad_{}[/tex]*The final angular speed is
[tex]\omega=0\text{ rad/s}[/tex]The angular acceleration of the flywheel is calculated by using the torque and moment of inertia relation as
[tex]\begin{gathered} T=I\alpha \\ \alpha=\frac{T}{I} \\ =\frac{-9.4}{3.45} \\ =-2.72rad/s^2 \end{gathered}[/tex]The formula for the initial angular speed of the flywheel is given by the rotational equation of motion as
[tex]\omega^2-\omega^2_0=2a\theta[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} (0)^2-\omega^2_0=2\times(-2.72)(2\pi) \\ \omega_0=\sqrt[]{2\times2.72\times2\pi} \\ =5.88\text{ rad/s} \end{gathered}[/tex]Hence, the initial angular speed of the flywheel is 5.88 rad/s
The source of the Sun’s heat and light energy is:A. combustion of helium gas.B. fusion of hydrogen nuclei.C. gravitational pressure.D. burning of fossil fuels.
Answers
To find
The source of the Sun’s heat and light energy is:
Explanation
The sun's core is very hot. So under pressure nuclear fusion takes place. Here hydrogen is changed to helium.
Conclusion
The correct option is
B. fusion of hydrogen nuclei.
When you apply a force F on an object of mass of m, it would produce acceleration a. If you apply the same force on another object of mass 2m, how would be the acceleration of the second object?
Answers
ANSWER
[tex]\frac{a}{2}[/tex]EXPLANATION
When a force is applied on an object of mass m, it produces an acceleration of a.
We can represent this relationship using Newton's second law of motion:
[tex]\begin{gathered} F=ma \\ \Rightarrow a=\frac{F}{m} \end{gathered}[/tex]Now, the same force is applied on an object with a mass of 2m.
Let the acceleration experienced by the object be a1. This implies that:
[tex]\begin{gathered} F=\left(2m\right)\left(a_1\right) \\ \Rightarrow a_1=\frac{F}{2m} \end{gathered}[/tex]We can write this new acceleration in terms of a as follows:
[tex]\begin{gathered} a_1=\frac{1}{2}\left(a\right) \\ a_1=\frac{a}{2} \end{gathered}[/tex]That would be the acceleration of the second object.
The answer is the third option.
Grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net. With what speed did he leave the floor?
Answers
Grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net, the speed from which he would have left the floor would be 4.64 m / s .
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2 × a × t²
v² - u² = 2 × a × s
As given in the problem grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net.
By using the third equation of the motion,
v² - u² = 2 × a × s
0 - u² = 2 × -9.81 × 1.10
u = 4.64 m / s
Thus, the speed from which he would have left the floor would be 4.64 m / s .
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What is the momentum of a 934 kg car moving 10 m/s?
Answers
ANSWER:
9340 kg*m/s
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 934 kg
Speed (v) = 10 m/s
The formula to calculate the momentum is as follows:
[tex]\begin{gathered} p=m\cdot v \\ \text{ we replacing} \\ p=934\cdot10 \\ p=9340\text{ kg*m/s} \end{gathered}[/tex]The momentum of the car is 9340 kg*m/s.
A bus is travelling along straight road at 100km/hr and the bus conductor walks a 6km/hr on the floor of the bus and in the same direction as the bus. Find the speed of the conductor relative to the road and relative to the bus. If the bus conductor now walks at the same rate but in opposite direction as the bus, find his new speed relative to the road.
Answers
Given data:
* The speed of the bus is 100 km/hr.
* The speed of the conductor is 6 km/hr.
Solution:
(a). If the bus and conductor are traveling in the same direction.
The net speed of the conductor relative to the road is,
[tex]v_1=v_b+v_c_{}[/tex]where v_1 is the velocity of the conductor relative to the road, v_b is the velocity of the bus, and v_c is the velocity of the conductor inside the bus,
Substituting the known values,
[tex]\begin{gathered} v_1=100+6 \\ v_1=106\text{ km/hr} \end{gathered}[/tex]Thus, the speed of the conductor relative to the road is 106 km/hr.
(b). The speed of the conductor relative to the bus is the speed of the conductor on the bus,
Thus, the speed of the conductor relative to the bus is 6 km/hr.
Using the work energy theorem, what is the final velocity of a roller coaster at the bottom of the hill. The coaster has a mass of 839 kg and starts at rest from the top of a hill that is 75 meters tall.
Answers
ANSWER
38.34 m/s
EXPLANATION
Given:
• The mass of the coaster, m = 839 kg
,• The initial height of the coaster, h = 75 m
,• The acceleration due to gravity, g = 9.8 m/s²
Find:
• The final velocity of the roller coaster at the bottom of the hill, v.
The roller coaster starts from rest, so at the top of the hill, it only has gravitational potential energy and no kinetic energy. Then, at the bottom of the hill, the roller coaster is in motion, to it has kinetic energy, and, because the difference of height with the reference - which is the bottom of the hill, is zero, it has no potential energy,
By the work-energy theorem, we have the equation,
[tex]KE_i+PE_i+W_{nc}=KE_f+PE_f_{}[/tex]As explained above, the initial kinetic energy is 0 and the final potential energy is also 0. If we assume that there is no friction, air resistance, or other external forces, then the work done by non-conservative forces is also 0,
[tex]PE_i=KE_f[/tex]Replace each kind of energy with the expression to obtain them,
[tex]m\cdot g\cdot h=\frac{1}{2}\cdot m\cdot v^2[/tex]The mass cancels out,
[tex]g\cdot h=\frac{1}{2}\cdot v^2[/tex]Solving for v,
[tex]v=\sqrt[]{2\cdot g\cdot h}[/tex]Replace with the known values and solve,
[tex]v=\sqrt[]{2\cdot9.8m/s^2\cdot75m}=\sqrt[]{1470m^2/s^2}\approx38.34m/s[/tex]Hence, the velocity of the roller coaster at the bottom of the hill is 38.34 m/s, rounded to the nearest hundredth.
Mass/energy equivalence is expressed mathematically in which of the following expressions?
Answers
Check each option to see how it relates to different concepts.
Option 1: E=hf
This equation tells the energy carried by an electromagnetic wave with frequency f.
Option 2: E=mc
This equation is not correct, since the left member is measured in units of energy and the right member does not.
Option 3: E=(1/2)mv^2
This equation relates the energy of a moving object with its mass and its velocity. It is known as kinetic energy.
Option 4: E=mc^2
Since c is a constant (the speed of light), this equation relates the energy of an object with its mass.
Therefore, the mass/energy equivalence is expressed mathematically in the equation:
[tex]E=mc^2[/tex]Two identical point charges exert a repulsive force of 0.500 N on one another when separated by 1.5 m. What is the magnitude of the net charge of either point charge?
Answers
Given,
The repulsive force exerted by the charges, F=0.500 N
The distance between the charges, d=1.5 m
From Coulomb's law,
[tex]F=\frac{\text{kqq}}{r^2}[/tex]Where q is the magnitude of the charge of each point charge and k is the coulomb's constant.
On rearranging the above equation,
[tex]\begin{gathered} F=\frac{kq^2}{r^2} \\ \Rightarrow q=\sqrt[]{\frac{F}{k}}r \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} q=\sqrt[]{\frac{0.5}{9\times10^9}}\times1.5 \\ =1.1\times10^{-5}\text{ C} \end{gathered}[/tex]Thus the magnitude of the charge of each point charge is 1.1×10⁻⁵ C
Therefore the correct answer is option B.
An electric oven has a resistance of 50.0 Ω and a voltage of 220 V. How much current does it draw?
Answers
Given
Resistance of the oven, R=50.0 ohm,
Voltage is V=220 V
To find
The current drawn
Explanation
Let the current be I
By Ohm's law,
[tex]V=RI[/tex]Putting the values,
[tex]\begin{gathered} 220=50I \\ \Rightarrow I=4.4A \end{gathered}[/tex]Conclusion
The current drawn is 4.4 A
some more disconnectiob
Answers
let the student report the tutor and check
answer is updated
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The position of a particle is F(t) = 4.01²î - 3.0ĵ +2.03 km. (a) What is the velocity of the particle at 0 s and at 1.0 s? (b) What is the average velocity between 0 s and 1.0
Answers
The velocity of the particle at 0 s and at 1.0 and the average velocity between 0 s and 1.0
a)[tex]\vec{V}(0)=0 m / s[/tex]
[tex]\vec{V}(1)=(8 \hat{\imath}+6 \hat{k}) m / s[/tex]
b)[tex](4 \hat{\imath}+2 \hat{k}) \mathrm{m} / \mathrm{s}[/tex]
This is further explained below.
What is the average velocity?Generally, the equation for the Position is mathematically given as
F(t) = 4.01²î - 3.0ĵ +2.03 km.
Therefore
[tex]\begin{aligned}&\vec{r}(t)=\left(4 t^2 \hat{l}-3 \hat{\jmath}+2 t^3 \hat{k}\right) m \\&\vec{V}(t)=\frac{d \vec{r}(t)}{d t}=\left(8 t \hat{\imath}+6 t^2 \hat{k}\right) m / s\end{aligned}[/tex]
For A
[tex]\begin{aligned}&\vec{V}(0)=0 m / s \\&\vec{V}(1)=(8 \hat{\imath}+6 \hat{k}) m / s\end{aligned}[/tex]
For B
[tex]\text { Average velocity } &=\frac{\text { Total displacement }}{\text { time interval }} \\[/tex]
[tex]=\frac{\vec{r}(1)-\vec{r}(0)}{1} \\&=4 \hat{\imath}-3 \hat{\jmath}+2 \hat{k}-\left(-3 \hat{\jmath}^{\prime}\right) \\&=(4 \hat{\imath}+2 \hat{k}) \mathrm{m} / \mathrm{s}\end{aligned}[/tex]
In conclusion, The speed of the particle at 0 seconds and at 1.0 seconds, as well as the average speed between 0 seconds and 1.0 seconds
a)[tex]\vec{V}(0)=0 m / s[/tex]
[tex]\vec{V}(1)=(8 \hat{\imath}+6 \hat{k}) m / s[/tex]
b)[tex](4 \hat{\imath}+2 \hat{k}) \mathrm{m} / \mathrm{s}[/tex]
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if i put my spoon in the microwave what will happen
Answers
If you put your spoon in the microwave, nothing will happen to it as long as it has round edges and it is not touching any sides of the microwave.
Putting a spoon in the microwaveSpoons are generally made from stainless steel. Stainless steels are iron and chromium. In some cases, other metals such as nickel are thrown into the mix.
Thus, being made of metals, these stainless steels are good conductors of heat and electricity.
A microwave works by heating foods put into it using electromagnetic radiation in the microwave frequency range. Electromagnetic radiation causes the molecules of food to rotate and produce thermal energy.
If the spoon touches the side of the microwave, sparks may result. Otherwise, it is totally fine because spoons usually have round edges. With rough edges, the waves may be reflected back and forth and create sparks.
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44) Find the x coordinate of the center of mass of the bricks shown.
Answers
We are asked to determine the x-coordinate of the center of mass of the given bricks. To do that, we will use the following formula:
[tex]\bar{x}=\frac{\Sigma x_im_i}{\Sigma m_i}[/tex]Where:
[tex]\begin{gathered} x_i=\text{ x-coordinate of the center of mass of each brick} \\ m_i=\text{ mass of each brick} \end{gathered}[/tex]Since we have three bricks, the formula expands to:
[tex]\bar{x}=\frac{x_1m_1+x_2m_2_{}+x_3m_3}{m_1+m_2+m_3}[/tex]Since we have three bricks with the same characteristics we will assume the three of them have the same mass:
[tex]\bar{x}=\frac{x_1m_{}+x_2m+x_3m_{}}{m_{}+m_{}+m_{}}[/tex]Taking "m" as a common factor and adding like terms in the denominator we get:
[tex]\bar{x}=\frac{m(x_1+x_2+x_3)}{3m}[/tex]Now we cancel out the "m":
[tex]\bar{x}=\frac{x_1+x_2+x_3}{3}[/tex]Now we determine the x-coordinates of each brick. Each brick is a parallelepiped, therefore, the x-coordinate is in the middle. Since each brick measures L, this means that the x-coordinate of the first brick is:
[tex]x_1=\frac{L}{2}[/tex]For the second brick, we have the L/2 of the separation from the first plus the L/2 of its length, therefore:
[tex]x_2=\frac{L}{2}+\frac{L}{2}=L[/tex]Now, for the third brick we have the L/4 of the separation from the second brick plus the L/2 of the separation of the second brick and the first brick and the L/2 of the length of the third brick, therefore:
[tex]x_3=\frac{L}{2}+\frac{L}{4}+\frac{L}{2}=\frac{5L}{4}[/tex]Now we substitute in the formula for the x-coordinate:
[tex]\bar{x}=\frac{(\frac{L}{2})+(L)+(\frac{5L}{4})}{3}[/tex]Adding like terms in the numerator:
[tex]\bar{x}=\frac{\frac{11L}{2}}{3}[/tex]Simplifying:
[tex]\bar{x}=\frac{11L}{6}[/tex]Therefore, the x-coordinate of the center of mass is located at 11L/6 from the origin.
Peter is trying to ignite the hotplate by turning the gas knob. Suppose thatthe minimum moment of couple about the center of the gas knob requiredto ignite the hotplate is 0.3 N m. Calculate the minimum force (F, and F2)that required to exert. Given that the diameter of gas knob is 5 cm.
Answers
the minimum force is 6 Newtons
Explanation
A moment of a force, or a torque, is a measure of a force's tendency to cause a body to rotate. The moment depends on both the force, and on the position at which the force acts, it is given by the expression
[tex]M=F\cdot d\text{ }[/tex]where F is the exerted force and d is the distance
Step 1
then, let
[tex]\begin{gathered} M=\text{0}.3\text{ Nm} \\ F=F \\ \text{distance}=\text{ 5 cm= }\frac{5}{100}m=0.05\text{ m} \\ \end{gathered}[/tex]now, replace in the formula
[tex]\begin{gathered} M=F\cdot d\text{ } \\ 0.3\text{ Nm=F}\cdot0.05\text{ m} \\ \text{divide both sides by 0.05 m} \\ \frac{0.3\text{ Nm}}{0.05\text{ m}}\text{=}\frac{\text{F}\cdot0.05\text{ m}}{0.05\text{ m}} \\ 6N=F \end{gathered}[/tex]therefore, the minimum force is 6 Newtons
I hope this helps you
On a standard day the speed of sound is 345 meters per second. A whistle whose frequency is 725 Hz is movingtoward an observer at a speed of 25.2 meters per second. What is the wavelength of the sound at the observer?(a) 0.367 m(b) 0.441 m(c) 0.511 m(d) 0.623 m
Answers
Take into account that this is a situation where the source moves presenting the Doppler effect.
In order to determine the wavelength of the sound generatd by the whistle at the observer, first calculate the frequency at the observer by using the following formula:
[tex]f=\frac{v}{v-v_s}f_s[/tex]where:
f: frequency at the observer = ?
fo: source frequency = 725 Hz
vs: source speed = 25.2 m/s
v: speed of sound = 345 m/s
replace the previous values of the parameters into the fomrula for f:
[tex]\begin{gathered} f=\frac{345m/s}{345m/s-25.2m/s}725Hz \\ f=782.1Hz \end{gathered}[/tex]Next, use the following formula to determine the wavelength of the sound at observer, by using the previous result:
[tex]\lambda=\frac{v}{f_s}=\frac{345m/s}{782.1Hz}=0.441m[/tex]Hence, the wavelength of the sound at the observer is 0.441 m
When a rubber band with a force constant of 58.7 N/m is stretched a certain distance,
there is 1.94 J of elastic potential NRG stored in it. How far has the band been stretched
Answers
According to the given statement the band has been stretched far is 0.25 m.
What is the meaning of elastic potential?The energy that is stored when a force is used to bend an elastic object is known as elastic potential energy. Until the force is released as well as the object springs returns to its original form, doing work in the process, the energy is retained. The object may be squeezed, stretched, or bent during the deformation.
What's an example of elastic potential energy?For example, if you pull a spring, it will return to its initial form when you release it (energy input equals energy output.) Elastic potential energy is rendered possible by this.
Briefing:U= 1.94 J
k = 58.7 N/m
U = 1/2 kΔ x²
1.94 = 1/2 kΔ x²
3.88/58.7 = Δ x²
x = √(3.88 /58.7)
x = 0.25 m
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You and some friends are duck pin bowling. You’re up and you roll your bowling ball (m = 1.6kg) down the lane. It collided with one pin (0.68kg) on the end in a perfectly elastic one dimensional collision. If the ball was moving at a velocity of 6 m/s just before it hits the pin, what is the velocity of the bowling pin after the collision?
Answers
Given:
The mass of the bowling ball is m1 = 1.6 kg
The mass of the pin is m2 = 0.68 kg
The initial velocity of the ball is
[tex]v_i=\text{ 6 m/s}[/tex]Required: Velocity of the bowling pin after the collision.
Explanation:
According to the conservation of momentum, the velocity after the collision will be
[tex]\begin{gathered} m1v_i+m2\times0\text{ =\lparen m1+m2\rparen v}_f \\ v_f=\frac{m1v_i}{m1+m2} \\ =\frac{1.6\times6}{1.6+0.68} \\ =4.21\text{ m/s} \end{gathered}[/tex]Final Answer: The velocity of the bowling pin after the collision is 4.21 m/s
A wooden sphere of mass 4.0 kg is completely immersed in water. A pushing force of 20. N is
applied.
21°
20 N
4.2 ms²
19⁰
At the moment shown in the diagram, the sphere is stationary and it experiences an
acceleration upwards and to the right as shown.
Calculate the size of the upwards force due to the water (upthrust) acting on the sphere.
Answers
The size of the upwards force due to the water (upthrust) acting on the sphere is 12.64 N.
What is upthrust?
Buoyancy or upthrust, is an upward force exerted by a fluid on an object immersed in the fluid due to the weight of the object
Thus, upthrust is the upward force acting on an object immersed in a liquid.
Fu - Df = F(net_u)
where;
Fu is the upward forceDf is the downward force applied on the objectF(net_u) is the net upward forceFu - F x sin(21) = ma x sin(19)
where;
m is the mass of the wooden spherea is the upward acceleration of the wooden sphereFu - 20 x sin(21) = (4 x 4.2) x sin(19)
Fu - 7.17 = 5.47
Fu = 12.64 N
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I dont understand this formula I need helpF = 6.67408 * 10^-11 * (1.5 * 10^5) (8.5 * 10^2) ------------------------------ 2500^2
Answers
Answer:
F = 1.36*10^-9
Explanation:
The given equation is
[tex]F=6.67\ast10^{-11}\ast\frac{(1.5\ast10^5)(8.5\ast10^2)}{2500^2}[/tex]To calculate the value of F, you need to multiply 1.5*10^5 by 8.5*10^2 as follows:
[tex]F=6.67\ast10^{-11}\ast\frac{1.275\times10^8}{2500^2}[/tex]Then, 2500² = 2500 x 2500 = 6.25 * 10^6, so replacing the value, we get:
[tex]F=6.67\ast10^{-11}\ast\frac{1.275\ast10^8}{6.25\ast10^6}[/tex]Now, we need to divide 1.275*10^7 by 6.25*10^6, so
[tex]F=6.67\ast10^{-11}\ast(20.4)[/tex]Finally, multiply 6.67*10^-11 by 20.4, so
[tex]F=1.36\ast10^{-9}[/tex]Newton’s Second Law states “The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.” Explain how your observations in both a and b support this Law.
Answers
Newton’s Second Law states “The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.”
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
Force = mass × acceleration
Assuming the force constant the acceleration is inversely proportional to the mass of the object.
Thus, acceleration is directly proportional to force and inversely proportional to the mass of the body.
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cassy can get more force on the bricks she breaks with a blow of her bare hand when _______.
Answers
Answer:Her hand is made to bounce from the bricks
Explanation:
Cassy can get more force on the bricks that she breaks with a blow of her bare hand when her hand is made to bounce from the bricks,
What is Force?A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's speed can vary, or accelerate, as a result of a force. Intuitively, a pull or a push can also be used to describe force. Being a vector quantity, a force also has magnitude and direction. The SI unit of newton is used to measure it (N). The letter F stands for force.
According to Newton's second law's original formulation, an object's net force is equal to the speed at which momentum is changing over time.
Objects' velocities can be altered by the concepts of push, drag, and torque. Thrust causes an object's velocities to increase, while torque causes an object's velocities to decrease. Each part of an extended body typically exerts pressures on its neighboring sections, and the internal mechanical force is determined by how these stresses are distributed.
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A rock sample has a mass of 6 kg and a volume of 0.002 m3. Calculate the density of this rock sample.
Answers
So 6/0.002=3,000 kg/m3.
The density of the rock sample is 3000 kg/m³.
Density is an important property of matter because it can be used to determine the weight of an object. For example, if you know the density of a rock sample and its volume, you can calculate its weight by multiplying the density by the volume. Density can also be used to determine the composition of a material. For example, if you know the densities of different materials, you can identify the material of a rock sample by measuring its density.
Density is a measure of how much mass is contained in a given volume.. The calculation of the density of the rock sample:
Density = Mass / Volume
Density = 6 kg / 0.002 m³
Density = 3000 kg/m³
As a result, the rock sample has a density of 3000 kg/m³.
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How much work does Scott do to push a 74 kg sofa 2.1 m across the floor at a constant speed? The coefficient of kinetic friction between the sofa and the floor is 0.23.
Answers
Work does Scott do to push a 74 kg sofa 2.1 m across the floor at a constant speed. The coefficient of kinetic friction between the sofa and the floor is 0.23 is 349 Nm.
given that :
mass = 74 kg
distance d = 2.1 m
coefficient of kinetic friction , μk = 0.23
work done is given as :
w = fd
f = μk m g
f = 0.23 × 74 × 9.8
f = 166 N
therefore ,
work = fd
w = 168 × 2.1
w = 349 Nm
Work does Scott do to push a 74 kg sofa 2.1 m across the floor at a constant speed. The coefficient of kinetic friction between the sofa and the floor is 0.23 is 349 Nm.
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Writing Simple ExpressionsChoose all of the TRUE statement(s).Add 6 and 5, then multiply by 4 is the same as 4(6 + 5).4 times greater than 80 + 25 is the same as 4 x (80 + 25).Subtract 15 from 42 is the same as 15 − 42.9 times greater than 11 + 12 is the same as 9 + 11 + 12.8 times greater than 21 + 15 is the same as 8(21 + 15).
Answers
Add 6 and 5, then multiply by 4 is the same as 4(6 + 5). TRUE
4 times greater than 80 + 25 is the same as 4 x (80 + 25). TRUE
Subtract 15 from 42 is the same as 15 − 42. TRUE
9 times greater than 11 + 12 is the same as 9 + 11 + 12. FALSE
9 (11+12)
8 times greater than 21 + 15 is the same as 8(21 + 15). TRUE
A 65.0-kg basketball player jumps vertically and leaves the floor with a velocity of 1.80 m/s upward. (a) What impulse does the player experience? (b) What force does the floor exert on the player before the jump? (c) What is the total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump?
Answers
(a) The impulse experienced by the player is 117 Ns.
(b) The force the floor exert on the player before the jump is 637 N.
(c) The total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump is 260 N.
What impulse does the player experience?
The impulse experienced by the player is the change in the momentum of the player.
J = ΔP
J = m(v - u)
where;
m is the mass of the playerv is the final velocity of the player = 0u is the initial velocity of the player = - 1.8 m/s (negative because of upward direction)J = 65 x (0 + 1.8)
J = 117 Ns
The force the floor exert on the player before the jump is calculated as;
F = mg
where;
g is acceleration is due to gravityF = 65 x 9.8
F = 637 N
The total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump is calculated as;
F = ma
F = m(v/t)
F = (mv)/t
F = (65 x 1.8) / 0.45
F = 260 N
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Four wires running through the corners of a square with sides of length 16.166 cm carry equal currents, 3.684 A. Calculate the magnetic field at the center of the square.
Answers
For practical reasons, we can consider each side of the square as an infinite wire. This can be seen on the following drawing:
This way, the field on the center will be the sum of the contribution of each wire. We can calculate the contribution of a single wire as:
[tex]B=\frac{\mu_0i}{2\pi d}=\frac{4\pi *10^{-7}*3.684}{2\pi(\frac{16.166*10^{-2}}{2})}=9.115*10^{-6}T[/tex]Then, the total field will be this, multiplied by the number of wires:
[tex]B_t=4*9.115*10^{-6}=36.46\mu T[/tex]Then, the resulting field will be Bt=36.46uT
Two Styrofoam blocks are brought near each other and are observed to repel each other. Each block has the same amount of charge on it. Which statement is true about this situation? A)the charge on each block must be positive B)the charge on each block must be negative C)the blocks are creating magnetic fields from non moving electric charges D)each block has the same kind of charge but we don't' know what kind it is
Answers
ANSWER:
D)each block has the same kind of charge but we don't' know what kind it is
STEP-BY-STEP EXPLANATION:
Two charges of the same type always repel each other.
In this case, it is not possible to know what type of load they are, only that they are the same.
Therefore, the correct answer is:
D)each block has the same kind of charge but we don't' know what kind it is
A sled of mass 26 kg has an 18 kg child on it. If big brother is pulling with a 30 N force to the right and 10 N up, and big sister is pushing with a 40 N force to the right and 16 N down, what is the normal force?
Answers
Given data:
* The mass of the sled is m_1 = 26 kg.
* The mass of the child is m_2 = 18 kg.
* The force in the upwards direction by the big brother is F_1 = 10 N.
* The force in the downwards direction by the big sister is F_2 = 16 N.
Solution:
The net mass on the sled along with the child is,
[tex]\begin{gathered} m=m_1+m_2 \\ m=26+18 \\ m=44\text{ kg} \end{gathered}[/tex]The net weight of the sled along with the child is,
[tex]\begin{gathered} w=mg \\ w=44\times9.8 \\ w=431.2\text{ N} \end{gathered}[/tex]The weight of the sled along the child is acting on the sled in the downwards direction.
Thus, the normal force acting on the sled (taking upward force as negative and downward force as positive) is,
[tex]\begin{gathered} N=w+F_2-F_1 \\ N=431.2+16-10 \\ N=437.2\text{ newton} \end{gathered}[/tex]Thus, the normal force acting on the sled is 437.2 N.