orce and Motion Unit TestUse the following scenario to answer the question.Taj and Micah chose to go bowling. Taj rolled the ball toward the pins first, knocking them all down.Which of the following is affecting these objects?point)O Gravity is affecting these objects.O An unbalanced force is affecting the objects.O Inertia is affecting these objects.O A balanced force is affecting the objects.
Answers
So lets go through all four answer choices.
The easiest to choose is whether gravity is affecting these objects. Assuming that there is some sort of gravity that would pull the pins down, gravity does affect these objects
Second is inertia.
We know that if an object has inertia, it will try to resist moving/coming to rest. In this case, we know that the pins have inertia because the pins fell over, so we know that the pins do have inertia
The last part is whether these objects have an unbalanced or balanced force. If a balanced force did exist, there would need to a force that would equally counteract the force of the bowling ball, which there isn't. Which means there is an unbalanced force affecting these objects.
Given that the user must choose one, the correct answer would be that an unbla
Related Questions
IThe alignment of electrons in a magnetic field does not contribute to the properties of magnets true or false?
Answers
To determine whether the given statement is true or false.
Explanation:
The alignment of electrons in a particular direction in a magnetic field form magnetic domains.
These magnetic domains are responsible for magnetic properties.
Thus the given statement is true.
Please do this step-by-step how do you do it when it’s between
Answers
Given:
• Mass of block A = 6.0 kg
,• Mass of block B = 7.0 kg
,• Mass of block C = 13.0 kg
,• Force, F = 13.0 N
Let's find the magnitude of the tension in the rope between B and C.
Let's first find the acceleration.
We have:
[tex]13-T_B+T_B-T_A+T_A=6a+7a+13a[/tex]Thus, we have:
[tex]\begin{gathered} 13=26a \\ \\ a=\frac{13}{26} \\ \\ a=0.5\text{ m/s}^2 \end{gathered}[/tex]To find the tension between blocks B and C, we have the equation:
[tex]\begin{gathered} F-T_B=M_C*a \\ \\ T_B=F-M_c*a \end{gathered}[/tex]Where:
F = 13 N
Mc is the mass of block C = 13 kg
a is the acceleration = 0.5 m/s²
Thus, we have:
[tex]\begin{gathered} T_B=13-13*0.5 \\ \\ T_B=13-6.5 \\ \\ T_B=6.5\text{ N} \end{gathered}[/tex]Therefore, the magnitude of the tension in the rope between blocks B and C is 6.5 N
ANSWER:
6.5 N
IBangkok, Thailand, and Lima, Peru, are located at opposite positions of Earth's surface. When a rock is dropped in both cities, observers would describe the motion of either rock with a word that means "down."
Where are the two rocks moving?
Responses
They are moving away from each other.
.
One rock is moving due east and the other is moving due west.
They are moving down, which is the same direction in both cities.
They are moving toward each other.
Answers
Answer :dawg you got snaids thats the answer awww man
Explanation:
answer is that of an answer
A jet engine applies a force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m, the kinetic energy gained by the rocket is
Answers
We will have the following:
[tex](300N)(2m)+(-200N)(2m)=200J[/tex]Then, we from the work-kinetic force theorem we will have that the total kinetic energy gained by the rocket was 200 Joules.
An arrangement of two pulleys, as shown in the figure, is used to lift a 64.3-kg crate a distance of 4.36 m above the starting point. Assume the pulleys and rope are ideal and that all rope sections are essentially vertical.What is the change in the potential energy of the crate when it is lifted a distance of 4.36 m? (kJ)How much work must be done to lift the crate a distance of 4.36 m? (kJ)What length of rope must be pulled to lift the crate 4.36 m? (m)
Answers
Given data:
* The mass of the crate is m = 64.3 kg.
* The height of the crate is h = 4.36 m.
Solution:
(a). The potential energy of the crate at the initial state is zero (as the height of the crate is zero at the initial state), thus, the change in the potential energy of the crate is,
[tex]\begin{gathered} dU=\text{mgh}-0 \\ dU=mgh \end{gathered}[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} dU=64.3\times9.8\times4.36 \\ dU=2747.4\text{ J} \\ dU=2.75\times10^3\text{ J} \\ dU=2.75\text{ kJ} \end{gathered}[/tex]Thus, the change in the potential energy is 2.75 kJ.
(b). The work done to lift the crate is equal to the change in the potential energy of the crate.
Thus, the work done on the crate is 2.75 kJ.
(c). As the single is pulling the two ropes to increase the height of the crate, thus, the length of the rope pulled in terms of the height of the crate is,
[tex]l=2h[/tex]Substituting the known values,
[tex]\begin{gathered} l=2\times4.36 \\ l=8.72\text{ m} \end{gathered}[/tex]Thus, the length of the rope pulled to lift the crate is 8.72 meters.
Suppose an elephant has a mass of 2750 kg.How fast, in meters per second, does the elephant need to move to have the same kinetic energy as a 66.5-kg sprinter running at 9.5 m/s?
Answers
Given:
The mass of the elephant is M = 2750 kg
The mass of the sprinter is m = 66.5 kg
The speed of the sprinter is v = 9.5 m/s
The kinetic energy of the sprinter is equal to the kinetic energy of the elephant.
Required: The speed of the elephant
Explanation:
The kinetic energy of the sprinter is equal to the kinetic energy of the elephant.
The speed of the elephant can be calculated by the formula
[tex]\begin{gathered} K.E._{elephant}=K.E._{sprinter} \\ \frac{1}{2}MV^2=\frac{1}{2}mv^2 \\ V=\sqrt{\frac{mv^2}{M}} \end{gathered}[/tex]On substituting the values, the speed of the elephant will be
[tex]\begin{gathered} V=\sqrt{\frac{66.5\times(9.5)^2}{2750}} \\ =1.48\text{ m/s} \end{gathered}[/tex]Thus, the speed of the elephant is 1.48 m/s
Final Answer: The speed of the elephant is 1.48 m/s
why free-fall acceleration can be regarded as a constant for objects falling within a few hundred miles of Earth’s surface.
Answers
Hello please could you help me with the physics ?
Answers
In order to calculate the amount of energy needed, we can use the formula below:
[tex]Q=m\cdot c\cdot\Delta T[/tex]Where Q is the amount of energy in Joules, m is the mass in kg, c is the specific heat coefficient and DeltaT is the change in temperature.
So, for m = 0.8, c = 4.2*10^3 and DeltaT = 85 (from 15°C to 100°C), we have:
[tex]\begin{gathered} Q=0.8\cdot4.2\cdot10^3\cdot85 \\ Q=285600\text{ J} \end{gathered}[/tex]Then, for the energy needed to evaporate the water, we have a similar formula:
[tex]Q=m\cdot L[/tex]Where L is the latent heat. So, for L = 2.3 * 10^6, we have:
[tex]\begin{gathered} Q=0.8\cdot2.3\cdot10^6 \\ Q=1840000\text{ J} \end{gathered}[/tex]A sea wave propagates with a speed of 12 cm/s and a length of 0.4 meters,find its period.A:0,1 minB:10sC:1sD:3,33s
Answers
Given:
The speed of the wave is v = 12 cm/s = 0.12 m/s
The wavelength of the wave is
[tex]\lambda\text{ = 0.4 m}[/tex]To find the period.
Explanation:
The time period can be calculated by the formula
[tex]T=\frac{\lambda}{v}[/tex]On substituting the values, the time period will be
[tex]\begin{gathered} T=\frac{0.4}{0.12} \\ =3.33\text{ s} \end{gathered}[/tex]Thus, the time period of the sea wave is 3.33 s
What is the average velocity of a car that travels 48 km north in 2.0 h?
Answers
Answer: 3.7 seconds. :)
Explanation:
Answer:
Explanation:
Given:
L = 48 km
t = 2.0 h
__________
V - ?
V = L / t = 48 / 2.0 = 24 km/h
You see a blue star directly over Avalon's equator. It appears to be moving north at 5.0 arcseconds per year. How far away is this star from Avalon (in parsecs - a parsec is ).
Answers
The distance of the star from the sun is 1.25 parsecs.
What quantity is measured in arcseconds?
The distance of the sun to other celestial bodies is measured in arcseconds.
The distance from the sun to a celestial object is the reciprocal of the angle, measured in arcseconds, of the object's apparent movement caused by parallax.
2 arcseconds = 0.5 parsecs
5 arcseconds = ?
= (5 arcseconds x 0.5 parsecs) / (2 arcseconds)
= 1.25 parsecs
Thus, the star appears to be moving north at 1.25 parsecs per year.
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Calculate the force between charges of 5.0x10^-8c and 1.0x10^-7c if they are .15m apart
Answers
Given data:
[tex]\begin{gathered} Q_1=5.0\times10^{-8}\text{ C} \\ Q_2=1.0\times10^{-7}\text{ C} \\ r=0.15\text{ m} \end{gathered}[/tex]The electrostatic force is given as,
[tex]F=\frac{KQ_1Q_2}{r^2}[/tex]Here, K is electrostatic force constant.
Substituting all known values,
[tex]\begin{gathered} F=\frac{9\times10^9\times5\times10^{-8}\times1\times10^{-7}^{}}{(0.15)^2} \\ =2\times10^{-3}\text{ N} \end{gathered}[/tex]Therefore, the force between the charges is 2×10^(-3) N. Hence, option (d) is the correct choice.
The block of a mass 10.2 kg is sliding at an initial velocity of 3.40 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.153.
Answers
m = mass = 10.2 kg
vo = initial velocity = 3.40 m/s
u = coefficient of kinetic friction = 0.153
g= gravity = 9.8m/s^2
a)
Fr = force of kinetic friction = u m g
Fr = 0.153 x 10.2 x 9.8 = -15.30N
b) Block's acceleration
Newton's second law of motion:
F = m*a
a = F/m = -15.30 / 10.2 = -1.5 m/s^2
c) USe the third equation of motion:
2as = vf^2 - vo^2
Where:
Vf= final velocity= 0 m/s
s = displacement
2 * -1.5 * s = 0^2 - 3.40^2
-3s = -11.56
s= -11.56/-3
s= 3.85 m
Physics
Hello how to solve this
Answers
2.15 mL of water will spill from the beaker when a 400mL glass beaker at room temperature is filled to the brim with cold water.
What is temperature and how come 2 mL of water spills off the beaker?Temperature as studied always is the measure for the degree of hotness or coldness.Here in the question is given a glass beaker of 400 mL which is at room temperature .Now the beaker is warmed up to 30 degree celsius, and then some of amount of water spilled.Change in volume = beta x v1 x change in temperature = 210 x 10^-4 x 400 x ( 30 - 4.4) = 2.15 mL.Hence the amount of water that will spill from the beaker would be 2.15 mL.To know more about temperature visit:
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1.Can money by happiness?why and Why not?
2.Are you happy with the possession you have? Yes/No Why?
Answers
Answer:
Explanation:
1) No it can't
2)Yes I am
Answer:
yes it's can be, but Don't put money first
Given the wave described by y(x,t)=5cos[π(4x-3t)], in meters. Find the following. Giveexact answers with units.
Answers
Answer:
a) 5 m
b) 0.667 s
c) 0.5 m
d) 0.75 m/s
e) -5 m
Explanation:
In an equation of the form
y(x, t) = Acos(kx - ωt)
A is the amplitude, ω = 2π/T where T is the period, and k = 2π/λ where λ is the wavelength. In this case, the equation os
y(x,t) = 5cos(π(4x - 3t)
y(x,t) = 5cos(4πx - 3πt)
So, A = 5, k = 4π, and ω = 3π. Then, we can find each part as follows
a) Amplitude
The amplitude is A, so it is 5 m.
b) the period
Using the equation ω = 2π/T and solving for T, we get:
[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{3\pi}=\frac{2}{3}=0.667\text{ s}[/tex]So, the period is 0.667 s
c) the wavelength.
using the equation k = 2π/λ and solving for λ, we get:
[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{4\pi}=0.5\text{ m}[/tex]So, the wavelength is 0.5 m
d) The wave speed
The wave speed can be calculated as the division of the wavelength by the period, so
[tex]v=\frac{\lambda}{T}=\frac{0.5\text{ m}}{0.667\text{ s}}=0.75\text{ m/s}[/tex]e) The height of the wave at (2, 1)
To find the height, we need to replace (x, t) = (2, 1) on the initial equation, so
[tex]\begin{gathered} y(x,t)=5\cos(\pi(4x-3t)) \\ y(2,1)=5\cos(\pi(4\cdot2-3\cdot1)) \\ y(2,1)=5\cos(\pi(8-3)) \\ y(2,1)=5\cos(\pi(5)) \\ y(2,1)=5\cos(5\pi) \\ y(2,1)=5(-1) \\ y(2,1)=-5 \end{gathered}[/tex]Then, the height of the wave is -5 m.
Therefore, the answers are
a) 5 m
b) 0.667 s
c) 0.5 m
d) 0.75 m/s
e) -5 m
You have a concave mirror with a focal length of 100.0 cm, and you want an image that is upright and 10.0 times as tall as the object. Where should you place the object?Select one:a.9.09 cmb.12.09 cmc.7.12 cmd.5.8 cm
Answers
ANSWER:
a. 9.09cm
STEP-BY-STEP EXPLANATION:
Given:
Focal length = 100 cm
Distance image = -10 cm
We use the following formula to calculate the value of the distance of the object:
[tex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/tex]We substitute each value and calculate the distance of the object just like this:
[tex]\begin{gathered} \frac{1}{d_o}+\frac{1}{-10}=\frac{1}{100} \\ \\ \frac{1}{d_{o}}=\frac{1}{100}+\frac{1}{10} \\ \\ \frac{1}{d_o}=\frac{11}{100} \\ \\ d_o=\frac{100}{11} \\ \\ d_o=9.09\text{ cm} \end{gathered}[/tex]Therefore, the correct answer is option a. 9.09cm
If an 800 kg roller coaster is at the top of its 50 m high track, it will have a potential energy 392,000 and a kinetic energy of 0J. This means the total mechanical energy is 392,000J. If the cart drops down to a new height of 10m, how much energy does the cart have now?
Answers
ANSWER:
313600 J
STEP-BY-STEP EXPLANATION:
We have that the gravitational potential energy is given by the following equation:
[tex]E_p=m\cdot g\cdot h[/tex]We substitute and calculate the potential energy, knowing that g is the acceleration of gravity and is equal to 9.8 m/s^2:
[tex]\begin{gathered} E_p=800\cdot9.8\cdot10 \\ E_p=78400\text{ J} \end{gathered}[/tex]We know that the total energy is 392,000 joules, so the energy it now carries would be the total minus the calculated potential energy:
[tex]\begin{gathered} E_k=392000-78400 \\ E_k=313600\text{ J} \end{gathered}[/tex]The energy carried by the cart is 313600 J
People weigh less on the moon than on the Earth because:the moon is smaller and has less gravity to pull on you.the Earth has less gravity than the Moon.people weigh the same on the moon as they do on Earth.the moon is magnetized.
Answers
Given
People weigh less on the moon than on the Earth
To find
The correct reason for the given statement
Explanation
The moon has much lesser mass than that of the earth. It is also just 60 percent as dense as the earth. So the gravitational pull on the moon is less than on the earth and thus people weigh less on the moon
Conclusion
The correct reason is
the moon is smaller and has less gravity to pull on you.
An archerfish squirts water with a speed 2 m/s at an angle 50 degrees above the horizontal, and aims for a beetle on a leaf 3cm above the water surface. (A) At what horizontal distance from the beetle should the archerfish fire if it is to hit its target in the least time? (B) How much time does the beetle have to react?
Answers
A ) The horizontal distance from the beetle, the archerfish should fire if it is to hit its target in the least time = 1.19 m
B ) The time beetle have to react = 0.93 s
T = ( [tex]u_{y}[/tex] + √ [tex]u_{y}[/tex]² - 2 g H ) / g
T = Total time taken
g = Acceleration due to gravity
H = Height above the ground
[tex]u_{y}[/tex] = Y-component of initial velocity
u = 2 m / s
θ = 50°
H = 3 cm
[tex]u_{y}[/tex] = u sin θ
[tex]u_{y}[/tex] = 2 * sin 50°
[tex]u_{y}[/tex] = 1.54 m / s
T = ( 1.54 + √ 1.54² - ( 2 * 9.8 * 3 ) ) / 9.8
T = ( 1.54 + √ 56.42 ) / 9.8
T = 9.1 / 9.8
T = 0.93 s
R = [tex]u_{x}[/tex] T
[tex]u_{x}[/tex] = u cos θ
[tex]u_{x}[/tex] = 2 * cos 50°
[tex]u_{x}[/tex] = 1.28 m / s
R = 1.28 * 0.93
R = 1.19 m
Therefore,
A ) The horizontal distance from the beetle, the archerfish should fire if it is to hit its target in the least time = 1.19 m
B ) The time beetle have to react = 0.93 s
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Philip jumps up with to a height of 3 m above the ground. What was Philip's initial velocity? round to the tenth.
Answers
The initial velocity of Philip was 7.66 m/s
Given data:
The vertical height is h=3 m.
Considering ground as the reference, then the initial potential energy of Philip is zero, i.e., PEi=0
The formula for the kinetic energy is given by,
[tex]KE_i=\frac{1}{2}mv^2[/tex]Here, m is mass and v is the velocity.
After reaching the height of 3 m Philip comes to a stop. It means the final kinetic energy is zero, i.e. KEf=0.
The final potential energy is given by,
[tex]PE_f=mgh[/tex]Here, g is the gravitational acceleration.
Applying the conservation of energy between initial position and final position.
[tex]\begin{gathered} KE_i+PE_i=KE_f+PE_f \\ \frac{1}{2}mv^2+0=0+mgh \\ v=\sqrt[]{2gh} \\ v=\sqrt[]{2\times9.8\times3} \\ v=7.66\text{ m/s} \end{gathered}[/tex]Thus, the initial velocity of Philip was 7.66 m/s.
Two capacitors of values of 1.0 μF and 0.50 μF are connected in parallel. The system is hooked up to a 100 V battery. Find the electrical potential energy stored in the 1.0 μF capacitor.Group of answer choices1.0x10-2 J1.9x10-3 J5.0x10-3 J6.5x10-3 J
Answers
5.0x10-3 J
Explanation
Step 1
find the equivalent capacitor,When capacitors are connected together in parallel the total or equivalent capacitance, CT in the circuit is equal to the sum of all the individual capacitors added together
so
[tex]\begin{gathered} C_{eq}=1.0\mu F+0.50\mu F \\ C_{eq}=1.50\mu F \end{gathered}[/tex]Step 2
We can proceed to use this formula and solve for the potential energy stored in the 1.0mF capacitor.
[tex]\begin{gathered} U=\frac{1}{2}cV^2 \\ U=\frac{1}{2}(0.000001c)(100^2) \\ U=0.005\text{ J} \\ U=5\cdot10^{-3\text{ }}J \end{gathered}[/tex]therefore, the answer is
5.0x10-3 J
I hope this helps you
A car initially at rest travels with a uniform acceleration of 8 m / s^2 . Calculate the distance covered by the car in 3 s .
Answers
We know that
• The initial velocity is zero because it starts from rest.
,• The acceleration is 8 m/s^2.
,• The time elapsed is 3 seconds.
Use a formula that relates initial velocity, acceleration, time, and distance.
[tex]d=v_0t+\frac{1}{2}at^2[/tex]Use the given magnitudes to find d.
[tex]\begin{gathered} d=0\cdot3\sec +\frac{1}{2}\cdot(8\cdot\frac{m}{s^2})(3\sec )^2 \\ d=4\cdot\frac{m}{s^2}\cdot9s^2 \\ d=36m \end{gathered}[/tex]Therefore, the distance covered is 36 meters.S=3-2t+3t^2
What is the instantaneous velocity and it’s acceleration at t=3s
At what time is the particle at rest
Answers
Answer:
Explanation:
Given:
X(t) = 3 - 2*t + 3*t²
t = 3 s
_______________
V(t) - ?
a(t) - ?
Speed is the first derivative of the coordinate, acceleration is the second.
1)
V(t) = X' = (3 - 2*t + 3*t²)' = 0 - 2 +6*t = 6*t - 2
V(3) = 6*3 - 2 = 16 m/s
2)
a(t) = X'' = V' = (6*t - 2)' = 6 m/s²
a(3) = 6 m/s²
3)
The body will stop (V = 0 ) in (t) seconds:
V(t) = 6*t - 2
0 = 6*t - 2
6*t = 2
t = 2/6 = 1/3 ≈ 0,33 s
You push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 m. What is the force of friction?
Answers
If you push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 meters, then the force of the friction would be 42 Newtons as per the concept of limiting friction.
What is friction?Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
As given in the problem if push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 m, then we have to find the force of the friction.
The force of the friction = The limiting friction force
If you apply 42N of force to a 2.1-kilogram item on a table. According to the theory of limiting friction, when the box gently slides to a halt over a distance of 2.2 meters, the force of friction would be 42 Newtons.
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A 4.0 kW clothes dryer is connected to a 220 V circuit. How much current does the dryer use?
Answers
Given:
The power take by the dryer is
[tex]\begin{gathered} P=4\text{ kW} \\ P=4000\text{ W} \end{gathered}[/tex]
The dryer is connected to source of
[tex]V=220\text{ V}[/tex]Required: the current use by the dryer
Explanation:
we know that power taken by any appliance is given by
[tex]P=VI[/tex]here, V is the voltage of the source and I is the current drawn by the electric appliance.
Plugging all the values in the above formula, we get
[tex]\begin{gathered} 4000\text{ W}=220\text{ V}I \\ I=\frac{4000\text{ W}}{220\text{ V}} \\ I=18.18\text{ A} \end{gathered}[/tex]Thus, the current used by the dryer is
[tex]18.18\text{ A}[/tex]A car is negotiating a flat circular curve of radius 50 m with a speed of 20 m/swithout slipping. The maximum centripetal force (provided by static friction) is 1.2 x10^4N. What is the mass of the car?1) 0.50 x 10^3 kg2) 1.0 x 10^3 kg3) 1.5 x 10^3kg4) 2.0 x 10^3 kg
Answers
We are given a car that is experiencing a centripetal force.
The formula for the force is given by:
[tex]F_c=\frac{mv^2}{r}[/tex]Where "m" is the mass, "v" is the velocity and "r" is the radius. Now we solve for the mass, first by multiplying both sides by r:
[tex]rF_c=mv^2[/tex]Now we divide by the velocity squared:
[tex]\frac{rF_c}{v^2}=m[/tex]Now we replace the known values:
[tex]\frac{(50m)(1.2\times10^4N)}{(20\frac{m}{s})^2}=m[/tex]Solving the operations:
[tex]1500kg=m[/tex]Therefore, the mass is 1500 kg.
part 2 of 2 ASSUME BOTH snowballs are thrown with the same initial speed 39.9 m/s. the first snowball is thrown at an angle of 51 degrees above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? how many seconds after the first snowball should you throw the second so that they arrive on target at the same time?
Answers
Explanation
Step 1
Let
a) for ball 1
[tex]\begin{gathered} \text{ Initial sp}eed=v_0=33.9\text{ }\frac{m}{s} \\ \text{ Angle=51 \degree} \end{gathered}[/tex]the formula for the distance is given by:
[tex]x=\frac{v^2_0\sin(2\theta)}{g}[/tex][tex]\begin{gathered} \text{hence, let v}_0=39.9,\text{ angle= 51 \degree , g=9.8 } \\ \text{replace to solve for x } \\ x=\frac{(39.9)^2\sin(2\cdot51)}{9.8} \\ x=158.9\text{ m} \\ \end{gathered}[/tex]hence, the horizontal distance reached by the ball 1 is 158.9 meters
Step 2
as the ball started from the same point at the same initial speed, the only way to make the second ball hits the same point as the first is thworing the second ball at the same angle, it is 51 °
Derek leaves his physics book on top of a drafting table that is inclined at a 35° angle. The table exerts a force on the book of 18 Newtons, and the force of gravity is 22 Newtons. If the Force of friction acting on the book is 11 Newtons, what is the Net Force on the book?
Answers
The net force on the book is equal to 1.61 N.
What is the net force?The net force on the object can be defined as the vector sum of all the forces together. Net force becomes resultant force and shows the same effect on the rotational motion as all actual forces.
Given, the angle of the book with the table, α = 35°
The force of friction, f = 11N
The normal force acting on the book, N = 18N
The force of gravity, W = 22 N
The net force along the table is F₁ = Wsinα - f
F₁= 22× sin35° - 11
F₁ = 1.6 N
The net force perpendicular to table is F₂ = W cosα -N
F₂= 22×cos35° -18
F₂ = 0.021 N
The net force on the book is [tex]F_{net} = \sqrt{F_1^2+F_2^2}[/tex]
[tex]F_{net} = \sqrt{(1.6)^2+(0.021)^2}[/tex]
[tex]F_{net} = 1.61N[/tex]
Therefore, the net force on the book is equal to 1.61 Newtons.
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A stone is thrown vertically upwards from a height of 1.5m and lands on the ground 6s later. What was the magnitude of the initial velocity?
Answers
Answer:
2.5m/s is the correct answer