0.We are asked to determine the location of an image formed by an 7.75mm tall object that is located a distance of 17.5 cm from a convex mirror.
First, we will calculate the focal length using the following formula:
[tex]f=-\frac{R}{2}[/tex]Where:
[tex]\begin{gathered} f=\text{ focal length} \\ R=\text{ radius} \end{gathered}[/tex]Substituting the values we get:
[tex]f=-\frac{9.40cm}{2}[/tex]Solving the operations:
[tex]f=-4.7cm[/tex]Now, we use the following formula:
[tex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/tex]Where:
[tex]\begin{gathered} d_0=\text{ distance of the object} \\ d_i=\text{ distance of the image} \end{gathered}[/tex]Now, we substitute the known values:
[tex]\frac{1}{17.5cm}+\frac{1}{d_i}=-\frac{1}{4.7cm}[/tex]Now, we solve for the distance of the image. First, we subtract 1/17.5 from both sides:
[tex]\frac{1}{d_i}=-\frac{1}{4.7cm}-\frac{1}{17.5cm}[/tex]Solving the operation:
[tex]\frac{1}{d_i}=-0.27\frac{1}{cm}[/tex]Now, we invert both sides:
[tex]d_i=\frac{1}{-0.27}cm=-3.7cm[/tex]Therefore. the location of the image is -3.7 centimeters.
The other parts are solved using the same procedure.
Part B. To calculate the size of the image we will use the following relationship:
[tex]\frac{h_i}{h_o}=-\frac{d_i}{d_0}[/tex]Where:
[tex]h_i,h_0=\text{ height of the image and height of the object}[/tex]Substituting we get:
[tex]\frac{h_i}{7.75mm}=-\frac{-3.7cm}{17.5cm}[/tex]Solving the operations on the right side:
[tex]\frac{h_i}{7.75mm}=0.21[/tex]Now, we multiply both sides by 7.75:
[tex]h_i=(7.75mm)(0.21)[/tex]Solving the operations:
[tex]h_i=1.64mm[/tex]Therefore, the height of the iamge is 1.64 mm.
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?
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
Which of the following water molecules have the greatest kinetic energy?Select one:a. Cool water.b. Warm water.c. Boiling water.d. They all have the same kinetic energy.
Since all the molecules in the boiling water will have more energy introduced by heat, then the molecules with the greatest kinetic energy are the ones from the boiling water.
A raindrop has a mass of 7.7 × 10-7 kg and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop.(a)Fraindrop= _________________ units ________(b)Fearth= _________________ units_____________
ANSWER:
a) Fraindrop
[tex]F=7.546\cdot10^{-6}N[/tex](b) Fearth
[tex]F=-7.546\cdot10^{-6}N[/tex]STEP-BY-STEP EXPLANATION:
(a)
We calculate the force, multiplying the value of the mass by gravity, just like this:
[tex]\begin{gathered} F=m\cdot a \\ F=7.7\cdot10^{-7}\cdot9.8 \\ F=7.546\cdot10^{-6}N \end{gathered}[/tex](b)
by newton's 3rd law they are are equal and opposite so:
[tex]F=-7.546\cdot10^{-6}N[/tex]Given the wave described by y(x,t)=5cos[π(4x-3t)], in meters. Find the following. Giveexact answers with units.
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
A cylinder of gas at room temperature has a pressure . To p_{1} what temperature in degrees Celsius would the temperature have to be increased for the pressure to be 1.5p_{1} ,
In order to calculate the temperature, we need to know that temperature and pressure are directly proportional, that is, if the pressure increases, the temperature (in Kelvin) also increases in the same proportion.
So, first let's convert the temperature from Celsius to Kelvin, by adding 273 units:
[tex]\begin{gathered} K=C+273 \\ K=20+273 \\ K=293 \end{gathered}[/tex]Then, let's calculate the proportion:
[tex]\begin{gathered} \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ \frac{p_1}{293}=\frac{1.5p_1}{T_2} \\ \frac{1}{293}=\frac{1.5}{T_2} \\ T_2=1.5\cdot293 \\ T_2=439.5\text{ K} \end{gathered}[/tex]Now, converting back to Celsius, we have:
[tex]\begin{gathered} C=K-273 \\ C=439.5-273 \\ C=166.5\text{ \degree{}C} \end{gathered}[/tex]So the temperature would be 166.5 °C.
why free-fall acceleration can be regarded as a constant for objects falling within a few hundred miles of Earth’s surface.
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
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
In terms of area, about how much more pizza is given if the diameter is 12 inches compared to one with a diameter of 8 inches?
B. 2.3 times more
Explanation:The pizza is circular in shape
The diameter of the large-sized pizza, d₁ = 12 inches
Tha area of the large sized pizza is calculated as:
[tex]\begin{gathered} A_1=\frac{\pi{d^2_1}}{4} \\ A_1=\frac{\pi{12^2}}{4} \\ A_1=\frac{\pi{144^{}}}{4} \\ A_1=36\pi\text{ in}^{2} \end{gathered}[/tex]The diameter of the small-sized pizza, d₂ = 8 inches
The area of the small-sized pizza is calculated as:
[tex]\begin{gathered} A_2=\frac{\pi{d^2_2}}{4} \\ A_2=\frac{\pi{8^2}}{4} \\ A_2=\frac{64\pi{}}{4} \\ A_2=16\pi\text{ in}^{2} \end{gathered}[/tex]Ratio of A₁ to A₂
[tex]\begin{gathered} \frac{A_1}{A_2}=\frac{36\pi{}}{16\pi} \\ \frac{A_1}{A_2}=2.25 \\ \frac{A_1}{A_2}=2.3(to\text{ the nearest 1 dp)} \end{gathered}[/tex]The 12 inches pizza is 2.3 times more than the 8 inches pizza
this is a 2 part question2) Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops byapplying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than,less than, or equal to the stopping distance of driver 2? (b) Choose the best explanation from among thefollowing: 1. Locking up the brakes gives the greatest possible braking force.2. The same tires on thesame road result in the same force of friction.3. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.
The maximum static friction between two surfaces is greater than the kinetic friction between them.
If the wheels of a car get locked, the surface of the wheel slides through the floor and kinetic friction acts to stop the car.
If the wheels of the car don't get locked, they may turn fast enough to prevent the surface of the wheel from sliding through the floor and static friction acts on the car.
Since the force acting on the car with its wheel locked is less than the force acting on the car with the turning wheels, then, the stopping distance is greater for driver 1 than for diver 2.
Therefore, the answers are:
a) The stopping distance of driver 1 is greater than the stopping distance of driver 2.
b) The best explanation is:
3. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.
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?
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
Assuming the jet slows with constant acceleration, find the magnitude and direction of its acceleration.
We are given that a jet is traveling with a speed of 78.6 m/s and travels a distance of 919m. We are asked to determine the constant acceleration when the jet stops. To do that we will use the following formula:
[tex]v^2_f=v^2_0+2ax[/tex]Where:
[tex]\begin{gathered} v_f=\text{ final speed} \\ v_0=\text{ initial speed} \\ a=\text{ acceleration} \\ x=\text{ distance traveled} \end{gathered}[/tex]Since the jet stops, this means that the final speed is zero. We will solve for the acceleration "a" in the formula. First, we will eliminate the term for the final speed since it is zero:
[tex]0=v^2_0+2ax[/tex]Now we will subtract the initial speed squared from both sides:
[tex]-v^2_0=2ax[/tex]Now we will divide by "2x" from both sides:
[tex]\frac{-v^2_0}{2x}=a[/tex]Now we replace the known values:
[tex]\frac{-(78.6\frac{m}{s})^2}{2(919m)}=a[/tex]Solving the operations:
[tex]-3.36\frac{m}{s^2}=a[/tex]Therefore, the magnitude of the acceleration is 3.36. Since the jet is deaccelerating in the direction due south, the direction of the acceleration is due north.
What is held in orbit by the gravitational pull of earth
The international space station.
The moon.
All TV satellites.
All weather satellites.
All GPS satellites.
More than 4000 other artificial satellites.
Thousands of pieces of "space junk"
Answer:
The Moon.
Explanation:
The earths gravity holds the moon in place.
A 1.4N friction force slows a block to a stop after sliding 7m. How much work was done by the friction force
Answer:
9.8J
Explanation:
The work done by the friction force can be calculated as
W = Fd
Where F is the friction force, and d is the distance that the block slide.
So, replacing F = 1.4 N and d = 7 m
W = (1.4N)(7 m)
W = 9.8 J
Therefore, the work done by the friction was 9.8J against the movement of the block
A small object of mass 0.500 kg is attached by a 0.440 m-long cord to a pin set into the surface of a frictionless table top. The object moves in a circle on the horizontal surface with a speed of 5.34 m/s.What is the magnitude of the radial acceleration of the object? What is the tension in the cord?
Given data:
* The mass of the object attached is m = 0.5 kg.
* The radius of the circle is r = 0.44 m.
* The speed of the object moving in circular motion is v = 5.34 m/s.
Solution:
(a). The radial acceleration of the object is also known as the centripetal acceleration of the object.
The value of centripetal acceleration in terms of the velocity of the object is,
[tex]a_c=\frac{v^2}{r}[/tex]Substituting the known values,
[tex]\begin{gathered} a_c=\frac{5.34^2}{0.44} \\ a_c=64.8ms^{-2} \end{gathered}[/tex]Thus, the radial acceleration of the object is 64.8 meters per second squared.
(b). The tension in the chord is equivalent to the centripetal force acting on the object which helps it to move in the circular motion.
Thus, the tension acting on the chord is,
[tex]F=ma_c[/tex]Substituting the known values,
[tex]\begin{gathered} F=0.5\times64.8 \\ F=32.4\text{ N} \end{gathered}[/tex]Thus, the tension acting in the chord is 32.4 N.
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.
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
Philip jumps up with to a height of 3 m above the ground. What was Philip's initial velocity? round to the tenth.
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.
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?
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 °
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?
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.
Learn more about friction here, refer to the link ;
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choose the 200 kg refrigerator. set the applied force to 400 n (to the right). be sure friction is turned off. what is the net force acting on the refrigerator?
Answer:
Explanation:
Ginen:
m₁ = 200 kg
F₂ = 400 N
g ≈ 10 m/s²
__________
R - ?
F₁ = m₁·g = 200· 10 = 2000 N
R = √ (F₁² + F₂²) = √ ( 2000² + 400²) ≈ 2040 N
The net force acting on the refrigerator having a mass of 200 kg and the applied force to 400 n (to the right) is 2040 Newtons.
What is force?Force is the influence of either pull or pushes in the body. Basically, gravitation forces, nuclear forces, and friction forces are the types of forces. For e.g. when the wall is hit by a hand then a force is exerted by the hand on the wall as well as the wall also exerts a force on the hand. There are different laws given to Newton to understand force.
Newton is a unit of force used by physicists that is part of the International System (SI). The force required to move a body weighing one kilogram one meter per second is known as a newton.
Given:
The mass of the refrigerator, m = 200 kg,
The force, F = 400 N,
Calculate the net force by the formula given below,
F = m × g
here, g is the gravitational acceleration.
Substitute the values,
F= 200 × 10 = 2000 N
[tex]R = \sqrt{F_1^2 +F_2^2}[/tex]
where R is the net force
[tex]R = \sqrt{2000^2 +400^2}[/tex]
R = 2040 Newton
Therefore, the net force acting on the refrigerator having a mass of 200 kg and the applied force to 400 n (to the right) is 2040 Newtons.
To know more about Force:
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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.
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
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
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.
Tall pacific coast redwood trees can reach heights of about 100 m. If air drag is negligibly small, how fast is a sequoia come moving when it reaches the ground if it dropped from the top of a 100 m tree?
Given data:
Height of the tree;
[tex]h=100\text{ m}[/tex]Initial velocity;
[tex]u=0\text{ m/s}[/tex]The velocity of sequoia when it reaches the ground is given as,
[tex]v=\sqrt[]{u^2+2gh}[/tex]Here, g is the acceleration due to gravity.
Substituting all known values,
[tex]\begin{gathered} v=\sqrt[]{(0\text{ m/s})^2+2\times(9.8\text{ m/s}^2)\times(100\text{ m})} \\ \approx44.27\text{ m/s} \end{gathered}[/tex]Therefore, sequoia will reach the ground with a velocity of 44.27 m/s.
What is the net force on an object with an applied force of 800N (right) and friction resisting at 750 N (left)?1 1550 N left2 1550 N right3 50 N left4 50 N right
Given,
The applied force, F=800 N
The friction, f=750 N
Friction is a force that opposes the motion of an object. Thus the net force will be equal to the difference between the applied force and the friction. As the applied force is greater than the frictional force, the net force will be in the same direction as the applied force, that is to the right.
Thus the net force is given by,
[tex]F_n=F-f[/tex]On substituting the known values,
[tex]\begin{gathered} F_n=800-750 \\ =50\text{ N} \end{gathered}[/tex]Therefore the net force on the object is 50 N to the right. Thus, the correct answer is option 4.
What is the average velocity of a car that travels 48 km north in 2.0 h?
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
7/21/22, 7:37 AMProblem Set ThreeNotes: Use 9.8 m/s 2 for the acceleration due to gravity. Formust be expressed in m/sLaw calculations, mass must be expressed in kg and velocity.A steady 45 N horizontal force is applied to a 15kg object on a table. The object slides against a friction force of 30 N. Calculate the acceleration of the object in m/s.
Given:
The mass of the object is
[tex]m=15\text{ kg}[/tex]The applied force on the object is
[tex]F=45\text{ N}[/tex]The frictional force on the object is
[tex]f=30\text{ N}[/tex]To find:
The acceleration of the object
Explanation:
The net force on the object is
[tex]\begin{gathered} F_{net}=F-f \\ =45-30 \\ =15\text{ N} \end{gathered}[/tex]The acceleration of the object is,
[tex]\begin{gathered} a=\frac{F_{net}}{m} \\ =\frac{15}{15} \\ =1\text{ m/s}^2 \end{gathered}[/tex]Hence, the acceleration is
[tex]1\text{ m/s}^2[/tex]A circular loop of wire with a diameter of 13.478 cm is in the horizontal plane and carries a current of 1.607 A counterclockwise, as viewed from above. What is the magnetic field, in microTeslas, at the center of the loop?
Given:
The number of the loops, n = 1
The diameter of the loop is d = 2r = 13.478 cm
The current in the loop is I = 1.607 A
To find the magnetic field in micro Tesla
Explanation:
The magnetic field can be calculated by the formula
[tex]B\text{ =}\frac{n\mu_0I}{2r}[/tex]Here, the value of the constant is
[tex]\mu_0=\text{ 12.57}\times10^{-7}\text{ H/m}[/tex]On substituting the values, the magnetic field will be
[tex]\begin{gathered} B=\frac{1\times12.57\times10^{-7}\times1.607}{13.478\times10^{-2}} \\ =1.499\text{ }\times10^{-5}\text{ T} \\ =14.99\times10^{-6}\text{ T} \\ =14.99\text{ }\mu T \end{gathered}[/tex]The magnetic field is 14.99 micro Tesla
A car is driving around a turn with a radius of 20.7 m. If the coefficient of friction between the tires and the road is 2.07, what is the maximum speed the car can maintain around the turn without slipping?
Answer:
A car completes a turn that has a radius of 20 meters. The coefficient of friction between the tires and road is 0.50. What maximum speed can the car safely maintain in order to complete the turn without skidding? (A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s (E) 25 m/s
An airplane traveling at 1008 meters above the ocean at 135 km/h is going to drop a box of supplies to shipwrecked victims below. How many seconds before the plane is directly overhead should the box be dropped?
The horizontal velocity of the airplane is,
[tex]v=135\text{ km/h}[/tex]The height of the airplane is,
[tex]h=1008\text{ m}[/tex]The vertical initial velocity of the box is zero as the airplane is moving in the horizontal direction.
let the time to reach the victim is t.
we can write,
[tex]\begin{gathered} h=\frac{1}{2}gt^2 \\ t=\sqrt[]{\frac{2h}{g}} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times1008}{9.8}} \\ =14.3\text{ s} \end{gathered}[/tex]Hence the required time is 14.3 s
What are the answers for a, b and c in MJ?
Given:
The orbital height of the satellite, h=94 km=94000 m
The mass of the satellite, m=1045 kg
The new altitude of the satellite, d=207 km=207000 m
To find:
a) The energy needed.
b) The change in the kinetic energy.
c) The change in the potential energy.
Explanation:
The radius of the earth, R=6.37×10⁶ m
The mass of the earth, M=6×10²⁴ kg
a) The orbital velocity is given by,
[tex]v=\sqrt{\frac{GM}{r}}[/tex]Where G is the gravitational constant and r is the radius of the satellite from the center of the earth.
Thus the initial orbital velocity of the earth,
[tex]\begin{gathered} v_1=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37×10^6+94000)}} \\ =7868.43\text{ m/s} \end{gathered}[/tex]The orbital velocity after changing the altitude is,
[tex]\begin{gathered} v_2=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37\times10^6+207000)}} \\ =7800.5\text{ m/s} \end{gathered}[/tex]Thus the total energy needed is given by,
[tex]E=(\frac{1}{2}mv_2^2-\frac{GMm}{(R+d)})-(\frac{1}{2}mv_1^2-\frac{GMm}{(R+h)})[/tex]On substituting the known values,
[tex]\begin{gathered} E=1045[(\frac{1}{2}\times7868.43^2-\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37\times10^6+207000)})-(\frac{1}{2}\times7800.5^2-\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37×10^6+94000)})] \\ =623\text{ MJ} \end{gathered}[/tex]b)
The change in the kinetic energy is given by,
[tex]\begin{gathered} KE=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 \\ =\frac{1}{2}m(v_2^2-v_1^2) \end{gathered}[/tex]On substituting the known values,
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)
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.