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3 years ago

A 1000-kg car moving at 10 m/s brakes to a stop in 5 s. the average braking force is

1 answer:

4 0

To calcculate the braking force of the car moving, we use Newton's second law of motion which relates the acceleration and the force of an object moving. The force of an object moving is directly proportional to its acceleration and the proportionality constant is the mass of the object. It is expressed as:

Force = ma

Acceleration is the rate of change of the velocity of a moving object. We calculate acceleration from the velocity and the time given above.

a = (10 m/s) / 5 s = 2 m/s^2

So,
Force = ma
Force = 1000 kg ( 2 m/s^2 )
Force = 2000 kg m/s^2 or 2000 N

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The magnitude of the voltage induced in a conductor moving through a stationary magnetic field depends on the _______ and the __

Virty [35]

The correct answer is: Option (D) length, speed

Explanation:

According to Faraday's Law of Induction:

ξ = Blv

Where,

ξ = Emf Induced

B = Magnetic Induction

l = Length of the conductor

v = Speed of the conductor.

As you can see that ξ (Emf/voltage induction) is directly proportional to the length and the speed of the conductor. Therefore, the correct answer will be Option (D) Length, Speed

3 0

3 years ago

Read 2 more answers

Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on

lys-0071 [83]

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

4 0

3 years ago

Make a graph of the data. You may use a graphing program. Think about what data should be on the y-axis and the x-axis. Be sure

Lynna [10]

Answer:

yes it was a constant speed and the car traveled 10 meters in 20 seconds.

Explanation:

8 0

3 years ago

Cross section of a solid circular rod is subject to a torque of t = 2.7 kn⋅m . if the diameter of the rod is d = 8 cm , what is

polet [3.4K]

Shear Stress= Force along the tangent/ Area
= 2700/π*r^2
=2700/3.14*16
=54

3 0

3 years ago

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

The locations are +5.9 cm or -5.9 cm

4 0

3 years ago