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4 years ago

3) Consider a rocket in deep space (so it's weightless) with thrusters that generate a forward

Physics

1 answer:

postnew [5]4 years ago

6 0

a). F= m a

500 N = (m) (5.00 m/s^2)

m = 100 kg

b). Accelerating body covers distance of

D = 1/2 a T^2

D = 1/2 (5.00 m/s^2) (100 s)^2

D = 25,000 meters

Note: I'm pretty sure the acceleration is typed wrong in the question. If so, then these answers won't be correct.

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What is the speed that the Earth orbits the Sun?

iris [78.8K]

Answer:

30 Kilometers per second

4 0

3 years ago

Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second. It hits stationary ball B and they undergo

blagie [28]

<span>If Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second, and It hits stationary ball B and they undergo elastic collision, thus the two balls have different masses, then the following statement which is true is the statement that stated that there was no y-momentum initially.</span>

7 0

4 years ago

A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N1 turns. A second toroidal solenoid w

lilavasa [31]

Answer:

Mutual inductance, $M=2.28\times 10^{-5}\ H$

Explanation:

(a) A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N₁ turns. A second thyroidal solenoid with N₂ turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius.

Mutual inductance is given by :

$M=\dfrac{\mu_oN_1N_2A}{2\pi r}$

(b) It is given that,

$N_1=550$

$N_2=290$

Radius, r = 10.6 cm = 0.106 m

Area of toroid, $A=0.76\ cm^2=7.6\times 10^{-5}\ m^2$

Mutual inductance, $M=\dfrac{4\pi \times 10^{-7}\times 550\times 290\times 7.6\times 10^{-5}}{2\pi \times 0.106}$

$M=0.0000228\ H$

$M=2.28\times 10^{-5}\ H$

So, the value of mutual inductance of the toroidal solenoid is $2.28\times 10^{-5}\ H$ . Hence, this is the required solution.

8 0

3 years ago

Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the

sdas [7]

Answer:

The new distance is d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value ( $q_{B}$ =q₂ = q₂ / 5) than new distance (d = n d₀)

dat

q₁ = $q_{A}$

q₂ = $q_{B}$

r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

F = k q₁ q₂ / d₀²

F = k q₁ (q₂ / 5) / (n d₀)²

.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

5 n² = 1

n = √ 1/5

n = 0.447

The new distance is

d = 0.447 d₀

6 0

3 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago