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lutik1710 [3]
4 years ago
15

Hey can someone send me the answer to this

Physics
1 answer:
bagirrra123 [75]4 years ago
5 0

Cars 'A' and 'C' look like they're moving at the same speed.  If their tracks are parallel, then they're also moving with the same velocity.

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8. True or False. Energy can move in waves.​
Basile [38]

Answer:

true

Explanation:

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3 years ago
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4. A car with a mass of 2,200 kg can accelerate from 0 to 30 m/s in 6 seconds.
wlad13 [49]

Answer:

The engine generates a force of 11000 Newtons.

Explanation:

To determine the force, first you will need to calculate the acceleration. Acceleration is change in velocity over change in time, so in this case:

a = \frac{\Delta v}{\Delta t}=\frac{(30-0)\frac{m}{s}}{(6 - 0 )s}=5 \frac{m}{s^2}

Force is then determined as

F=ma = 2200kg\cdot 5\frac{m}{s^2}=11000N

This is the forward force, i.a. one acting in the direction of the car's acceleration. we are told that there is no inefficiency in the engine and transmission, therefore this same force is generated by the engine.  

4 0
3 years ago
What is the acceleration of a boy on a skateboard if the unr
Alex777 [14]

Answer:

0.26 m/s^2

Explanation:

We're asked to find the acceleration of a body given the net force acting on it and it's mass.

To do this, we use Newton's second law.

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3 years ago
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A wheelbarrow is an example of what type of lever
grin007 [14]

A wheelbarrow is a 2nd class lever. Your hands on the handles give the effort. The load inside the wheelbarrow is the resistance. The wheel is the fulcrum. When you lift up the handles, the load goes up in the same direction.

3 0
3 years ago
A point charge q1 = -8.9 μC is located at the center of a thick conducting shell of inner radius a = 2.8 cm and outer radius b =
musickatia [10]

Answer:

1) Ex(P) = -8.34602 N/C

2) E_y(P) = -5.23850174216 N/C

3) Question (3) is a similar question to (1)

4) E_y(P) = -5.23850174216 N/C

5) \sigma _b ≈ 1.041466 C/m²

6) σₐ ≈ 2.2330413 C/m²

Explanation:

The given parameter of the point charge located at the center of a conducting shell

The charge of the point charge, q₁ = -8.9 μC

The inner radius of the shell, a = 2.8 cm

The outer radius of the shell, b = 4.1 cm

The charge of the conducting shell, q₂ = 2.2 μC

Therefore, we have;

1) The point P(8.5, 0)

V = k \cdot \dfrac{q_1 + q_2 }{r^2}

By plugging in the values, we have;

For

R₁ < R₂ < r, for the electric field at the point, 'P', we have;

E_x(P) = 9 \times 10^9 \times \dfrac{-8.9 \ \times 10^{-6} + 2.2 \ \times 10 ^{-6} }{(0.085 \ )^2} = -8.34602

Ex(P) = -8.34602 N/C

2) For the point given with coordinates (8.5, 0), the distance of the y-component of point from the center = 0

The y-component of the electric field = 0 N/C

4) For r = 1.4 cm, along the y-axis we have;

R₁ < r < R₂

Therefore, we have;

E = k \cdot \left( \dfrac{q_1 }{r} + \dfrac{q_2}{R_2}\right)

Substituting the values, we get;

E_y(P) = 9 \times 10^9 \times \left( \dfrac{-8.9 \times 10^{-6}}{0.014} + \dfrac{2.2 \times 10^{-6}}{0.041}\right) = -523850174216

E_y(P) = -5.23850174216 N/C

5) The charge density, \sigma _b, is given as follows;

\sigma_b = \dfrac{Q}{A}  = \dfrac{2.2 \times 10^{-6} }{4\times \pi  \times 0.041^2 } \approx 1.041466 \ C/m^2

6) Similarly, we have;

\sigma_a = \dfrac{Q}{A_a}  = \dfrac{2.2 \times 10^{-6} }{4\times \pi  \times 0.028^2 } \approx 2.2330413 \ C/m^2

5 0
3 years ago
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