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vichka [17]
3 years ago
11

Can someone help asap with this

Physics
1 answer:
kondaur [170]3 years ago
7 0

Answer:

B

Explanation:

i had a test on this and got it correct

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Calculate the acceleration of the car for each set of conditions using the formula a = (v2 – v1) / (t2 – t1) where v2 and v1 are
Arisa [49]

Answer:  1.  0.19

               2.  0.33

               3.  0.47

               4.  0.62

Explanation:  just did it & these are the correct answers

4 0
3 years ago
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In which of the following materials is the velocity of light greatest?
just olya [345]
<span>The velocity of light is by far the greatest in air. There is nothing in air to stop or impede the movement of light in air. When light travels in air it will not stop unless running into clouds, smoke, smoke, droplets, or impurities. Any other substance is harder for light to travel through.</span>
8 0
3 years ago
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A helicopter pulls upward by means of a rope on a 250 kg crate to lift it UNIFORMLY. What is the net force on the crate?
Cloud [144]

Answer:

The net force = 0

Explanation:

The given information includes;

The mass of the crate = 250 kg

The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)

In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.

The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate

The weight of the crate, F_w↓ = 250 kg × 9.81 m/s² = 2,452.5 N

The force the helicopter should provide to just lift the crate, F_{(helicopter)}↑ = The weight of the crate = 2,452.5 N

The net force, F_{(net)} = F_{(helicopter)}↑ - F_w↓ = 2,452.5 N - 2,452.5 N = 0

The net force = 0.

3 0
2 years ago
A car is traveling at 21.0 m/s. It slows to a stop at a constant rate over 5.00s. How far does the car travel during those 5.00
Doss [256]

Answer:

d = 105 m

Explanation:

Speed of a car, v = 21 m/s

We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :

d = v × t

d = 21 m/s × 5 s

d = 105 m

So, it will cover 105 m before it stops.

5 0
2 years ago
A charge +1.9 μC is placed at the center of the hollow spherical conductor with the inner radius 3.8 cm and outer radius 5.6 cm.
Archy [21]

To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.

PART A)

The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:

Q_{int}=-Q1=-1.9*10^{-6} C. This is the total charge on the inner surface of the conducting shell.

PART B)

The positive charge (of the same value) on the external surface of the conducting shell is:

Q_{ext}=+Q_1=1.9*10^{-6} C

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,

Q_{ext, Total}=Q_2+Q_{ext}

Q_{ext, Total}=1.9+3.8

Q_{ext, Total}=5.7 \mu C

5 0
3 years ago
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