All categories

2 years ago

To drive a car at a constant velocity, you

Physics

1 answer:

kipiarov [429]2 years ago

8 0

Answer:

the answer is C

Explanation:

The car, first is at rest and if you don't accelerate it won't move. When to hit the gas it will accelerate from rest

You might be interested in

Why does it take double the applied force to move a mass double the size?

Ratling [72]

56-999999999999999999999-4 is the best for my mom

5 0

3 years ago

If 30ml of a fluid has a mass of 63g what is it’s density

murzikaleks [220]

Answer:

The answer to your question is 2.1 g/ml

Explanation:

Data

volume = 30 ml

mass = 63 g

density = ?

Process

Density is defined as the mass per unit volume. The units of density are g/ml or kg/m³.

Formula

Density = mass / volume

Substitution

Density = 63 / 30

Result

Density = 2.1 g/ml

7 0

3 years ago

A pulley has a mechanical advantage of 1. What does this tell you about the size and direction of the input and output forces?

Nadya [2.5K]

Answer:

The number of input force is the same as output. ... If it equals once, then both numbers are equal making it the same.Explanation:

3 0

3 years ago

Read 2 more answers

A fluid is a substance that can _____________ and takes the _____________ of its container.

mezya [45]

Evaporate
Shape

I wish there were provided answer choices. Hope this helped

7 0

3 years ago

A conducting sphere with a radius of 0.25 m has a total charge of 5.90 mC. A particle with a charge of −1.70 mC is initially 0.3

notsponge [240]

Explanation:

The given data is as follows.

$r_{1}$ = 0.25 m, q = 5.90 mC = $5.90 \times 10^{-3} C$

$r_{2}$ = 0.35 m, q = 1.70 mC = $1.70 \times 10^{-3} C$

(a) Now, we will calculate the electric potential as follows.

V = $k \frac{q}{r}$

First, we will calculate the initial and final electric potential as follows.

$V_{i} = 9 \times 10^{9} \times \frac{5.90 \times 10^{-3}}{0.25 m}$

= $212.4 \times 10^{6}$

or, = $2.124 \times 10^{8}$

$V_{f} = 9 \times 10^{9} \times \frac{1.70 \times 10^{-3}}{0.35 m}$

= $43.71 \times 10^{6}$

or, = $4.371 \times 10^{8}$

Hence, the value of change in electric potential is as follows.

$\Delta V = V_{f} - V_{i}$

= $4.371 \times 10^{8} - 2.124 \times 10^{8}$

= $2.247 \times 10^{8}$ V

Therefore, the difference in electric potential energy is $2.247 \times 10^{8}$ V.

(b) Now, we will calculate the potential energy as follows.

P.E = qV

= $-1.70 \times 10^{-3}C \times 2.247 \times 10^{8}$ V

= $-3.8199 \times 10^{5}$

Therefore, the change in the system's electric potential energy is $-3.8199 \times 10^{5}$ .

4 0

3 years ago