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Rom4ik [11]
3 years ago
10

2. A 1750 kg car accelerates at a rate of 4.0 m/s2. How much force is the cars engine producing?

Physics
1 answer:
8090 [49]3 years ago
6 0

Answer:

<h2>7000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 1750 × 4

We have the final answer as

<h3>7000 N</h3>

Hope this helps you

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If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled
mash [69]

Explanation:

The electric field of an isolated charged parallel-plate capacitor is given by :

E=\dfrac{q}{A\epsilon_o}........(1)

Where

q is the electric charge

A is the area of cross section of parallel plate

It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.

So, the correct option is (E) i.e. "none of the above".

5 0
2 years ago
Q7:<br> A 4 kg toy is lifted off the ground and falls at 3 m/s. What is the toy's energy?
alexandr1967 [171]

Answer:

The toy's energy is 18 J.

Explanation:

We have, a 4 kg toy is lifted off the ground and falls at 3 m/s. It is required to find toy's energy.

The toy will have kinetic energy due to its motion. The energy is given by :

E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 4\times 3^2\\\\E=18\ J

So, the toy's energy is 18 J.

7 0
2 years ago
A 10 ohms resistor is powered by a 5-V battery. The current flowing<br> through the source is:
mario62 [17]
  • Resistance=R=10ohm
  • Voltage=V=5V
  • Current=I

Applying ohm's law

\\ \sf\longmapsto \dfrac{V}{I}=R

\\ \sf\longmapsto I=\dfrac{V}{R}

\\ \sf\longmapsto I=\dfrac{5}{10}

\\ \sf\longmapsto I=0.5A

4 0
2 years ago
The electric field is measured for points at distances r from the center of a uniformly charged insulating sphere that has volum
RoseWind [281]

For the electric field is measured for points at distances r,Electric field  is mathematically given as

E=19.9*10^{-5}c/m^3

<h3>What is the Electric field?</h3>

Generally, the equation for the  electric field uniform charged is mathematically given as

E=\frac{lr}{3e0}

Therefore

E=lr/3E0=3Ee0/r

Therefore

E=3*6*10^4*8.85*10^{-12}

E=19.9*10^{-5}c/m^3

In conclusion, E=19.9*10^{-5}c/m^3

E=19.9*10^{-5}c/m^3

Read more about Electric field

brainly.com/question/9383604

CQ

Complete question attached below

4 0
2 years ago
Three wires meet at a junction. wire 1 has a current of 0.40 aa into the junction. the current of wire 2 is 0.73 aa out of the j
Mnenie [13.5K]

The magnitude of the current in wire 3 is (I₃)= 0.33A

<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>

To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,

I₁ + I₂ + I₃ = 0

Where we are given,

I₁ = current in wire 1

=0.40 A.

I₂ = current in wire 2

= -0.73 A.

We have to calculate the magnitude of the current in wire 3, I₃

Now we put the known values in above equation, we get,

I₁ + I₂ + I₃ = 0

Or, I₃ = -.(I₁ + I₂)

Or, I₃ = -.(0.40 - 0.73)

Or, I₃ = 0.33 A

From the above calculation, we can conclude that the current in wire 3 is  I₃ = 0.33 A

Learn more about current:

brainly.com/question/25537936

#SPJ4

7 0
1 year ago
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