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3 years ago

Force exerted by a person or object is called

1 answer:

6 0

Force exerted by a person or thing is called _____ force. applied. A change in the speed or direction of an object is called. acceleration. Force is a vector.

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A car travels from Boston to Hartford in 4 hours. The two cities are 240 kilometers apart. What was the average speed of the car

Y_Kistochka [10]

Answer:

Average speed is 60 km/hour

Explanation:

When we need to calculate average speed, we use this equation:

$V = \frac{x_{f} - x_{o}}{t_{f} - t_{o}}$

Where: $x_{o} = 0 km$ position at the beginning

$x_{f} = 240 km$ at the end

$t_{o} = 0 hours$

$t_{f} = 4 hours$

Then: $V = \frac{240 km - 0 km}{4 hours - 0 hours}$

$V = \frac{240 km}{4 hours}$

Finally V = 60 km/hour

6 0

3 years ago

5. 3 women push a stalled car. Each woman pushes with a 400N force. What is the mass of the car if the car accelerates at 0.85 m

iris [78.8K]

Answer:

Explanation:

The the mass of the car can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{400}{0.85} \\ = 470.58823...$

We have the final answer as

Hope this helps you

7 0

3 years ago

An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th

neonofarm [45]

Answer:

Electric force, $F=2.24\times 10^{-14}\ N$

Explanation:

It is given that,

Charge on an electron is $-1.6\times 10^{-19}\ C$

Electric field, $E=1.4\times 10^5\ N/m$

We need to find the magnitude of the electric force on this electron due to this field. The electric force is given by :

$F=qE\\\\F=1.6\times 10^{-19}\times 1.4\times 10^5\\\\F=2.24\times 10^{-14}\ N$

So, the electric force is $2.24\times 10^{-14}\ N$ .

6 0

3 years ago

Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

kobusy [5.1K]

Answer:

$\Delta \lambda=14.3\ nm$

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, $\Delta Y=2.98\ mm = 0.00298\ m$

Let d is the slit width of the grating,

$d=\dfrac{1}{N}$

$d=\dfrac{1}{900\ cm}$

$d=1.11\times 10^{-5}\ m$

For the first wavelength, the position of maxima is given by :

$y_1=\dfrac{L\lambda_1}{d}$

For the other wavelength, the position of maxima is given by :

$y_2=\dfrac{L\lambda_2}{d}$

So,

$\Delta \lambda=\dfrac{\Delta y d}{l}$

$\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}$

$\Delta \lambda=1.43\times 10^{-8}\ m$

$\Delta \lambda=14.3\ nm$

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0

3 years ago

Driving Zone 7 corresponds with

stepan [7]

Your driving zone refers to the areas of space around your car, it refers to all the area around your car as far as your eyes can see.
Each car has seven zones numbered from 1 to 7. Driving zone 7 corresponds with THE SPACE YOUR VEHICLE IS OCCUPYING. The other zones are as follows:
zone 1 = area directly infront of your car
zone 2 = your left lane
zone 3 = your right lane
zone 4 = left rear of your car
zone 5 = right rear of your car
zone 6 = area directly behind your car.
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3 0

3 years ago