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3 years ago

If a 100 N net force acts on a 50 kg car, what will the acceleration of the car will be?

Physics

1 answer:

Allushta [10]3 years ago

6 0

According to newton's law Force = mass * acceleration
so , 100 = 50 * a
so , a= 2 m/s^2

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A train is moving with a velocity of 70 km/h. If north is taken as the positive direction, what is the direction of movement of

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2 years ago

A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and t

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Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

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Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

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3 years ago

A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the

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Answer with Explanation:

We are given that

Mass of squirrel,m=530 g= $\frac{530}{1000}=0.530 kg$

1kg=1000 g

Area=A= $935 cm^2=935\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Height,h=4.4 m

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Width of rectangular prism,b=11.6 cm= $\frac{11.6}{100}=0.116 m$

1 m=100 cm

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Area= $l\times b=0.116\times 0.232=0.0269 m^2$

Terminal velocity, $v_t=\sqrt{\frac{2mg}{\rho CA}}$

Where $g=9.8 m/s^2$

Using the formula

$v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}$

$v_t=17.86 m/s$

The velocity of person, $v=\sqrt{2gh}$

Using the formula

$v=\sqrt{2\times 9.8\times 4.4}$

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3 years ago

Because of the curvature of the earth, the maximum distance D that you can see from the top of a tall building of height h is es

mojhsa [17]

Answer:

D = 9.9 10⁶ mi

Explanation:

In the exercise they give the expression for maximum viewing distance

D = 2 r h + h²

Ask for this distance for a height of 1100 feet

Let's calculate

D = 2 3960 1100 + 1100²

D = 8.712 10⁶ + 1.21 10⁶

D = 9.92 10⁶ mi

D = 9.9 10⁶ mi

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3 years ago