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aalyn [17]
4 years ago
10

calculate the acceleration of a 520 kg car that had 5050N of net force exerted on it by the engine. explain how you solve the pr

oblem please
Physics
1 answer:
elena55 [62]4 years ago
3 0

I'll explain first:

You use the formula for Newton's second law of motion.
The formula is so simple, and it solves so many problems,
that it's actually a good idea to memorize it.
(I learned it 58 years ago, and I still remember it.)

The formula is                 F  =  M A

It means                    Force = (mass) x (acceleration)

In this question, you know the force, and you know the mass,
so calculating the acceleration is a piece-o-cake.

                                                     Force = (mass) x (acceleration)

                                                (5,050 N) = (520 kg) x (acceleration)

Divide each side by  520 kg :  (5,050 N / 520 kg) = acceleration

                                                 Acceleration = 9.712 m/s² (rounded)
 
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Answer:

v = 5.9 x 10⁷ m/s

Explanation:

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K.E = eV--------------- equation (1)

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K.E = \frac{1}{2}mv^{2}\\--------- equation (2)

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = ?

Therefore, comparing equation (1) and equation (2), we get:

\\\frac{1}{2}mv^{2} = eV\\\\\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)v^{2} = (1.6\ x\ 10^{-19}\ C)(9900\ volts)\\\\v = \sqrt{34.81\ x\ 10^{14}} \\

<u>v = 5.9 x 10⁷ m/s</u>

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A tall, open container is full of glycerine. At what depth h below the surface of the glycerine is the pressure 2370 Pa greater
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Answer:

So, at the depth of 24 cm below the surface of the glycerine the pressure is  2970 Pa. Hence, this is the required solution.

Explanation:

Given that,

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The density of glycerine,  

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h = 0.24 meters

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h = 24 cm

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Answer:

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3 years ago
Which change is an example of transforming potential energy to kinetic energy?.
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Answer:

Kinetic energy is energy an object has because of its motion. A ball held in the air, for example, has gravitational potential energy. If released, as the ball moves faster and faster toward the ground, the force of gravity will transfer the potential energy to kinetic energy.

Explanation:

there hope this helps

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2 years ago
A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball
adell [148]

Answer:

Part a)

T = 2\sqrt{\frac{R}{3g}}

Part b)

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

v_y = \sqrt{Rg/3}

Part d)

v = \frac{1}{2}\sqrt{13Rg}

Part e)

\theta_i = 33.7 degree

Part f)

H = \frac{13R}{8}

Part g)

X = \frac{13R}{4}

Explanation:

Initial speed of the launch is given as

initial speed = v_i

angle = \theta_i degree

Now the two components of the velocity

v_x = v_i cos\theta_i

similarly we have

v_y = v_i sin\theta_i

Part a)

Now we know that horizontal range is given as

R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}

maximum height is given as

H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}

so we have

v_i sin\theta = \sqrt{Rg/3}

time of flight is given as

T = \frac{2v_isin\theta_i}{g}

T = \frac{2\sqrt{Rg/3}}{g}

T = 2\sqrt{\frac{R}{3g}}

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

R = v_x T

R = v_x 2\sqrt{\frac{R}{3g}}

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

Initial vertical velocity is given as

v_y = v_i sin\theta_i

v_i sin\theta = \sqrt{Rg/3}

Part d)

Initial speed is given as

v = \sqrt{v_x^2 + v_y^2}

so we will have

v = \sqrt{Rg/3 + 3Rg/4}

v = \frac{1}{2}\sqrt{13Rg}

Part e)

Angle of projection is given as

tan\theta_i = \frac{v_y}{v_x}

tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}

\theta_i = 33.7 degree

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

H = \frac{v^2}{2g}

H = \frac{13R}{8}

Part g)

For maximum range the angle should be 45 degree

so maximum range is

X = \frac{v^2}{g}

X = \frac{13R}{4}

3 0
3 years ago
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