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3 years ago

A compact car moving at 60 m/s has 320,000 J of kinetic energy. What is its mass?

Physics

1 answer:

joja [24]3 years ago

8 0

Answer:

320,000J

Explanation:

mass=320,000J

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3 years ago

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A man throws a ball straight up to his friend on a balcony who catches it at its highest point. The ball was thrown with an init

Ede4ka [16]

Answer:

The maximum height reached by the ball is 16.35 m.

Explanation:

Given;

initial velocity of the ball, u = 17.9 m/s

the final velocity of the ball at the maximum height, v = 0

The maximum height reached by the ball is given by;

v² = u² + 2gh

During upward motion, gravity is negative

v² = u² + 2(-g)h

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u² / 2g

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h = 16.35 m

Ttherefore, the maximum height reached by the ball is 16.35 m.

3 0

3 years ago

A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =

natali 33 [55]

With acceleration

$\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j$

and initial velocity

$\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i$

the velocity at time t (b) is given by

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

We can get the position at time t (a) by integrating the velocity:

$\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du$

The particle starts at the origin, so $\mathbf x(0)=\mathbf0$ .

$\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du$

$\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}$

$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at t = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

$\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j$

so the particle is located at (x, y) = (64.0, 64.0).

Get the speed at t = 8.00 s by evaluating $\mathbf v(t)$ at the same time:

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

This is the velocity at t = 8.00 s. Get the speed by computing the magnitude of this vector:

$\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}$

5 0

3 years ago