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3 years ago

A 2-kg book falls off a shelf. it hits a student traveling 2 m/s. how much kinetic energy does the book have?

Physics

1 answer:

jekas [21]3 years ago

5 0

Kinetic energy = 1/2 * m * V^2

K = 1/2 * 2kg * 4

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In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys

son4ous [18]

Answer:

2500 J

Explanation:

We can solve the problem by using the first law of thermodynamics:

$\Delta U =U_f - U_i =Q-W$

where

Uf is the final internal energy of the system

Ui is the initial internal energy

Q is the heat added to the system

W is the work done by the system

In this problem, we have:

Q = +1000 J (heat that enters the system)

W = +500 J (work done by the system)

Ui = 2000 J (initial internal energy)

Using these numbers, we can re-arrange the equation to calculate the final internal energy:

$U_f = U_i + Q-W=2000 J+1000 J-500 J=2500 J$

3 0

3 years ago

Give an example of Newton's 1st law

Vesna [10]

Answer:

A ball moving until gravity pulls it back down to the ground

Explanation:

5 0

3 years ago

A circular loop of radius 11.9 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane

djverab [1.8K]

Answer:

(a). The strength of the magnetic field is 0.1933 T.

(b). The magnetic flux through the loop is zero.

Explanation:

Given that,

Radius = 11.9 cm

Magnetic flux $\phi=8.60\times10^{-3}\ T m^2$

(a). We need to calculate the strength of the magnetic field

Using formula of magnetic flux

$\phi=BA\cos\theta$

$\phi=BA\cos0$

$\phi=BA$

$B=\dfrac{\phi}{A}$

$B=\dfrac{\phi}{\pi r^2}$

Put the value into the formula

$B=\dfrac{8.60\times10^{-3}}{\pi\times(11.9\times10^{-2})^2}$

$B=0.1933\ T$

(b). If the magnetic field is directed parallel to the plane of the loop,

We need to calculate the magnetic flux through the loop

Using formula of flux

$\phi=BA\cos\theta$

Here, $\theta=90^{\circ}$

$\phi=BA\cos90$

$\phi=0$

Hence, (a). The strength of the magnetic field is 0.1933 T.

(b). The magnetic flux through the loop is zero.

3 0

3 years ago

Read 2 more answers

I need helpppppppppppppppppp!

allsm [11]

Answer:

i think its B

Explanation:

but check it before do it

3 0

3 years ago

Read 2 more answers

Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0

3 years ago