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vladimir1956 [14]
3 years ago
11

What is the meaning of joule​

Physics
2 answers:
SOVA2 [1]3 years ago
7 0

A joule is a unit of energy.

slega [8]3 years ago
7 0

Answer:

The work done by a force of one Newton is known as joule

Explanation:

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A 4.33 kg cat has 41.7 J of KE How fast is the cat moving?
balandron [24]

Answer:

The answer to your question is:

Explanation:

Data

mass = 4.33 kg

E = 41.7 J

v = ?

Formula

Ke = (1/2)mv²

Clear v from the equation

v = √2ke/m

Substitution

v = √2(41.7)/4.33

v = 19.26 m/s          Result

7 0
3 years ago
A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,
shusha [124]

Answer:

v=d\sqrt{\frac{k}{m}}

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

U_{0}+K_{0}=U_{f}+K_{f}

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

U_{0}=K_{f}

The initial potential energy of the spring is given by the equation:

U_{0}=\frac{1}{2}kd^{2}

the Kinetic energy of the block is then given by the equation:

K_{f}=\frac{1}{2}mv_{f}^{2}

so we can now set them both equal to each other, so we get:

=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

kd^{2}=mv_{f}^{2}

so now we can solve this for the final velocity, so we get:

v=d\sqrt{\frac{k}{m}}

6 0
3 years ago
A person pulls a box across the floor with a rope. The rope makes an angle of 40 degrees tot he horizontal, and a total of 125 n
RSB [31]

Answer:

The angle formed of the rope with the surface = 40°

Force applied = 125Newtons

The displacement covered by the box =25metres

W= FDcos theta

[125×40×cos(40°) ] Joules

= [ (3125×0.76604444311)]Joules

= 2393.88888472 joules(ans)

Hope it helps

3 0
2 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
A ball is dropped from rest from the top of a cliff that is 15.0 m high. From ground level, a second ball is thrown straight upw
lesya [120]

Answer:

3.75 meters below the top of the cliff.

7 0
3 years ago
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