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Andreas93 [3]
3 years ago
13

Someone who is a girl friend me because I’m trying to get simp record

Physics
2 answers:
Lelu [443]3 years ago
6 0

Answer:

okay then

Explanation:

timurjin [86]3 years ago
3 0

Answer:

wdym

Explanation:

You might be interested in
Which is the momentum of a 100 kilogram boulder rolling south down a hill at a velocity of 5 meters per second
adell [148]
P = m * v
P = 100 * 5 Kg-m/s towards south
P = 500 Kg-m/s towards south

In short, your answer would be 500 Kg-m/s towards south

Hope this helps!
6 0
3 years ago
1. A dog of mass 5 kg jumps on a table of mass 30 kg. As the dog walks around on the table, what is the average force that the t
Oksana_A [137]

Answer:

50N

The table is managing to keep the dog up, the dog is not moving up or down so the force that the table applies to the dog must be equal to the dogs weight - I draw this conclusion from Newton's 3rd law.

W= mg

W = 5 × 9.8

W = 49N

round to.one significant figure = 50N

5 0
3 years ago
The part of the shore that represents the water's edge and migrates back and forth with the tide is known as the ________.
Dennis_Churaev [7]

Answer :It is know as the shore line

Explanation:

7 0
3 years ago
A 5 kg book falls from a shelf. If it lands at a speed of 5.7 m/s, from what height did it fall?
Pachacha [2.7K]

Answer:

57m

Explanation:

speed = 5.7m/s

or 57/10 m/s

so definitely the time taken will be 10 secs

so speed = distance/time taken

5.7= distance/10

distance = 5.7× 10

Distance = 57m

7 0
3 years ago
A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to
Naily [24]

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

\tau = I\alpha

Where \tau is the torque

I is the moment of inertia

\alpha is the angular acceleration

But, the angular acceleration is given by

\alpha = \frac{\omega}{t}

Where \omega is the angular speed

and t is time

Then, we can write that

\tau = \frac{I\omega}{t}

Hence,

\omega = \frac{\tau t}{I}

Now, to determine the angular speed, we first determine the Torque \tau and the moment of inertia I.

Here, The torque is given by,

\tau = rF

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ \tau = 3.00 \times 195

\tau = 585 Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

I = \frac{1}{2}MR^{2}

Where M is the mass and

R is the radius

∴I = \frac{1}{2} \times 325 \times (3.00)^{2}

I = 1462.5 kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

\omega = \frac{\tau t}{I}

\omega = \frac{585 \times 2.05}{1462.5}

\omega = 0.82 rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

6 0
3 years ago
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