All categories

3 years ago

HELP!! AM I CORRECT?? PLS TELL ME, IF U ANSWER PROPERLY I'LL GIVE U BRAINLIEST!!

Physics

2 answers:

sweet [91]3 years ago

7 0

Answer:

HON, i'm going to be honest, i'm not sure but it sounds like it is the Universal Expansion from what I learned about space

I'm sure this helps but if it doesn't DON'T GIVE ME BRAINLIEST, if it does then do

S_A_V [24]3 years ago

3 0

Answer:ur right I think

Explanation:

You might be interested in

Which free body diagram is in equilibrium?

bearhunter [10]

A. because everything is balanced.

6 0

3 years ago

the two rotating systems shown in the figure differ only in that the two identical movable masses are positioned at different di

svetoff [14.1K]

Answer: the block at the right lands first

Explanation:

7 0

3 years ago

What is sound reverberation?

otez555 [7]

Sound reverberation is created when sound or signal is reflected.

3 0

3 years ago

A 13.0-g wad of sticky clay is hurled horizontally at a 110-g wooden block initially at rest on a horizontal surface. The clay s

Alisiya [41]

Answer:

$v_{ic}=92.53 m/s$

Explanation:

We need to apply conservation of momentum and energy to solve this problem.

Conservation of momentum

$p_{i}=p_{f}$

$m_{c}v_{ic}=(m_{c}+m_{w})V$ (1)

m(c) is the mass of stick clay
m(w) is the mass of the wooden block
v(ic) is the initial velocity of clay
V is the final velocity of the system clay plus wood.

Conservation of total energy

The change in kinetic energy is equal to the change in internal energy, in our case it would be the energy loss due to the friction force. Let's recall the definition of work, it is the dot product between force and displacement, Therefore:

$\Delta E=W$

$\frac{1}{2}(m_{c}+m_{w})V^{2}=F_{friction}*d$

$\frac{1}{2}(m_{c}+m_{w})V^{2}=\mu (m_{c}+m_{w})gd$

We can find V from this equation:

$V=\sqrt{2\mu gd}=\sqrt{2*0.65*9.81*7.5}=9.78 m/s$

Now, let's put V into the equation (1) and find v(ic)

$v_{ic}=\frac{(m_{c}+m_{w})V}{m_{c}}=\frac{123*9.78}{13}=92.53 m/s$

I hope it helps you!

8 0

3 years ago

A small 0.13 kg metal ball is tied to a very light (essentially massless) string that is 0.70 m long. The string is attached to

beks73 [17]

Answer:

Explanation:

Let l be th length of pendulum

loss of height

= mg ( l - l cos50)

= mg l ( 1-cos50)

1/2 mv² = mgl ( 1-cos50)

v = √[2gl( 1- cos50)]

= √( 2 x 9.8 x .7 x ( 1-cos50)

= 2.2 m / s

speed at the bottom = 2.2 m /s

b )

centripetal acceleration

= v² / r

= 2.2 x 2.2 / .7

= 6.9 m /s²

C )

If T be the tension

T - mg = mv² / r

T = mg + mv² / r

= .13 X 9.8 + .13 X 6.9

= 2.17 N

7 0

3 years ago