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Karo-lina-s [1.5K]
3 years ago
14

According to Einstein's theory of relativity, what happens to the mass of an object as the object approaches the

Physics
2 answers:
Anuta_ua [19.1K]3 years ago
8 0

Answer:

B

Explanation:

please mark me brainiest

irina [24]3 years ago
7 0
The answer is b. Because that is one of the characteristics of light
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3. Suppose you take a pendulum with length L and mass m having a period T to a
natta225 [31]

(C)

Explanation:

t = 2\pi \sqrt{ \frac{l}{g} }

If g is only 1/6 on another planet, then

t = 2\pi \sqrt{ \frac{l}{ \frac{g}{6} } } =  2\pi \sqrt{ \frac{6l}{g} }

=  \sqrt{6} \: (2\pi \sqrt{ \frac{l}{g} } ) = 2.4  \times t(on \: earth)

6 0
3 years ago
Find the acceleration if a 32.5 N force is<br> used on an object that has a mass of<br> 128.6 kg.
Sholpan [36]

Answer:

Acceleration=3.95

Explanation:Use the formula a=m/f

a=128.6/32.5

a=3.95

3 0
3 years ago
Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birm
forsale [732]

Answer:

they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta

Explanation:

use in your own words teachers know when your not trust me.

4 0
3 years ago
The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

3 0
3 years ago
What is the magnitude of the momentum change of two gallons of water (inertia about 7.3 kg ) as it comes to a stop in a bathtub
aliya0001 [1]

We know that the change in momentum is equals to the product of force and time that is impulse (  F \times t). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,

s=ut+\frac{1}{2} gt^2

Here, u is initial velocity which is zero.

s= \frac{1}{2} gt^2 \\\\ t = \sqrt{\frac{2s}{g} }.

Thus, impulse

= F \times \sqrt{\frac{2s}{g} }

From Newton`s second law,

F =mg

Therefore, impulse

= mg \times \sqrt{\frac{2s}{g} } = m \sqrt{2gs}

Given,  m = 7.3 kg and s = 2.0 m

Substituting these values, we get

Change in momentum = impulse  

= 7.3 \ kg \sqrt{2 \times 9.8 \ m/s^2 \times 2.0 \ m } = 45 .7 \ Ns.

8 0
3 years ago
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