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Bess [88]
3 years ago
14

10. What is momentum of the water bottle before the collision? (read the

Physics
1 answer:
Alona [7]3 years ago
8 0

Answer:

O ksm/s

Explanation:

before collision,

Velocity =0

So,momentum of the bottle before collision=mass ×velocity

=mass×0

=0 kgm/s

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A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k
rewona [7]

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

3 0
3 years ago
A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of the E
Sveta_85 [38]

Answer:

995 N

Explanation:

Weight of surface, w= 4000N

Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

m= 4000/9.81= 407.7472

Using radius of orbit of 6371km

The force of gravity of satellite in its orbit, F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}

Where G=6.67*10^{-11} and M=5.94*10^{24}

F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}

F= 995.01142 then rounded off

F=995N

6 0
3 years ago
Which of the following statements about X-rays and radio waves is not true? Which of the following statements about X-rays and r
Lapatulllka [165]

Answer:

X-rays travel through space faster than radio waves.

Explanation:

Electromagnetic waves consist of oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave.

All electromagnetic waves travel in a vacuum always at the same speed, the speed of light, whose value is:

c=3.0\cdot 10^8 m/s

Electromagnetic waves are classified into 7 different types, according to their wavelength/frequency. From shortest to longest wavelength (and so, from highest to lowest frequency), we have:

Gamma rays

X rays

Ultraviolet

Visible light

Infrared radiation

Microwaves

Radio waves

Now we can analyze the 4 statements:

X-rays and radio waves are both forms of light, or electromagnetic radiation --> TRUE. They are both types of electromagnetic waves.

X-rays have higher frequency than radio waves. --> TRUE, as we can see from the table above.

X-rays have shorter wavelengths than radio waves. --> TRUE, as we can see from the table above.

X-rays travel through space faster than radio waves. --> FALSE: all electromagnetic waves travel in space at the same speed, the speed of light.

8 0
3 years ago
A car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. When the brakes are applied, it takes the ca
Mariulka [41]

i believe it is C....tell me if im right please<3

4 0
3 years ago
Read 3 more answers
Rearrange the equation to solve for the permeability of free space (μ0). Remember that the slope is the ratio of magnetic field
IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %

Answer:

The permeability of free space is \mu_0 = 1.32*10^{-6} \ Tm/A

The percentage error is  % error = 5.25%

Explanation:

From the question we are told that

            The slope is  s= 3.9\  G/A , and

Generally  1 \ gauss =  10^{-4 } tesla

               So  s = 3.9 *10^ {-4} T /A

From the relation given in the question

                        \mu_0 = \frac{2R}{N} [\frac{B}{I} ]

 Where R is the radius of the coil  =\frac{Diameter \ of \ coil }{2} = 0.017m

             N is the number of loops of the coil = 10

Now from the question we are told that

                  s = \frac{B}{I}

substituting into the equation above

               \mu_0 = \frac{2R }{N} s

Substituting values

              \mu_0= \frac{2 * 0.017}{10 } * 3.9*10^{-4}

                   = 1.326 *10^ {-6} \ Tm/A

Generally the % error is mathematically represented as

                    %error = \frac{Measured \  value - accepted \ value}{accepted \  value } *100

Given that the accepted value is \mu_o_ {acc} = 1.26 *10 ^{-6} \ T \cdot m /A

      Hence substituting values

                      %error  = \frac{(1.326-1.26)*10^{-6}}{1.26 *10 ^ {-6}} *100

                                  = 5.24

6 0
3 years ago
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