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AleksandrR [38]
3 years ago
7

Suppose an object is moving in a straight line at 50 miles/hr. According to Newton's first law of motion, the object will ______

___.a. continue to move in the same way forever, no matter what happens. b. continue to move in the same way until it is acted upon by a force. c. eventually slow down and come to a stop. d. continue to move in a straight line forever if it is in space, but fall to the ground if it is on Earth.
Physics
1 answer:
Setler [38]3 years ago
8 0

Answer:

B

Explanation:

Newton's first law of motion states that a body will remain in its state of rest or if its in motion will continue to move in a straight line, unless its acted upon by an external force.The ability of an object to stay at rest or in motion if its in motion is known as inertia.

Hence the correct option is B.

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If we double the mass of the sun, what would happen to the gravitational force between the sun and earth?
Sphinxa [80]
It would increase
because the force is directly proportional to the Value of masses given
4 0
4 years ago
In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small spher
boyakko [2]

Answer:

  L₀ = L_f ,  K_f < K₀

Explanation:

For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.

This means that the angular momentum before and after the collision changes.

Initial instant. Before the crash

        L₀ = I₀ w₀

Final moment. Right after the crash

        L_f = (I₀ + mr²) w

we treat the clay sphere as a point particle

how the angular momentum is conserved

       L₀ = L_f

       I₀ w₀ = (I₀ + mr²) w

       w = \frac{I_o}{I_o + m r^2}   w₀

having the angular velocities we can calculate the kinetic energy

       

starting point. Before the crash

        K₀ = ½ I₀ w₀²

final point. After the crash

        K_f = ½ (I₀ + mr²) w²

sustitute

        K_f = ½ (I₀ + mr²)  ( \frac{I_o}{I_o + m r^2}   w₀)²

        Kf = ½  \frac{I_o^2}{ I_o + m r^2}   w₀²

we look for the relationship between the kinetic energy

        \frac{K_f}{K_o}=   \frac{I_o}{I_o + m r^2}

       \frac{K_f}{K_o } < 1

      K_f < K₀          

we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision

6 0
3 years ago
Where is the switch located on this diagram?
Fed [463]
For this case, the switch is located at point B of the diagram.
 Remember that point D is the universal symbol for resistance.
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 Therefore, the answer for this case is B.
6 0
3 years ago
An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.12 kV . Then it passes through a regi
son4ous [18]

Answer:

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

\frac{1}{2}*m*v^{2}  =e*V

where :

m is the mass of electron

v is the velocity

V is the potential difference

v=\sqrt{\frac{2*e*V}{m} }    eq 1

Radius of electron moving in magnetic field is given by:

R=\frac{m*v}{q*B}       eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}

B=\sqrt{\frac{2*m*V}{e*R^{2} } }

B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3})  }{(1.60*10^{-19})*(0.170)^{2}  } }

B=9.1397*10^-4 Tesla

3 0
3 years ago
A Foucault pendulum consists of a brass sphere with a diameter of 31.0 cm suspended from a steel cable 11.0 m long (both measure
kozerog [31]

Answer:

43.7 °C

Explanation:

\alpha_b = Coefficient of linear expansion of brass = 18\times 10^{-6}\ ^{\circ}C

\alpha_s = Coefficient of linear expansion of steel = 11\times 10^{-6}\ ^{\circ}C

L_{0b} = Initial length of brass = 31 cm

L_{0s} = Initial length of steel = 11 m

\Delta L = Total change in length = 3 mm

Total change in length would be

\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C

\Delta T=23.7\\\Rightarrow T_f-T_i=23.7\\\Rightarrow T_f=23.7+T_i\\\Rightarrow T_f=23.7+20\\\Rightarrow T_f=43.7\ ^{\circ}C

The final temperature is 43.7 °C

6 0
4 years ago
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