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Orlov [11]
2 years ago
9

What happens to the period of a spring-cart system motion when the system is in orbit inside the international space station?

Physics
1 answer:
lozanna [386]2 years ago
8 0
The period of a spring-cart system motion when the system is in orbit inside the international space station will not change because in conditions of weightlessness it is not possible to measure the mass of a body from its weight and the period depends on the mass of the body.
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A 7.0-kilogram cart, A, and a 3.0-kilogram cart, B, are initially held together at rest on a horizontal, frictionless surface. W
7nadin3 [17]
For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively

(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s

<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>
7 0
2 years ago
A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106 hz and 1hz=
AURORKA [14]
The frequency of the radio station is
f=88.7 fm= 88.7 MHz = 88.7 \cdot 10^6 Hz

For radio waves (which are electromagnetic waves), the relationship between frequency f and wavelength \lambda is
\lambda= \frac{c}{f}
where c is the speed of light. Substituting the frequency of the radio station, we find the wavelength:
\lambda= \frac{3 \cdot 10^8 m/s}{88.7 \cdot 10^6 Hz}=3.38 m
6 0
2 years ago
Read 2 more answers
A student throws a 0.22 kg rock horizontally at 20.0 m/s from 10.0 m above the ground. Find the initial kinetic energy of the ro
LekaFEV [45]

Answer:

44J

Explanation:

Given parameters:

Mass of rock  = 0.22kg

Initial velocity  = 20m/s

Distance moved  = 10m

Unknown:

Initial kinetic energy of the rock  = ?

Solution:

To solve this problem, we need to understand that kinetic energy is the energy due to the motion of a body.

It is mathematically expressed as;

     Kinetic energy  = \frac{1}{2} m v²

m is the mass

v is the velocity

   Kinetic energy  =  \frac{1}{2} x 0.22 x 20²   = 44J

6 0
1 year ago
To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Kryger [21]

Answer:

H = 10.05 m

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

so this is the speed at the top of flag pole

now we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

now the height of flag pole is given as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
2 years ago
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

\frac{dA}{dt} = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0
2 years ago
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