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tia_tia [17]
2 years ago
11

A student attaches a block to a vertical spring so that the block-spring system will oscillate if the block-spring system is rel

eased from rest at a vertical position that is not the system’s equilibrium position. The system oscillates near Earth’s surface. The system is then taken to the Moon’s surface, where the gravitational field strength is nearly 16 that of the gravitational field strength near Earth's surface. Which of the following claims is correct about the period of oscillation for the system?
Physics
1 answer:
vodka [1.7K]2 years ago
4 0

Answer:

Time period of the motion will remain the same while the amplitude of the motion will change

Explanation:

As we know that time period of oscillation of spring block system is given as

T= 2\pi\sqrt{\frac{M}{k}}

now we know that

M = mass of the object

k = spring constant

So here we know that the time period is independent of the gravity

while the maximum displacement of the spring from its mean position will depends on the gravity as

mg = kx

x = \frac{mg}{k}

so we can say that

Time period of the motion will remain the same while the amplitude of the motion will change

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the water behind hoover dam in nevada is 206 m higher than the colorado river below it. at what rate must water pass through the
Nitella [24]

Answer:

the required mass flow rate is 49484.37 kg/s

Explanation:

Given the data in the question;

we first determine the relation for mass flow rate of water that passes through the turbine;

so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;

W_{net} = m ( Δ P.E )

so we substitute (gh) for ( Δ P.E );

W_{net} = m (gh)

m = W_{net} / gh

so we substitute our given values into the equation

m = 100 MW / ( 9.81 m/s²) × 206 m

m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m

m = 10 × 10⁷ / 2020.86

m = 49484.37 kg/s

Therefore, the required mass flow rate is 49484.37 kg/s

5 0
2 years ago
A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a
IgorC [24]

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0
2 years ago
Many birds can attain very high speeds when diving. Using radar, scientists measured the altitude of a barn swallow in a vertica
scoray [572]

Answer:

0.109

Explanation:

8 0
2 years ago
In an experiment, Mary compares a collision of two 1-kilogram steel balls with a collision of two 1-kilogram foam rubber balls.
IgorC [24]
1. Greater then
<span>2. Inelastic</span>
8 0
2 years ago
Read 2 more answers
To support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground. T
podryga [215]

Answer:

The wire meet the ground at an angle of 56.4 degrees

Explanation:

It is given that,

To support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground.

The hypotenuse is, H = 12 foot

The perpendicular distance is, P = 10 feet

The angle between the tree and the ground is 90 degrees

Using Pythagoras theorem as :

sin\theta=\dfrac{P}{H}

sin\theta=\dfrac{10}{12}

\theta=56.4^{\circ}

So, the wire meet the ground at an angle of 56.4 degrees. Hence, the correct option is (d).                                                    

6 0
1 year ago
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