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siniylev [52]
3 years ago
11

Compared to that of a plane on the ground, Earth’s gravitational pull on a plane 7 miles above Earth is approximately A) zero. B

) the same. C) half as much. D) twice as much.
Physics
2 answers:
NeTakaya3 years ago
7 0
I am proply right that it is B
Nezavi [6.7K]3 years ago
6 0
The correct answer is 

B) the same
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A little help please?
Tpy6a [65]
A. Made of the marble.

the mass remains constant when you drop the marble but the rest of the variables change as the marble is dropped, therefore, the only constant variable is its mass.
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3 years ago
A 7kg object is dropped on Earth,. Assuming it has not yet hit the ground, what is the velocity of the object after 2 seconds of
kirza4 [7]
The answer would be 9.8 meters/ sec^2.

7 0
3 years ago
Read 2 more answers
4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two
Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:

p_{o}=p_{f} (1)

Where:

p_{o}=mV_{o}+MU_{o} (2)

p_{f}=(m+M)V_{f} (3)

m=1200 kg is the mass of the first car

V_{o}=20 m/s is the velocity of the first car, to the North

M=1400 kg is the mass of the second car

U_{o}=-22 m/s is the mass of the second car, to the South

V_{f} is the final velocity of both cars after the collision

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg} (6)

Finally:

V_{f}=-2.61 m/s (7) This is the resulting velocity of the wreckage, to the south

7 0
3 years ago
A medicine ball has a mass of 5kg and is thrown with a speed of 3 m/sec what is it's kinetic energy
marissa [1.9K]
KE = 1/2mv^2

KE = 1/2 (5kg)(3m/s)

KE = 22.5 J
8 0
2 years ago
A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential
BARSIC [14]

Answer:

1.88\cdot 10^{-5} A

Explanation:

The capacitance of a parallel plate capacitor is given by:

C=\frac{\epsilon_0 A}{d} (1)

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

The charge stored on the capacitor is given by

Q=CV (2)

where C is the capacitance and V is the voltage across the capacitor.

The displacement current in the capacitor is given by

J=\frac{Q}{t} (3)

where t is the time elapsed

Substituting (1) and (2) into (3), we find an expression for the displacement current:

J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}

where we have

A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2

d = 0.58 mm = 5.8\cdot 10^{-4} m

\frac{V}{t}=500,000 V/s

Substituting into the equation, we find

J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A

6 0
3 years ago
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