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Bad White [126]
3 years ago
8

Describe a solar eclipse. Be sure to include the positions of the sun, moon, and earth.

Physics
1 answer:
Alex777 [14]3 years ago
7 0
It’s the type of eclipse that occurred when the moon passes between the sun and earth, and when the moon fully or partially blocks the sun.
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Find the mass of a flying discus that has a net force of 1.05 newtons and accelerates at 3.5 m/s^2
Ilya [14]
F=ma
m=1.05/3.5= 0.3kg
8 0
3 years ago
A monkey (mass m) is swinging on a vine of length L while carrying a bunch of bananas (a large bunch, mass m/2). His swinging mo
dmitriy555 [2]

Answer:

Explanation:

The period of oscillation will remain unchanged because the period of oscillation of a pendulum does not depend upon the mass of the bob  . Here monkey along with bunch of banana represents bob .

When the monkey and banana were at height h /2 , they have potential energy as well as kinetic energy . banana is separated from the system . It carried its total energy along with it . But the energy of monkey remained intact with it . So it will keep on moving as usual . So it will attain the same maximum height as before .

Hence the amplitude of oscillation too will remain unchanged .

6 0
3 years ago
How long must a tow truck apply a force of 600 N to increase the speed of a 1,500 kg car at rest to 2 m/s?
slega [8]

The time the truck must apply the given force to increase its speed to given value is 5 s.

The given parameters;

  • <em>applied force, F = 600 N</em>
  • <em>mass of the truck, m = 1,500 kg</em>
  • <em>speed of the truck, v = 2 m/s</em>

The force applied to the truck is determined by Newton's second law of motion; <em>which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.</em>

F = ma

F = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{1500 \times 2}{600} \\\\t = 5 \ s

Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.

Learn more here:brainly.com/question/1988795

7 0
2 years ago
An archer puts a 0.30-kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m. Assuming th
vovikov84 [41]

Answer:

41.74 m/s

Explanation:

The energy used to draw the bowstring = the kinetic energy of the arrow.

Fd = 1/2mv²................................ Equation 1

Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.

make v the subject of the equation

v = √(2Fd/m)...................... Equation 2

Given: F = 201 N, m = 0.3 kg, d = 1.3 m.

Substitute into equation 2

v = √(2×201×1.3/0.3)

v = √(1742)

v = 41.74 m/s.

Hence the arrow leave the bow with a speed of 41.74 m/s

3 0
3 years ago
Read 2 more answers
If an egg person starts from rest then falls directly downward and hits the ground with a velocity of 12 m/s but
Anni [7]

Answer:

Explanation: Determine the gravitational acceleration. ...

Decide whether the object has an initial velocity. ...

Choose how long the object is falling. ...

Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt

6 0
2 years ago
Read 2 more answers
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