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NikAS [45]
3 years ago
11

An astronaut drops a hammer from 2.0 meters above the surface of the Moon. If the acceleration due to gravity on the Moon is 1.6

2 meters per second2, how long will it take for the hammer to fall to the Moon’s surface?
A. 0.62 s
B. 1.2 s
C. 1.6 s
D. 2.5 s
Physics
1 answer:
Elenna [48]3 years ago
8 0

Answer:

option C

Explanation:

given,

height of the hammer, h = 2 m

acceleration due to gravity of the moon, g = 1.62 m/s²

time taken by the hammer to reach at bottom = ?

initial speed of the hammer, u = 0 m/s

using equation of motion for the time calculation

s = ut + \dfrac{1}{2}gt^2

2 = 0+\dfrac{1}{2}\times 1.62\times t^2

    t² = 2.469

     t = 1.57 s ≅ 1.6 s

hence, the time taken by the hammer to reach the bottom is equal to 1.6 s.

The correct answer is option C

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Calculate the force on an object with mass of 50kg and gravity of 10​
GarryVolchara [31]
Answer: 500 N

Explanation:

The formula to find the force exerted by a mass, we may use F = mg, where g, the gravity, and a, the acceleration, can be interchangeable in the formula.

1) F = 50 x 10
2) F = 500 N

Hope this helps, brainliest would be appreciated :)
6 0
3 years ago
Please help me out as soon as you can. Really need help.
olga nikolaevna [1]
The answer would be 6 because 2.0x3= 6

(newton’s 2nd law)

mark me brainliest
4 0
3 years ago
How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s
solong [7]

Answer:

71 rpm

Explanation:

Given that:

Angular momentum (L) = 0.26

Diameter = 25cm = 0.25 cm

Radius, r = (d/2) = 0.125m

Mass = 5.6 kg

Moment of inertia (I) = 2mr² / 5

I = (2 * 5.6 * 0.125^2) / 5

= 0.175

= 0.175 / 5

= 0.035 kgm²

Angular speed (w) ;

w = L / I

w = 0.26 / 0.035

= 7.4285714

= 7.429 rad/s

w = (7.429 * 60/2π)

w = 445.74 / 2π rpm

w = 70.941724

Angular speed = 70.94 rpm

= 71 rpm

5 0
3 years ago
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
IrinaK [193]

Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

6 0
3 years ago
A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of wor
joja [24]

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is \Delta U_1=222 J

Work done on the system W_1=-150 J

For second step

change in internal Energy of the system is \Delta U_2=123 J

Work done on the system W_2=-195 J

Work done on the system is considered as Positive and vice-versa.

and from first law of thermodynamics

Q=\Delta U+W

for first step

Q_1=222-150=72 J

Q_2=123-195=-72 J

overall heat added=Q_1+Q_2

Q_{net}=72-72 =0

For overall Process Heat added is 0 J

8 0
3 years ago
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