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Julli [10]
3 years ago
14

Lf a dropped hammer's velocity increases from 0.Omis to 15.0 m/s in 4.04s

Physics
1 answer:
Slav-nsk [51]3 years ago
6 0

Answer: The acceleration due to gravity on the surface of the Moon is approximately 1.625 m/s2, about 16.6% that on Earth's surface or 0.166 ɡ.

Heres your answer.

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Suppose Earth's mass increased but Earth's diame-
navik [9.2K]

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

F = G \frac{m1m2}{r^2}

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

7 0
3 years ago
Plz help An X-ray technician will place a lead cloth over parts of your body when taking an X-ray to protect your cells from bei
gayaneshka [121]
Letter d. Hope it helps.
6 0
3 years ago
Read 2 more answers
A 0.155 kg arrow is shot upward
Soloha48 [4]

Answer: 244.05 J

Explanation:

To find speed  at 30 m above the ground use equation:

V²=Vo²-2Gs

V0=31.4m/s

s=30m

G=9.81m/s²

-----------------------

V²=31.4²-2*9.81*30

V²=985.96+588.6

V²=1574.56

V=39.68m/s ---speed of arrow on 30 m obove the ground

Use equation for kinetic enrgy:

Ke=mV²/2

m=0.155kg

V=39.68m/s

-------------------------

Ke=0.155kg*(39.68m/s)²/2

Ke=0.155*1574.5/2

Ke=244.05J

6 0
3 years ago
Help on 9 and 10! Please help me!
Ivan

9 is D I believe, I don't know about 10

6 0
3 years ago
Student bikes to school by traveling first dN = 1.10 miles north, then dW = 0.300 miles west, and finally dS = 0.200 miles south
melisa1 [442]
Taking the vertical component of the displacement
1.1 - 0.2 = 0.9 mile
The horizontal component of the displacement
-0.3 mile

The magnitude of the displacement is
√[ (0.9)² + (-0.3) ] = 0.95 mile

The direction is
θ = tan-1 (-0.3/0.9)
θ = 161.57 degrees.
8 0
3 years ago
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