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Masteriza [31]
1 year ago
6

A ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally aw

ay from and below the point of release. What is the magnitude of its velocity just before it strikes the ground?​
Physics
1 answer:
astraxan [27]1 year ago
5 0

26.54 m/s  is the magnitude of its velocity just before it strikes the ground

h=100m,v=20m/s,g=9.8m/s

time it takes to reach the ground,

[t=\sqrt2h/g],[=\sqrt2*100/9.8=4.51s]

x= 120m

t= 4.52

v= x/t

v= 120/4.52

v= 26.54 m/s

The "speed at which an object changes its location" can be expressed using a vector number called velocity. Consider a person who moves swiftly while taking two steps forward and two steps back while remaining in one location. Velocity is a vector quantity. Therefore, velocity is cognizant of direction. The direction must be taken into account when determining an object's velocity. A speed of 55 mph is not enough information. The direction must be used to appropriately depict the item's velocity. Simply said, the direction of the velocity vector indicates the direction of motion of an object.

To know more about  velocity visit : brainly.com/question/16379705

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The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold
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Answer:

L_{o}=0.1224m

Explanation:

Given data

Force F=2 N

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Spring Constant k=42 N/m

To find

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Solution

From Hooke's Law we know that

F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m

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3 years ago
The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the
Inga [223]

Answer:

27.5 days

0.92 month

Explanation:

r = radius of the orbit of moon around the earth = 3.85\times10^{8} m

M = Mass of earth = 5.98\times10^{24} m

T = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

T^{2} = \frac{4\pi ^{2} r^{3}  }{GM}

inserting the values, we get

T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3}  }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days

1 month = 30 days

T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month

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3 years ago
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