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JulsSmile [24]
3 years ago
12

a ball is thrown down at 25 m/s from a 500m tall building. how fast is it traveling when it hits the ground?

Physics
1 answer:
Mrrafil [7]3 years ago
3 0

Answer:

The speed of the ball when it hits the ground is 102.1 m/s

Explanation:

Given;

initial velocity of ball, u = 25 m/s

distance traveled by the ball = height of the building = h = 500 m

when the ball hits the ground, the final velocity, v = ?

The final velocity of the ball is given by;

v² = u² + 2gh

where;

g is acceleration due to gravity = 9.8 m/s²

v² = (25)² + 2(9.8)(500)

v² = 10425

v = √10425

v = 102.1 m/s

Therefore, the speed of the ball when it hits the ground is 102.1 m/s

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Answer:

1.7×10^5 ms-1

Explanation:

From

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heterotrophs

Explanation:

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The term heterotroph comes from the Greek, language in which the prefix hetero means different and trophies means food. In this way, the heterotroph is one that feeds on elements other than one, which takes elements from nature, from the surrounding space to feed. While autotrophic beings have the ability to synthesize inorganic elements such as light, water, carbon dioxide and convert them into food; Heterotrophic beings do not have that capacity, so they must consume plants (in the case that they are herbivores) or animals that have already consumed those plants (that is, in the case that they are carnivorous). In other words, animals and humans always need to feed on other living beings, they could never do so only from inorganic elements such as water.

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A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
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Answer: 1.51 km

Explanation:

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Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

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\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

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