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3 years ago

Science: Look at the image and answer the question correctly.

Physics

1 answer:

Flauer [41]3 years ago

4 0

I believe it is D I’m sorry if it’s wrong

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The first few hundred million years of the solar system's history were the time of the __________, during which Earth suffered m

azamat

Answer:

heavy bombardment

Explanation:

5 0

3 years ago

The height of an object dropped from the top of a 64-foot building is given by h(t)=-16t^2+64. How long will it take the object

mojhsa [17]

Answer:

1.86 s

Explanation:

Given the expression

h(t) = -16t²+ 64...................... Equation 1

Where h = height of the object, t = time it will take the object to hit the ground.

Given: h = 64 foot.

We have to concert from foot to meters

If 1 foot = 0.3048 meters

Then, 64 foot = 0.3048×64 = 19.51 meters.

We substitute the value of h into equation

119.51 = -16t²+64

-16t² = 199.51-64

-16t² = 55.51

t² = 55.51/-16

t² = 3.469

t = √3.469

t = 1.86 s.

Hence it will take the object 1.86 s to hit the ground.

7 0

3 years ago

9. Kokio dydžo Archimedo jėga veikia 0,5 m² tūrio medinį rastą vandenyje?

ICE Princess25 [194]

Answer:

i dont knowwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

6 0

3 years ago

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

tekilochka [14]

Answer: a) It will take more time to return to the point from which it was released

Explanation: To determine how long it takes for the ball to return to the point of release and considering it is a free fall system, we can use the given formula:

$d=v_{0}.t + \frac{1}{2} .a.t^{2}$ , where:

d is the distance the ball go through;

v₀ is the initial velocity, which is this case is 0 because he releases the ball;

a is acceleration due to gravity;

t is the time necessary for the fall;

Suppose <em>h</em> is the height from where the ball was dropped.

On Earth:

h=0.t + $\frac{1}{2}.10.t^{2}$

h = 5t²

$t_{T}$ = $\sqrt{\frac{h}{5} }$

On the other planet:

h = 0.t + $\frac{1}{2}.30.t^{2}$

h = 15.t²

$t_{P}$ = $\sqrt{\frac{h}{15} }$

Comparing the 2 planets:

$\frac{t_{T} }{t_{P} } = \frac{\sqrt{\frac{h}{5} } }\sqrt{{\frac{h}{15} } }$

$\frac{t_{T} }{t_{P} }$ = $\sqrt{3}$ or $t_{T} = \sqrt{3}.t_{P}$

Comparing the two planets, on the massive planet, it will take more time to fall the height than on Earth. In consequence, it will take more time to return to the initial point, when it was released.

5 0

3 years ago

1) Halving the distance (i.s., decreasing by a factor of two) between two charged objects will cause the electrical force betwee

Vedmedyk [2.9K]

Answer:

Explanation:

For an electric force, F the formula:

F = kQq/r^2

Given:

r2 = 1/2 × r1

F1 × r1 = k

F1 × r1 = F2 × r2

F2 = (F1 × r1^2)/(0.5 × r1)^2

= (F1 × r1^2)/0.25r1^2

= 4 × F1.

7 0

4 years ago