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Alina [70]
3 years ago
6

Hi! Can you help me? A rock is dropped off a tall building and takes 4 seconds to hit the ground. How fast was it going? Using y

our last answer, how tall was the building? Thanks! :)
Physics
1 answer:
AURORKA [14]3 years ago
4 0
<span><span>Free fall distance h=<span /></span></span>78.4532 m<span><span><span>
</span></span><span>Free fall velocity v=</span></span>39.2266 m/s<span><span><span>
</span></span><span>= </span></span>141.21576 km/h<span><span><span>
</span></span></span>
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As the mass of object increased it is density increased
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Answer:

True.

Explanation:

The density of an object is given by its mass divided by its volume. It can be given as follows :

d=\dfrac{m}{V}

It can be seen that the density of an object is directly proportional to its mass. It means if the mass of an object increase, its density will also increase. Hence, the given statement is true.

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Why does light behave as shown in the image when it passes from air to glass?
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A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe
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Answer: Here is the complete question:

A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 103 N/C. If the ball is in equilibrium when the string makes a 30 angle with the vertical, what is the net charge on the ball?

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Elsie is finishing second grade. If she goes to school 147 day per year and she have 10 years of school left, how many days of s
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6 0
3 years ago
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0
2 years ago
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