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zalisa [80]
3 years ago
10

A brick is released with no initial speed from the roof of a building and strikes the ground in 1.80 s , encountering no appreci

able air drag. how tall, in meters, is the building?
Physics
1 answer:
Mars2501 [29]3 years ago
8 0
<span>Kinematics is used in this problem. The mass does not matter here because the question is mass independent. vi = 0 vf = x d = ? d = vi + 1/2 a t^2 d = 0 + 1/2 (9.8) (1.8)^2 d = 15.9 m (counting sig figs)</span>
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If the displacement of a horizontal mass-spring system was doubled, the elastic potential energy in the system would change by a
bixtya [17]

Answer:

K'=4K

Explanation:

The electric potential energy is given by :

E=\dfrac{1}{2}kx^2

Where

k is spring constant

x is compression or extension in the spring

If the displacement of a horizontal mass-spring system is doubled, x'= 2x

New elastic potential energy :

E'=\dfrac{1}{2}kx'^2\\\\E'=\dfrac{1}{2}k(2x)^2\\\\=\dfrac{1}{2}k\times 4x^2\\\\K'=4\times \dfrac{1}{2}kx^2\\\\K'=4K

So, new elastic potential energy 4 times the initial elastic potential energy.

3 0
2 years ago
A young man and woman are sitting on opposite sides of a park bench (1m). If the young man has a mass of 70kg and the woman has
ZanzabumX [31]

Answer:

130N

Explanation:

F<em>=</em><em>(</em><em>M1+</em><em>M</em><em>2</em><em>)</em><em>V</em>

<em>F=</em><em> </em><em>(</em><em>7</em><em>0</em><em>+</em><em>6</em><em>0</em><em>)</em><em>*</em><em>1</em>

<em>F=</em><em>1</em><em>3</em><em>0</em><em>*</em><em>1</em>

<em>F=</em><em>1</em><em>3</em><em>0</em><em>N</em><em>/</em><em>/</em>

6 0
2 years ago
A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?
pochemuha
10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv
7 0
3 years ago
A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as t
salantis [7]

Answer:

2.07

Explanation:

Since you didn't supply the drawing, here is what I assumed:

A is the corner opposite the axis of rotation

B is one of the remaining two corners

L1 is the side between A & B

Centripetal acceleration is given by:

ac = v^2 / r = (v / r) * (v / r) * r…………1

Also angular speed is

w = v / r,………….2

Substituting (2) in (1) gives:

ac = (v / r) * (v / r) * r……….3

= (v / r)^2 * r

= w^2 * r

Therefore, the angular acceleration at A and at B are given by:

acA = w^2 * rA……..4

acB = w^2 * rB……..5

It is given that:

acA = n * acB…………6

Substituting (4) and (5) into (6) gives:

w^2 * rA = n * w^2 * rB ……….7==>

rA = n * rB……..8

In terms of the sides L1 and L2:

rA = sqrt (L1^2 + L2^2)…….9

and

rB = L2…………10

Considering (8):

n * L2 = sqrt (L1^2 + L2^2)………11

Squaring both sides:

n^2 * L2^2 = L1^2 + L2^2……….12

Dividing by L2^2:

n^2 = L1^2 / L2^2 + L2^2 / L2^2…….13

= (L1 / L2)^2 + 1 ==>

n^2 - 1 = (L1 / L2)^2 ………14==>

L1 / L2 = sqrt (n^2 - 1) ………15

= sqrt (2.30^2 - 1)

= 2.07. . . . . . <<<=== the value of the ratio L1 / L2 when n = 2.30

8 0
2 years ago
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus
babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

I = m\cdot (\vec{v}_{2} - \vec{v_{1}}) (1)

Where:

I - Impulse, in kilogram-meters per second.

m - Mass, in kilograms.

\vec{v_{1}} - Initial velocity of the hockey park, in meters per second.

\vec{v_{2}} - Final velocity of the hockey park, in meters per second.

If we know that m = 0.2\,kg, \vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right] and \vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right], then the impulse applied by the stick to the park is approximately:

I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]

I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

8 0
3 years ago
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