Ask question

Ask question

All categories

3 years ago

15

What provides the force necessary to start a building or bridge oscillating?

1 answer:

Andrew [12]3 years ago

4 0

Usually, the forces that start the oscillation of buildings are the wind and microearthquakes.

You might be interested in

I need the right answer ASAP NO LINKS!!!

serious [3.7K]

Answer:

Water and power come from external sources.

Explanation:

3 0

3 years ago

If a 4kg Bird is pushed by the window with a force of 60 N how fast is the bird accelerate?

LiRa [457]

Force=60N
Mass=4kg

$\\ \ast\sf\hookrightarrow Force=Mass\times Acceleration$

$\\ \ast\sf\hookrightarrow Acceleration=\dfrac{Force}{Mass}$

$\\ \ast\sf\hookrightarrow Acceleration=\dfrac{60}{4}$

$\\ \ast\sf\hookrightarrow Acceleration=15m/s^2$

3 0

3 years ago

Shalnov [3]

Answer:

Explanation:

We will use the KE equation you wrote here and fill in what we are given:

$36=\frac{1}{2}m(12)^2$ and isolating the m:

$m=\frac{2(36)}{12^2}$ which gives us

m = .50 kg

3 0

3 years ago

A crate rests on the flatbed of a truck that is initially traveling at 15 m/s on a level road. The driver applies the brakes and

lina2011 [118]

Answer:0.3

Explanation:

Given

velocity of car=15 m/s

truck brought to halt in a distance of 38 m

We know

$v^2-u^2=2as$

Final velocity (v)=0

$0-(15)^2=2(a)(38)$

$a=\frac{-225}{76}$

$a=-2.96 m/s^2$ (deceleration)

Therefore minimum coefficient of friction \mu will be

$\mu \times g=a$

$\mu =\frac{a}{g}$

$\mu =\frac{2.96}{9.8}=0.302$

7 0

4 years ago

A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her heigh

eduard

To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

$V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}$

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

$P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}$

$P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}$

$P^2 = \frac{P^2V^2}{c^2}+m^2V^2$

$P^2 = V^2 (\frac{P^2}{c^2}+m^2)$

$V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}$

$V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}$

$V = 2.81784*10^8m/s$

Therefore the height with respect the observer is

$l = l_0*\sqrt{1-\frac{V^2}{c^2}}$

$l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}$

$l = 0.56m$

Therefore the height which the observerd measure for her is 0.56m

8 0

3 years ago