If the car went 30 km west in 25 min. and then 40 km east in 35 min., what would be its total displacement?

alexgriva [62]

It is the branch of science, in which we study different phenomena of atmosphere including climate and weather.

5 0

3 years ago

In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the ancho

melisa1 [442]

Answer:

largest lead = 3 m

Explanation:

Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.

So, first we need to calculate the velocities of both the anchorman

given data:

Distance = d = 100 m

Time arrival for A = 9.8 s

Time arrival for B = 10.1 s

Velocity of anchorman A = D / Time arrival for A

=100/ 9.8 = 10.2 m/s

Velocity of anchorman B = D / Time arrival for B

=100/10.1 = 9.9 m/s

As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use

d = vt

= 9.9 x 9.8 = 97 m

So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.

largest lead = 100 - 97 = 3 m

So if his lead no more than 3 m anchorman A win the race.

5 0

3 years ago

What is the force holding you down?

Snezhnost [94]

Answer:

The force holding you down is gravity.

Explanation:

Gravity is a force between two objects with mass. It pulls things together. You have mass, and the Earth has mass, so gravity tries to pull you and the Earth together. The gravitational force is much bigger for more massive objects.

3 0

3 years ago

Read 2 more answers

A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at

Vesna [10]

Answer:

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0 So B=0

-16 B-32 A+16B=6 So A=-3/16

y=-3/16 cos 4t

Now to find the CF of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

$m^2+8m+16=0$

So $CF =(C_1+tC_2)e^{-2t}$

So the general equation

$Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t$

Given that t=0 Y=0 So

$C_1=\dfrac{3}{16}$

t=0 Y'=0 So

$C_2 =\dfrac{3}{8}$

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

The above equation is the general equation for motion.

3 0

3 years ago

A tin can collapses if all air inside it is taken out why

Veseljchak [2.6K]

That only happens when the tin can is IN air.

In the familiar, comfy part of Earth's atmosphere where we live, the normal pressure of air is around 14.6 pounds on every square inch of everything. That's a big part of the reason why we're built with bodies that generate that same amount of pressure on the INSIDE pressing OUT. That way, we always have the same pressure pushing in both directions, so we know that we won't get crushed or blow up like balloons.

But we have to be careful with our bodies or other things when they're in places where the atmospheric pressure on the outside is NOT normal.

-- When a deep-sea diver goes hundreds of feet down in the ocean, and the pressure of the water is much GREATER than normal air.

-- When an astronaut has to go outside ... where there's NO air ... and fix something on the International Space Station.

When the pressure on the outside becomes very unusual, we have to wear special suits to protect our bodies from the unusual conditions.

The tin can in the story is a lot like our bodies. As long as it has air inside and air outside, the pressure is the same in both directions, so there's no particular force trying to deform the can. But ...

-- If you seal the can with the air inside it, take the can into a vacuum chamber, and pump the air out of the vacuum chamber, then the can only has pressure inside. It'll expand, and eventually spring a little hole in the metal, and all the air inside will blow out.

-- If you take all the air OUT of the can (so the can is REALLY 'empty'), then the pressure on it is all from the outside. In that situation, the can simply collapses, because there's nothing inside to provide pressure in the outward direction.

One more little thing to think about:

When you want some toothpaste to come drizzling out of the tube onto your brush, what do you do ? Do you perhaps squeeze the tube, and increase the pressure on the outside ?

4 0

3 years ago