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2 years ago

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You drop a 30 g pebble down a well. You hear a splash 2.7 s later. What was the pebble’s speed just as it reaches the water? Ass

ume g = 9.8 m/s2

1 answer:

Flauer [41]2 years ago

7 0

vf=vi+at

No need for rearranging because it is already set up for Vf (final velocity)

a= -9.8m/s² (because it is falling)

vi= 0

t= 2.7

0 + -9.8(2.7) = vf

vf = -26.5 (-26.46)

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Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the man

v is the speed after jumping

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s$

In the acceleration phase, we know that the initial velocity is

$u=0$

And the force exerted on the floor is 2.3 times the gravitational force, so

$F=2.3 mg$

This means the net force on you is

$F_{net} = F-mg=2.3mg-mg=1.3 mg$

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

$a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g$

Now we can use the following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

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

where s is the quantity we want to find. Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m$

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

$\Delta h = -0.77 m$

This means that the lowest point reached by the center of mass above the floor during the crouch is

$h'=h+\Delta h = 1.0 - 0.77 = 0.23 m$

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0

2 years ago

Paula is studying two different animals. Both animals are classified within the same genus, but they are different species. Base

Ksivusya [100]

Answer:c

Explanation:

i think

6 0

2 years ago

Read 2 more answers

A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106

defon

Answer:

$4.7\cdot 10^{-14}N$

Explanation:

For a charge moving perpendicularly to a magnetic field, the force experienced by the charge is given by:

$F=qvB$

where

q is the magnitude of the charge

v is the velocity

B is the magnetic field strength

In this problem,

$q=1.6\cdot 10^{-19} C$

$v=6.5\cdot 10^6 m/s$

$B=4.5\cdot 10^{-2} T$

So the force experienced by the electrons is

$F=(1.6\cdot 10^{-19}C)(6.5\cdot 10^6 m/s)(4.5\cdot 10^{-2} T)=4.7\cdot 10^{-14}N$

3 0

2 years ago

__________ curves help lessen the effect of the force of the forward motion on your vehicle as it enters the curve.

Rainbow [258]

Answer:

Banked

Explanation:

Banked curves are formed when the inner edge is below the outer edge.

It is done in order to ensure the reliability of the frictional force as it varies when the road is wet wet or oily. Thus in order to avoid these problems the curved roads are banked.

Banking of the curve provides the necessary centripetal force, i.e., the horizontal component of the normal reaction force to keep the vehicle i motion and thus helps in reducing the effect of the forward motion force on the vehicle.

5 0

2 years ago

I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =

Helen [10]

Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)

7 0

2 years ago