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Mumz [18]
3 years ago
12

Help Please

Physics
1 answer:
densk [106]3 years ago
3 0

Answer:

Energy=3.1times 10^-17 J

Rest mass: 6.2 kg

Speed: 47.5 m/s

Wavelength: 2.659 times 10^-6

Momentum: 67.3 kg(m/s)

Explanation:

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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
3 years ago
. The naturally occurring charge on the ground on a fine day out in the open country is –1.00nC/m2 . (a) What is the electric fi
JulijaS [17]

Answer:

a) -113 N/C

b) -423.71 V

Explanation:

The Step by step explanation and solution has been attached in the attachment section.

4 0
3 years ago
We are usually not aware of the electric force acting between two everyday objects because
Semmy [17]
Most everyday objects have as many plus charges as they have minus charges
5 0
3 years ago
Read 2 more answers
A rock is dropped from a tower 70.0 m high. How long does it take for the rock to hit the ground?​
Stella [2.4K]

i assume it will be 5mins i hope thi shelp u out

5 0
3 years ago
The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
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