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Anna007 [38]
3 years ago
10

13. How many feet high do you think you could jump on the moon?

Physics
1 answer:
Bas_tet [7]3 years ago
3 0
More than on Earth bc the gravitational force isn’t as strong
You might be interested in
804 n of force are applied to a 51.7 kg. What is the acceleration that the object experiences?
Andreyy89

We can use Newton II here  (where F=m*a), that F is the net (or resultant) force on the object, m is the mass of the object and a is the acceleration the object experiences.

This means, in this case there would be no friction and absolutely no other force which gives a component in the plane of motion, only then can you assume that F=804N.

Now using F= m*a

804 = 51.7*a

Therefore a = 804/51.7 = 15.55 m/s²


7 0
3 years ago
Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir
strojnjashka [21]

Answer:

1.475\times 10^{-13}\ C/m^3

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

A = Area

h = Altitude = 600 m

Electric flux through the top would be

-110A (negative as the electric field is going into the volume)

At the bottom

120A

Total flux through the volume

\phi=120-110\\\Rightarrow \phi=10A

Electric flux is given by

\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0

Charge per volume is given by

\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3

The volume charge density is 1.475\times 10^{-13}\ C/m^3

7 0
3 years ago
A crate filled with delicious junk food rests on a horizontal floor. Only gravity and the support force of the floor act on it,
stealth61 [152]

The words "... as shown ..." tell us that there's a picture that goes along
with this question, and you decided not to share it.  That's sad and
disappointing, but I think the question can be answered without seeing
the picture.

The net force on the crate is zero.  Evidence for this is that fact that
the crate is just sitting there.  If the net force on an object is not zero,
then the object is accelerating ... it's either speeding up, slowing down,
or its the direction of its motion is changing.  If none of these things is
happening, then the net force on the object must be zero.

4 0
3 years ago
A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

V=30.1m/s

C_d=0.45

A=3.3m^2

\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

F_D=807.25N

We need calculate now the energy lost through a time T, then,

W = F_D d

But we know that d is equal to

d=vt

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0
3 years ago
A children's roller coaster is released from the top of a track. If its maximum speed at ground level is 3 m/s, find the height
san4es73 [151]

Answer:

h = 0.46 m

Explanation:

According to the law of conservation of energy:

Potential Energy Lost by Roller Coaster = Kinetic Energy Gained by Roller Coaster

mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\h = \frac{v^2}{2g}

where,

h = height = ?

v = speed at bottom = 3 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

h = \frac{(3\ m/s)^2}{(2)(9.81\ m/s^2)}

<u>h = 0.46 m</u>

4 0
3 years ago
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