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2 years ago

The force of gravity on a car driving on the surface of a planet is called

Physics

2 answers:

nata0808 [166]2 years ago

8 0

The answer should be A) Mass.

Katena32 [7]2 years ago

4 0

C) Weight is correct. The force of gravity acting upon anything is weight.
A) mass is how much stuff is in something
B) relates to freefall, after the resistance in the air prevents any acceleration
D) freefall=object falling

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In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because _____.

pickupchik [31]

In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because of friction.
AMA (Actual mechanical advantage) is found by dividing output force by effort force. The actual mechanical advantage will always be less than the ideal mechanical advantage. The ideal mechanical advantage assumes perfect efficiency which doesn't account for friction, while actual mechanical advantage does. Therefore; the IMA is always greater than the actual mechanical advantage because all machines must overcome friction.

5 0

3 years ago

A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$

6 0

3 years ago

Three examples that prove newton's law of motion

s344n2d4d5 [400]

If you have a skateboard and you skate into a tree on accident the same amount of force you put onto that tree when you was on the skateboard will come back at you when you bounce back

7 0

2 years ago

PLEASE HELP URGETNT

Angelina_Jolie [31]

Answer:

I think (d) is right answer

6 0

2 years ago

Which formula can be used to find the tangential speed of an orbiting object?

Luden [163]

The correct formula for calculating the tangential speed of an orbiting object is V(t)=wr.
V(t)= Tangential Speed
w= Angular Velocity
r= Radius of the Path

Hope this helps.

6 0

3 years ago

Read 2 more answers