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bulgar [2K]
2 years ago
14

HELP ME PLEASE WITH MY HOMEWORK ILL GIVE BRAINLIEST

Physics
1 answer:
nadezda [96]2 years ago
8 0

Answer:  Mass

Explanation:

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HELP!!! The planet Mars has a mass about one-tenth the mass of Earth. Even though Mars has two moons, their tidal forces have a
vladimir2022 [97]

Answer:

Say: Mars has a much weaker gravity effect than it does because it is smaller and cannot have as much gravity effect than it does on earth.

Explanation:

4 0
3 years ago
Can someone help me out?
MArishka [77]

Answer:

B). Rarefaction :) _____

8 0
3 years ago
Read 2 more answers
A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a
MrRissso [65]

Answer:

h>\dfrac{5}{2}R

Explanation:

Given that

Height = h

Radius = R

From energy conservation

KE_A+U_A=KE_B+U_B

At point B

The minimum speed to complete the   the circle

V_B=\sqrt{gR}\ m/s

So the kinetic energy at point B

KE_B=\dfrac{1}{2}mV^2

KE_B=\dfrac{1}{2}mgR

KE_A+U_A=KE_B+U_B

0+mgh=\dfrac{1}{2}mgR+2mgR

Without falling off at the top (point B)

0+mgh>\dfrac{1}{2}mgR+2mgR

mg(h-2R)>\dfrac{1}{2}mgR

g(h-2R)>\dfrac{1}{2}gR

h>\dfrac{5}{2}R

6 0
3 years ago
Please help me
qaws [65]

-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
y-component of position.

-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.

Now, the position function may look big and ugly in the picture.  But with the
exception of  't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation.  The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.

From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t  -  6 (a₀ / 30t₀⁴ ) t⁵  =  (a₀ - g) t  -  (a₀ / 5t₀⁴ ) t⁵

There's your velocity . . . /\ .

Second derivative . . . y'' (t) = (a₀ - g) -  5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) -  (a₀ /t₀⁴ ) t⁴

and there's your acceleration . . . /\ .
That's the one you're supposed to graph.

a₀ is the acceleration due to the model rocket engine thrust
     combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀  is how long the model rocket engine burns

Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.

The big name in model rocketry is Estes.  Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.


6 0
3 years ago
According to the quantum theory of radiation,
masya89 [10]

Answer:

Option B

Explanation:

According to quantum theory of radiation electromagnetic field or electromagnetic  radiation( like light) produce by accelerated charge object and the quantum of EM radiations is photon which has discrete energy. So, EM field can have only certain values of total energy and no other value due the discrete nature of the energy of photon. Hence option B is correct

4 0
3 years ago
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