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3 years ago

HELP ME PLEASE IM DESPERATE Many inland areas around Florida contain extinct marine fossils.

Physics

1 answer:

chubhunter [2.5K]3 years ago

8 0

Answer:b

Explanation:

thats what my teacher said

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Select ALL of the following statements that provide evidence that there is friction acting on a cart moving along a level track.

Sati [7]

Answer:

1st and 4th one................

6 0

3 years ago

A bucket of mass M (when empty) initially at rest and containing a mass of water is being pulled up a well by a rope exerting a

Naily [24]

Answer:

$V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT$

Explanation:

Given that

Constant rate of leak =R

Mass at time T ,m=RT

At any time t

The mass = Rt

So the total mass in downward direction=(M+Rt)

Now force equation

(M+Rt) a =P- (M+Rt) g

$a=\dfrac{P}{M+Rt}-g$

We know that

$a=\dfrac{dV}{dt}$

$\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g$

$\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt$

$V=\dfrac{P}{R}\ ln\dfrac{M+RT}{M}-gT$

$V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT$

This is the velocity of bucket at the instance when it become empty.

6 0

4 years ago

Consider two identical insulated metal spheres, A and B. Sphere A initially has a charge of -6.0 units and sphere B initially ha

Oksi-84 [34.3K]

Answer:

<em>-2 units of charge</em>

Explanation:

charge on A = Qa = -6 units

charge on B = Qb = 2 units

if the spheres are brought in contact with each other, the resultant charge will be evenly distributed on the spheres when they are finally separated.

charge on each sphere will be = $\frac{Qa + Qb}{2}$

charge on each sphere = $\frac{-6 + 2}{2}$ = $\frac{-4}{2}$ = <em>-2 units of charge</em>

8 0

3 years ago

Please help asap

posledela

Answer:

Explanation:

the object's mass determines the speed of the object and its kinetic energy

4 0

3 years ago

Read 2 more answers

If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?

Zielflug [23.3K]

The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
$g=9.81 m/s^2$

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
$y(t) = \frac{1}{2}gt^2$
And if we substitute t=2.7 s, we find the distance covered:
$y(t)= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m$

6 0

3 years ago