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Mice21 [21]
3 years ago
12

A coin is thrown with a velocity of 0 m/s down a dry well and hits bottom in 1.2s, what’s the depth of the well?

Physics
1 answer:
pentagon [3]3 years ago
5 0

Answer:

The well is 7.1 meters deep.

Explanation:

The formula to use here is the distance in a uniformly accelerated motion:

d = \frac{1}{2}at^2+v_0t+d_0

where d stands for distance, t for time, a for acceleration, v0 and d0 for initial velocity and distance, respectively. Since the initial distance and velocity are both zero, we are left with the first term. The coin is in free fall and so it is accelerated by gravity:

d = \frac{1}{2}at^2= \frac{1}{2}gt^2=\frac{1}{2}9.8\frac{m}{s^2}1.2^2s^2=7.1m

The well is 7.1 meters deep.

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3 years ago
A baseball player slides into third base with an initial speed of 4.0 m/s. If the coefficient of
musickatia [10]
The frictional force is given by F = μmg 

<span>where μ is the coeficient of friction. </span>

<span>Work done by frictional force = Fd = μmgd </span>

<span>Kinetic energy "lost" = 1/2 mv² </span>


<span>Fd = μmgd = 1/2 mv² </span>

<span>The m's cancel μgd = v² / 2 </span>



<span>d = v² / 2μg </span>

<span>d = 8² / 2(0.41)(9.8) </span>

<span>d = 32 / (0.41)(9.8) </span>

<span>d = 7.96 </span>

<span>Player slides 8 m . </span>



<span>Note. In your other example μ = 0.46 and v = 4 m/s </span>



<span>d = v² / 2μg </span>




<span>= 4² / 2(0.46)(9.8) </span>

<span>= 8 / (0.46)(9.8) </span>

<span>= 1.77 or 1.8 m.
</span>
Hope i Helped :D
3 0
3 years ago
Read 2 more answers
An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

3 0
2 years ago
A cell phone is released from the top with the speed of 10ms what is the speed 3s after?
sergeinik [125]

Answer:

30ms

Explanation:

you need to multiple the 10ms by 3s which gives you 30ms

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While on vacation, a student picks up surface rocks from around the world to add to her rock collection. The composition of her
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I want to say that they will be primarily flat but I honestly don't know
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