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4 years ago

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]4 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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Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer sec

GaryK [48]

Mary and her younger brother Alex decide to ride the carousel at the State Fair, Mary's and Alex's angular speed M and tangential speed vM is mathematically given as

Mary's and Alex's angular speed=1.43

Tangential speed mary=3.22 m/s

Tangential speed alex =2.260m/s

<h3>What is Mary's and Alex's angular speed M and tangential speed vM?</h3>

Generally, the equation for angular speed is mathematically given as

$w=2\pi /T\\\\Therefore\\\\w=2\pi/3.9$

w = 1.61 rev/see 3.9

Centripetal acc mary = v^2/r

Centripetal acc mary = w^2r

Centripetal acc mary = w^2x 2m

Centripetal acc. of Alex = w²x L.u

Therefore

$\frac{(ac) mary }{(ac) plex}= 1.43$

Hence

tang. speed V=Wr

tang. speed of mary = 1.61x2 = 3.22 m/s

tang. speed of Alex: 1.61X1·4 =2.260m/s

Read more about Speed

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8 0

2 years ago

M/s

SashulF [63]

Answer:

a. Final velocity, V = 2.179 m/s.

b. Final velocity, V = 7.071 m/s.

Explanation:

Given the following data;

Acceleration = 0.500m/s²

a. To find the velocity of the boat after it has traveled 4.75 m

Since it started from rest, initial velocity is equal to 0m/s.

Now, we would use the third equation of motion to find the final velocity.

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.

Substituting into the equation, we have;

$V^{2} = 0^{2} + 2*0.500*4.75$

$V^{2} = 4.75$

Taking the square root, we have;

$V^{2} = \sqrt {4.75}$

Final velocity, V = 2.179 m/s.

b. To find the velocity if the boat has traveled 50 m.

$V^{2} = 0^{2} + 2*0.500*50$

$V^{2} = 50$

Taking the square root, we have;

$V^{2} = \sqrt {50}$

Final velocity, V = 7.071 m/s.

8 0

3 years ago

What is an unbalanced force

Butoxors [25]

There's no such thing as "an unbalanced force".

If all of the forces acting on an object all add up to zero, then we say that
the group of forces is balanced. When that happens, the group of forces
has the same effect on the object as if there were no forces on it at all.

An example:
Two people with exactly equal strength are having a tug-of-war. They pull
with equal force in opposite directions. Each person is sweating and straining,
grunting and groaning, and exerting tremendous force. But their forces add up
to zero, and the rope goes nowhere. The group of forces on the rope is balanced.

On the other hand, if one of the offensive linemen is pulling on one end of
the rope, and one of the cheerleaders is pulling on the other end, then their
forces don't add up to zero, because even though they're opposite, they're
not equal. The group of forces is unbalanced, and the rope moves.

A group of forces is either balanced or unbalanced. A single force isn't.

7 0

3 years ago

A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from

o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

$\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}$