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3 years ago

Where do I start to solve: A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 74.0 m high

1 answer:

5 0

I believe the answer to that question is A. If you can obtain the time it takes to reach the bottom of the cliff, then you can solve for the other two values.

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By which factor is the sound intensity decreased at 3 meters?

Nostrana [21]

Answer:

what should be done io protect forests

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3 years ago

a 77 N net force is applied to a box which slides horizontally across a floor for 6.7 m. What amount of work is done on the box

ArbitrLikvidat [17]

W=fd
W=77*6.7
W=515.9 J

7 0

3 years ago

A student made the claim that a 4 gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy a

vovikov84 [41]

This question involves the concept of kinetic energy.

The student's claim is "right".

<h3>Kinetic Energy</h3>

The energy possessed by a body, by the virtue of its motion is called kinetic energy. Mathematically it is given by the following formula:

$K.E =\frac{1}{2}mv^2$

where,

K.E = Kinetic energy
m = mass
v = velocity

Therefore,

For the paintball:

$K.E = \frac{1}{2}(4\ g)(90\ m/s)^2$

K.E = 16200 J

For the pellet:

$K.E = \frac{1}{2}(1\ g)(180\ m/s)^2$

K.E = 16200 J

Hence, both paintball and pellet will have same kinetic energy. The student is right.

Learn more about kinetic energy here:

brainly.com/question/12669551

#SPJ1

7 0

3 years ago

An object is moving with a constant velocity of 278 m/s. How long will it take it to travel 7500 m, using the formula Delta X=Vt

rusak2 [61]

If the velocity is constant then the acceleration of the object is zero.
$a=0 (m/s^2)$
Thus when we apply the equation
$\Delta X=vt+(at^2/2)$
It remains
$\Delta X =vt$
or equivalent
$t=(\Delta X/v) =7500/278 =26.98 (seconds)$

7 0

4 years ago

An archer shoots an arrow with a mass of 45.0 grams from bow pulled

Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 0

3 years ago