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DENIUS [597]
3 years ago
8

How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will

oscillate at 1.00 Hz?A) 39.5 kgB) 2.00 kgC) 1.00 kgD) 1.56 kgE) 6.29 kg
Physics
1 answer:
nordsb [41]3 years ago
6 0

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

The frequency of oscillation, f = 1 Hz

The frequency of oscillation is given by the formula as formula as follows :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg

So, the mass that is attached to the spring is 1 kg.

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An object is thrown with a horizontal velocity of 20 m/s from a cliff that is 125 m above level ground. If air resistance is neg
Serjik [45]

Given that,

Horizontal velocity of the object, v = 20 m/s

Height of the cliff, h = 125 m

We need to find the time that it takes the object to fall to the ground from the cliff is most nearly. It can be calculated using second equation of motion. Let us consider that the initial speed of the object is 0. So,

h=ut+\dfrac{1}{2}at^2

Here, a = g and u = 0

h=\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 125}{10}} \\\\t=5\ s

So, the object will take 5 seconds to fall to the ground from the cliff.

5 0
3 years ago
An attacker at the base of a castle wall 3.65 m high throws a rock straight up with speed 7.4m/s from a height of 1.55m above th
Natali5045456 [20]

a) Yes, the rock will reach the top

b) The final speed is 3.7 m/s

c) The change in speed is 2.4 m/s

d) The change in speed in the two situations do not agree

e) Because the kinetic energy depends quadratically on the speed, K\propto v^2

Explanation:

a)

The mechanical energy of the rock at the moment it is thrown from the ground is equal to the sum of its kinetic energy and its potential energy:

E=KE_i + PE_i = \frac{1}{2}mu^2 + mgh_i

where

m is the mass of the rock

u = 7.4 m/s is the inital speed

g=9.8 m/s^2 is the acceleration of gravity

h_i = 1.55 m is the initial height of the rock

Substituting, we find the initial mechanical energy of the rock

E=\frac{1}{2}m(7.4)^2 + m(9.8)(1.55)=42.6m [J]

In order to reach the top of the castle, the rock should have a mechanical energy of at least

E' = mgh'

where

h' = 3.65 m is the heigth of the top

Substituting,

E'=m(9.8)(3.65)=35.6m [J]

Since E > E', it means that the rock has enough mechanical energy to reach the top.

b)

The final mechanical energy of the rock at the top is

E=mgh'+ \frac{1}{2}mv^2 (1)

where:

v is the final speed of the rock at the top

Since the mechanical energy is conserved, this should be equal to the initial mechanical energy:

E=42.6 m [J] (2)

Therefore, equating (1) and (2), we can find the final speed of the rock:

mgh' + \frac{1}{2}mv^2 = 42.6m\\v=\sqrt{2(42.6-gh')}=\sqrt{2(42.6-(9.8)(3.65))}=3.7 m/s

c)

Since the motion of the rock is a free fall motion (constant acceleration equal to the acceleration of gravity), we can use the following suvat equation:

v^2 - u^2 = 2as

where

v is the final speed, at the bottom

u = 7.4 m/s is the initial speed of the rock, at the top

a=9.8 m/s^2 is the acceleration of gravity

s = 3.65 - 1.55 = 2.1 m is the vertical displacement of the rock

Solving for v, we find the final speed:

v=\sqrt{u^2+2as}=\sqrt{7.4^2 + 2(9.8)(2.1)}=9.8 m/s

Therefore, the change in speed is

\Delta v = v-u = 9.8 - 7.4 =2.4 m/s

d)

In the first situation (rock thrown upward), we have:

u = 7.4 m/s (initial speed)

v = 3.7 m/s (final speed)

So the change in speed is

\Delta v = v-u =3.7 - 7.4 = -3.7 m/s

While the change in speed in the second situation (rock thrown downward) is

\Delta v = 2.4 m/s

Therefore, we see that their magnitudes do not agree.

e)

In both situations, the change in kinetic energy of the rock is equal in magnitude to the change in gravitational potential energy, since the total mechanical energy is conserved.

The change in gravitational potential energy in the two situations is the same (because the change in height is the same), therefore the change in kinetic energy in the two situations is also the same.

However, the kinetic energy of the rock is not directly proportional to the speed, but to the square of the speed:

K\propto v^2

Since the initial speed is the same for both situation (7.4 m/s), but the change in kinetic energy has opposite sign in the two situations (negative when the rock is thrown upward, positive when thrown downward), the situation is not symmetrical, therefore in order to have the same magnitude of change in the kinetic energy, the change in speed must be larger when the kinetic energy involved is lower, so in the first situation.

Learn more about kinetic energy and about potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

6 0
4 years ago
What is a method to predict an earthquake?
Pie

Measurements of microtremors, slope of the land, and motion of tectonic plates predict earthquakes. But we can't predict WHEN accurately enough yet to make the prediction useful to the general public.

8 0
3 years ago
If a bowling ball is moving with acceleration of 15 m/s2 and has a mass of 10kg what is the force of the ball as it knocks down
Lina20 [59]

Answer:

150 newtons

Explanation:

F = ma

force = mass * acceleration

the mass is 10 kg, and the accelration is 15 m/s^2

10*15 = 150

the force unit that matches with meters/second and kg is newtons

F =ma

150 newtons = 10 kg * 15 m/s^2

6 0
3 years ago
Imagine rubbing a balloon on your hair or clothes and putting extra electrons on the balloon to give it a negative charge. If yo
3241004551 [841]

Answer:

Because here is a number of electrons more applied to the balloon

6 0
3 years ago
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