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3 years ago

What are some similarity’s and differences between potential and kinetic energy

1 answer:

6 0

Answer: posteintal is like something that is about to happen and kinetic is like the energy while something is moving

Explanation:

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A parallel-plate air capacitor is connected to a constant-voltage battery. If the separation between the capacitor plates is dou

Oliga [24]

Answer:

Drop to half of the previous value

Explanation:

Energy stored in capacitor is inversly propotional to the distance between the plates.

3 0

4 years ago

19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,

sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

$t=15.2 s$ is the time it takes to the child to reach the bottom of the slope

$V_{o}=0$ is the initial velocity (the child started from rest)

$\theta=15\°$ is the angle of the slope

$d$ is the length of the slope

Now, the Force exerted on the sled along the ramp is:

$F=ma$ (1)

Where $m$ is the mass of the sled and $a$ its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

$F=mg sin \theta$ (2)

Where $g=9.8 m/s^{2}$ is the acceleration due gravity

Then:

$ma=mg sin \theta$ (3)

Finding $a$ :

$a=g sin \theta$ (4)

$a=9.8 m/s^{2} sin(15\°)$ (5)

$a=2.5 m/s^{2}$ (6)

Now, we will use the following kinematic equations to find $d$ :

$V=V_{o}+at$ (7)

$V^{2}=V_{o}^{2}+2ad$ (8)

Where $V$ is the final velocity

Finding $V$ from (7):

$V=at=(2.5 m/s^{2})(15.2 s)$ (9)

$V=38 m/s$ (10)

Substituting (10) in (8):

$(38 m/s)^{2}=2(2.5 m/s^{2})d$ (11)

Finding $d$ :

$d=288.8 m$

6 0

3 years ago

What is the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25

alexandr1967 [171]

The mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.

HOW TO CALCULATE MASS:

The mass of an object can be calculated by dividing the force applied to the object by its acceleration.

According to this question, a bus can accelerate from rest to 15.5 m/s over 8.25s. The acceleration can be calculated as follows:

a = (v - u)/t

a = 15.5 - 0/8.25

a = 15.5/8.25

a = 1.88m/s²

The mass of the bus = 7500N ÷ 1.88m/s²

The mass of the bus = 3989.4kg

Therefore, the mass of a school bus if it can accelerate from rest to 15.5 m/s over 8.25s with 7,500 N of force is 3989.4kg.

Learn more about mass at: brainly.com/question/20259048?referrer=searchResults

3 0

3 years ago

A scaffold of mass 77 kg and length 5.0 m is supported in a horizontal position by a vertical cable at each end. A window washer

Lynna [10]

Answer:

A) T2 = 912.88 N

B) T1 = 607.12 N

Explanation:

First of all, we see that the sum of the tensions is equal to the total weight.

Now, for the scaffold, weight; W_s = 77 x 9.8 = 755.6 N

For the window washer, Weight; W_w = 78 x 9.8 = 764.4 N

Total weight;W_t = W_s + W_w

W_t = 755.6 N + 764.4 N = 1520 N

Thus,

T1 + T2 = 1520

Where T1 and T2 are the tensions in farther and nearer cables respectively.

Now, we need to do a torque problem.

The window washer is 1.8m from the right end of the scaffold and so the weight of the scaffold is at its center. This is 2.5 m from either end. Let the pivot point be at right end of the scaffold.

For the window washer, counter clockwise torque = 764.4 x 1.5 = 1146.6 N.m

For the scaffold, counter clockwise torque = 755.6 x 2.5 = 1889 N.m

Total Torque; T = 1146.6 + 1889 = 3035.6 N.m

For the cable at the left end of the cable, clockwise torque = T1 x 5

Set this equal to the total counter clockwise torque and solve for T1.

Thus,

T1 x 5 = 3035.6

T1 = 3035.6 ÷ 5 = 607.12 N

T2 = 1520 – 607.12 = 912.88 N

6 0

4 years ago

In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water

kotykmax [81]

My work for your question

3 0

4 years ago