Question
In order for work to be done, what three things are necessary

The answer is 3.53400621

Reduce the expression, if possible, by cancelling common factors.

Answer: 3.53400621

Explanation: Hope this helped! :)

Answer: A. a straight line inclined to the time axis

Explanation:

Answer:

350 N

Explanation:

F=ma

[tex]f = force \\ m = mass \\ a = acceleration[/tex]

[tex]m = 2(55kg) + 240kg \\ a = 1.0 \frac{m}{ {s}^{2} } [/tex]

Force = 350 Newtons

The net force acting on the elevator would be 350 Newtons as it accelerates upward at 1.00 m/s2.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

The mathematical expression for Newton's second law is as follows

F = ma

As given in the problem two people each have a mass of 55 kg. They are both in an elevator that has a mass of 240 kg. When the elevator begins to move, the people and the elevator have an upward acceleration of 1.00 m/s2, then we have to find the net force acting on the elevator,

The net force acting on the elevator,

F = ma

F =(2×55 + 240)×1

  = 350 Newtons

Thus, the net force acting on the elevator would be 350 Newtons as it accelerates upward at 1.00 m/s2

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Answer:

Explanation:

The question is incomplete. Here is the complete question.

A ball with mass m kg is thrown upward with initial velocity 22 m/s from the roof of a building 17 m high. Neglect air resistance. Use g=9.8 m/s2. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. xmax= meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground.

a) Using the equation of motion formula;

v² = u²+2gH where;

u is the initial velocity

v is the final velocity

theta is the angle of launch

g is the acceleration due to gravity.

H is the maximum height reached by the ball

Since the ball is thrown upwards, the acceleration due to gravity will be negative. The equation then becomes;

v² = u²-2gH

Given

v = 0m/s

u = 22m/s

g = 9.8m.s²

0² = 22²-2(9.8)H

-22² = -19.6H

H = -22²/-19.6

H = 24.69m

If the biuliding is 17m high, the maximum height above the ground that the ball reaches will be;

Hmax = 24.69+17

Hmax = 41.69m

b) The time it takes to hit the ground can be expressed using the formula

v = u-gt

0 = 22-9.8t

-22 = -9.8t

t = -22/9.8

t = 2.45secs

Answer:

15 m/s

Explanation:

s=d/t

s=60/4

s=15 m/s

Answer:

The answer would be 15 m/s

Explanation:

s=d/t

s=60/4

s= 15 m/s

Have a beautiful day beautiful

You welcome.

Answer:

Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects, usually denser than water, to float and slide on a water surface.

Answer: ummm arent you in my physical fundations class lol

Explanation:

are you or am i tripping? and sorry i dont have the answer :(

Common reasons people give for not being active include not having enough time, finding physical activity inconvenient, lacking self-motivation, etc.

Any consensual bodily mobilization generated by skeletal muscles that necessitate caloric expenditure is characterized as physical activity.

Physical activity includes all activities of any frequency, at any moment of day or night. It incorporates both strength training and incidental activity into the daily routine.

Physical activity can improve your brain health, help you manage your weight, lower your risk of disease, strengthen your bones and muscles, and improve your ability to do everyday tasks.

Adults who sit less and engage in moderate-to-vigorous physical activity reap health benefits.

Internal barriers were classified into three types namely, a lack of energy, a lack of motivation, and a lack of self-efficacy.

External barriers were also classified into three categories namely a lack of resources, a lack of social support, and a lack of time.

Thus, these are some reasons that may not prepare to be safe and comfortable when they participate in physical activity

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Answer:

The temperature rises because for a given volume of gas, a rise pressure of the gas in pressure results in a proportionate rise in the temperature of the gas

Similarly when the handle is raised to draw air causes a fall in pressure that results in proportionate fall in temperature, for a given volume of gas

Explanation:

From Gay-Lussac's law, states that the pressure of a given mass of gas is directly proportional to its Kelvin temperature, provided that the volume is held constant

Mathematically, the law states that Pressure ∝  Temperature, at constant Volume

Therefore;

P₁/T₁ = P₂/T₂

Similarly, by kinetic theory of gases, we have;

The

[tex]P = \dfrac{n \cdot MW \cdot v_{rms}^2}{3 \cdot V}[/tex]

[tex]v_{rms} = \sqrt{\dfrac{3 \cdot R \cdot T}{MW} }[/tex]

Therefore, as in order for the hand pump to inflate the bicycle tires, the air in the pump has to be compressed to force it into the tire, thereby increasing the pressure, of the air in a given volume of the pump which results in the raising of the temperature of the air in the pump, which raises the temperature of the wall of the pump.

The temperature of the air in the pump also falls as the pressure in the pump is reduced by raising the pump handle, to reduce the air pressure inside the pump and and allow air to be taken into the pump.

The question is incomplete. Here is the complete question.

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity when the intensity as measured in W/m² changes by a multiplicative factor. The number of decibels increase by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is

[tex]\beta=10log(\frac{I}{I_{0}} )dB[/tex],

where [tex]I_{0}[/tex] is a reference intensity. for sound waves, [tex]I_{0}[/tex] is taken to be [tex]10^{-12} W/m^{2}[/tex]. Note that log refers to the logarithm to the base 10.

Part A: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity, i.e. [tex]I=10I_{0}[/tex]? Express the sound intensity numerically to the nearest integer.

Part B: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity, i.e. [tex]I=100I_{0}[/tex]? Express the sound intensity numerically to the nearest integer.

Part C: Calculate the change in decibels ([tex]\Delta \beta_{2},\Delta \beta_{4}[/tex] and [tex]\Delta \beta_{8}[/tex]) corresponding to f = 2, f = 4 and f = 8. Give your answer, separated by commas, to the nearest integer -- this will give an accuracy of 20%, which is good enough for sound.

Answer and Explanation: Using the formula for sound intensity level:

A) [tex]I=10I_{0}[/tex]

[tex]\beta=10log(\frac{10I_{0}}{I_{0}} )[/tex]

[tex]\beta=10log(10 )[/tex]

β = 10

The sound Intensity level with intensity 10x is 10dB.

B) [tex]I=100I_{0}[/tex]

[tex]\beta=10log(\frac{100I_{0}}{I_{0}} )[/tex]

[tex]\beta=10log(100)[/tex]

β = 20

With intensity 100x, level is 20dB.

C) To calculate the change, take the f to be the factor of increase:

For [tex]\Delta \beta_{2}[/tex]:

[tex]I=2I_{0}[/tex]

[tex]\beta=10log(\frac{2I_{0}}{I_{0}} )[/tex]

[tex]\beta=10log(2)[/tex]

β = 3

For [tex]\Delta \beta_{4}[/tex]:

[tex]I=4I_{0}[/tex]

[tex]\beta=10log(\frac{4I_{0}}{I_{0}} )[/tex]

[tex]\beta=10log(4)[/tex]

β = 6

For [tex]\Delta \beta_{8}[/tex]:

[tex]I=8I_{0}[/tex]

[tex]\beta=10log(\frac{8I_{0}}{I_{0}} )[/tex]

β = 9

Change is

[tex]\Delta \beta_{2},\Delta \beta_{4}[/tex], [tex]\Delta \beta_{8}[/tex] = 3,6,9 dB

Following are the calculation to the given question:

Given:

Please find the given question.

To find:

value=?

Solution:

With terminal voltage, no force equals zero.

So,

[tex]\to mg = 6\times \pi \times \eta \times r\times \vartheta \\\\[/tex]

As [tex]35\ V[/tex] is weight neutralizing voltage

[tex]\to mg= q\times v \times r\\\\[/tex]

So,

[tex]\to q\times v \times r= 6 \times \pi \times \eta \times r \times \vartheta \\\\\to q\times v = 6 \times \pi \times \eta \times \vartheta \\\\\to q =\frac{6 \times \pi \times \eta \times \vartheta}{v} \\\\[/tex]

      [tex]=\frac{6 \times 3.14 \times \eta \times 2.64 \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 18.84 \times \eta \times 2.64 \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 49.7376 \times \eta \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 49.7376 \times \eta \times 10^{-5}\ \frac{m}{s}}{35} \\\\ =3.2 \times 10^{-10}\ coulombs[/tex]

Therefore

[tex]\to q= n \times e\\\\[/tex]

Hence n comes to be [tex]2\times 10^9\[/tex] electrons.

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Answer: true

Explanation: a force can be anything that effects an object, as long as the object moves

Answer:

[tex]Velocity = \frac{s(8)}{8}[/tex]

Explanation:

Given

[tex]0 \leq t \leq 12[/tex]

Required

Determine the velocity when t = 8

This type of velocity is referred to as an instantaneous velocity.

In this case, it is calculated using

[tex]Velocity = \frac{Distance\ at\ 8 second}{t = 8}[/tex]

Given that s(t) models the distance;

s(8) = distance at 8 seconds

So;

[tex]Velocity = \frac{s(8)}{8}[/tex]

Option B answers the question

Answer:

The satellite that is 7500 km above earth's surface is 1.24 times faster than the satellite at 15000 km above earth's surface

Explanation:

Generally the gravitational force acting on each satellite is mathematically represented as

[tex]F = \frac{G * M_e * m_2 }{r^2}[/tex]

Here [tex]M_e[/tex] is the mass of the earth

[tex]r = r_e + R [/tex]

So [tex] r_e [/tex] is the radius of the satellite and R is the radius of earth with value R = 6371 km

This force is also equivalent to the centripetal force acting on each satellite which is mathematically represented as

[tex]F = \frac{mv^2}{ r^2}[/tex]

So

[tex]\frac{mv^2}{ r^2} = \frac{G * m_1 * m_2 }{r^2}[/tex]

=> [tex]v = \sqrt{ \frac{GM_e}{r} }[/tex]

So for [tex]r = 7500+ 6371 = 13871 \ km [/tex]

We have that

[tex]v_1 = \sqrt{ \frac{GM_e}{13871} }[/tex]

and [tex]r = 15000+ 6371 = 21371 \ km [/tex]

We have that

[tex]v_2 = \sqrt{ \frac{GM_e}{21371} }[/tex]

So

[tex]\frac{v_1}{v_2} = \frac{\sqrt{\frac{GM_e}{13871} } }{ \sqrt{\frac{GM_e}{21371}} }[/tex][tex]v = \sqrt{ \frac{GM_e}{r} }[/tex]

 =>  [tex]\frac{v_1}{v_2}  = \sqrt{ \frac{21371}{13871} }[/tex]

 =>   [tex]v_1    = 1.24 v_2[/tex]

So satellite that is 7500 km above earth's surface is 1.24 times faster than the satellite at 15000 km above earth's surface

Answer:

Now e is due to the ring at a

So

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

So

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

So the direction here is away from the ring

Answer:

b) 40.5 m

Explanation:

Given:

v₀ = 0 m/s

a = 9 m/s²

t = 3 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (3 s) + ½ (9 m/s²) (3 s)²

Δx = 40.5 m

Answer:

0.5 m

Explanation:

The following data were obtained from the question:

Force (F) applied = 5 N

Work done (W) = 2.5 J

Distance (s) =?

Work done (W) can be defined as the product of force (F) and distance (s) moved in the direction of the force. From the above definition, work done (W) can be represented mathematically as:

Work done (W) = Force (F) × Distance (s)

W = F × s

With the above formula, we can obtain the distance moved by the book as shown below:

Force (F) applied = 5 N

Work done (W) = 2.5 J

Distance (s) =?

W = F × s

2.5 = 5 × s

Divide both side by 5

s = 2.5/5

s = 0.5 m

Therefore, the book moved a distance of 0.5 m.

Answer:

The answer is 21

Explanation:

5 + 2 + 6 + 8 = 21

Answer:

2.13 m/s^2.

Find the remaining answer in the explanation

Explanation:

Given that a sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 41 km/h at the 61-m mark. He then maintains this speed for the next 83 meters before uniformly slowing to a final speed of 34 km/h at the finish line.

1.) The maximum horizontal acceleration will be when she attained her maximum speed. That is,

Speed = 41 km/h

Convert km/h to m/s

(41×1000) / 3600

Speed = 11.39 m/s

Using speed formula to calculate time

Speed = distance/time

11.39 = 61/t

t = 61/11.39

t = 5.36s

Where the distance = 61 m

Maximum acceleration = velocity /time

Maximum acceleration = 11.39/5.36

Maximum acceleration = 2.13 m/s^2

The maximum acceleration value occurs when the sprinter is starting from rest and attaining the maximum speed.

Answer:

Unbalanced

Explanation:

It would be speeding and under the influence of an imbalanced force if it were rounding a curve. The train must be subject to an unbalanced force in order for it to continue moving at a constant speed.

Newton's first law is only this one. Unless operated upon by an imbalanced force, an item at rest remains at rest and an object in motion maintains constant speed in the same direction.

The car tends to maintain its travel at a steady speed and direction because the two forces balance each other out and cancel each other out.

The forces are out of balance when an object's motion changes. Equal in size and directed in the opposite direction, balanced forces are. When forces are evenly distributed, motion remains unchanged.

Therefore, One of your scenarios from the previous section involved pushing or pulling an object with the equal amount of force in opposite directions.

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Explanation:

distance = rate*time

time = distance divided by rate

now just substitute

time = distance (60 m) divided by rate (34.8 m/sec)

Answer: 5 m

Explanation:

X^2= 9+16

X = 5

According to pythagoras law

Answer:

The Answer Is B not C

Explanation:

Answer:b

Explanation:

Answer:

average acceleration is 1.365 × [tex]10^{4}[/tex] m/s²

Explanation:

given data

initila speed  u = -32.2 m/s

final speed v = 21.6 m/s

time taken t = 0.00394 s

solution

we get here average acceleration that will be express as

v = u + at    ..........................1

put here value and we get

21.6 = -32.2 + a × 0.00394

solve it we get

a = 1.365 × [tex]10^{4}[/tex] m/s²

so average acceleration is 1.365 × [tex]10^{4}[/tex] m/s²