PHYSICS HIGH SCHOOL

What is the mass of a falling rock if it produces a force of 147 N with an acceleration of 9.8 m/s2?

Answers

Answer 1

Answer:

Answer:

15 kg

Explanation:

The mass of the object can be found by using the formula

$m = (f)/(a) \n$

f is the force

a is the acceleration

From the question we have

$m = (147)/(9.8) \n$

We have the final answer as

15 kg

Hope this helps you

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Why is the efficiency of a machine always less than 100 percent

Answers

Explanation :

The efficiency of a machine is always less than 100 %. This is because some of the energy gets lost in the form of heat energy, sound energy and frictional energy.

Mathematically it is defined as :

$\eta=(W_o)/(W_i)$

Where,

$\eta$ is the efficiency

$W_o$ is output work

$W_i$ is input work

Hence, this is the required explanation.

energy is lost due to friction or air resistance, through sound and heat energy.

15) What is the frequency of a pendulum that is moving at 30 m/s with a wavelength of .35 m?show step by step

Answers

We know that there is a formula velocity = frequency x wavelength for all types of waves.

If we assume one complete oscillation of a pendulum to be wavelength we can apply the above formula for the pendulum too.

So as v = fλ and f = v/λ we can just plug in the values to get our answer of frequency.

So frequency = 30/0.35 which is equal to 85.17 Hertz (Hz).

I think you're trying to take the formulas for speed, wavelength, and
frequency of a wave, and apply them to a pendulum. You can't do that.
It doesn't work.

A pendulum is moving in 'simple harmonic motion', not wave motion.
It's speed is continuously changing, from zero at both ends of its swing,
to maximum as it passes through the 'rest' position at the bottom. And
there's no wavelength defined for a pendulum ... if you're thinking that
it could be the distance from end to end of its swing, or maybe half of
that, you should know that the frequency of an ideal simple pendulum
is not related to that distance at all.

Finally, in the real world, the numbers in this question really kind of
don't make any sense. You have a structure where some part of it is
roughly a foot long (0.35m = 13.8 inches), and at least at some point
during its swing, something is moving at 30 m/s ... about 67 mph !
If something like that could even stay on the table, and IF its frequency
were (speed/wavelength) ... like a wave's frequency is ... then its frequency
would be (30 / 0.35) = 85.7 Hz ! ! The thing would be wiggling back and
forth every 0.017 second ! It would need to be operated only inside
a bomb shelter, with all personnel withdrawn beyond a safe perimeter
before it flies apart and scatters shrapnel everywhere.

What is the potential energy of a 1-kilogram ball is thrown into the air with an initial velocity of 30m/sec?

Answers

the ball thrown from height=0
potental energy(PE) =0
kinetic energy(KE) = 0.5mv^2 = 0.5(1)(30^2) = 450

at the highest point the ball does not moved, v=0
potential energy at its maximum
kinetic energy = 0

Energy is conserved then
total energy before = after
PE1 + KE1 = PE2 + KE2
0 + 450 = PE2 + 0

conservation of energy is fun fact

mass=1kg
g=10m/s^2 (assuming)
u=30m/s
height(h)=u^2/2g
=900/20=45m
P.E.=mgh
=1×10×45=450 joules

How does the color of an object affect its ability to emit or absorb thermal radiation?

Answers

Answer:

The darker the object, the better it emits heat, because it's a better absorber of light. On the other hand, a white object appears white because it reflects all the different wavelengths and absorbs little to no light. It doesn't absorb much energy, then, and puts off little to no heat.

Explanation:

A 7.80-g bullet moving at 575 m/s strikes the hand of a superhero, causing the hand to move 5.50 cm in the direction of the bullet’s velocity before stopping. (a) Use work and energy considerations to find the average force that stops the bullet. (b) Assuming the force is constant, determine how much time elapses between the moment the bullet strikes the hand and the moment it stops moving.

Answers

Answer:

a)40.77N

b)0.11seg

Explanation:

The bullet has a kinetic energy:

$K=1/2*m*v^(2) = 1/2*0.0078kg*575m/s=2.2425J$

When the bullet stops this energy goes to zero, as the energy must be conserved the work done by the head of the superhero must be equal to the original energy of the bullet:

$W=K$

Considering an average constant force, the work can be calculated as:

$W=F*d=K$

Solving for F:

$F=K/d=2.2425J/0.055m=40.77N$

The acceleration(on the opposite direction of motion) on the bullet will be:

$a=F/m=40.77N/0.0078kg=5226.9 m/s^(2)$

The velocity of the bullet considering a constant acceleration can be calculate as:

$V=V_(0) - a*t$

It stops moving when V=0 so:

$V_(0) - a*t=0=>t=V_(0)/a=(575m/s)/(5226.9m/s^(2) ) =0.11seg$

Answer:

a) 23444.3 N

b) 0.191 ms

Explanation:

Given:

Mass of the bullet m = 7.8 g = 7.8 * 10⁻³ Kg

Initial speed u = 575 m/s

Final speed v = 0

Distance covered s = 5.50 cm = 0.055 m

(a)

According to work energy theorem

Work done W = Change in Kinetic energy

hence, force*distance = 1/2*mass*velocity²

that is, F*S = (1/2)mu²

F*0.055 = 0.5 * 7.8 * 10⁻³ * (575)^2

F = 23444.3 N

(2)

From Impulse definition

force*time = change in momentum

where momentum = mass * velocity

F*t = Change in momentum = m* u

23444.3t = 7.8*10⁻³ * 575

Time, t = 0.191 ms

The Earth’s rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 2006 the Earth took about 0.840 s longer to complete 365 revolu- tions than it did in the year 1906. What was the average angular acceleration, in rad>s2 , of the Earth during this time?

Answers

Answer:

Approximately $\rm 6.1 * 10^(-22)\; rad \cdot s^(-2)$ .

Explanation:

Angular acceleration is equal to $\displaystyle \frac{\text{Change in angular speed}}{\text{Time taken}}$ .

Apparently, for this question, the time taken is $100\; \text{years} \approx 100 * 365.24* 24 * 3600 \; \text{seconds}$ . The challenge is to find the change in angular speed over that period of time.

Let the time (in seconds) it took to complete $365$ revolutions be $t$ in the year 1906. In 2006 that number would be $(t + 0.840)$ .

Each revolution is $2\pi$ radians. $365$ revolutions will be an angular displacement of $365 * 2\pi$ in radians. Angular speed is equal to $\displaystyle \frac{\text{Angular Displacement}}{\text{Time Taken}}$ .

The average angular speed in 1906 could thus be written as $\displaystyle (365* 2\pi)/(t)$ .

Similarly, the average angular speed in 2006 could be written as $\displaystyle (365* 2\pi)/(t + 0.840)$ .

The difference between the two will be equal to:

$\begin{aligned} & \; \Delta \omega \cr = &\; (365* 2\pi)/(t) - (365* 2\pi)/(t + 0.840)\cr =& \; 365 * 2\pi * \left.((t + 0.840) - t)/(t(t + 0.840))\right. \cr =& \;365 * 2\pi * \left.(0.840)/(t(t + 0.840))\end{aligned}$ .

Since the value of $t$ (about the same as the number of seconds in 365 days) is much bigger than $0.840\; \rm s$ , apply the approximation $t + 0.840 \approx t$ .

$\begin{aligned} &\;365 * 2\pi * \left.(0.840)/(t(t + 0.840)) \cr \approx &\; 365 * 2\pi * \left.(0.840)/(t^2) \cr \approx & \; 365 * 2\pi *(0.840)/((365 * 24 * 3600)^2) \cr \approx &\; 1.9370* 10^(-12)\; \rm rad \cdot s^(-1)\end{aligned}$ .

And that's approximately the average change in angular velocity over a period of 100 years. Apply the formula for average angular acceleration:

$\displaystyle (\Delta \omega)/(\Delta t) = \rm (1.9370* 10^(-12)\; rad \cdot s^(-1))/(100 * 365.24 * 24 * 3600\; s) \approx 6.1* 10^(-22)\; rad \cdot s^(-2)$ .

The average angular acceleration of the Earth during this time was approximately -7.42 × $10^{-22$ rad/s², indicating that the Earth's rotation is slowing down.

To calculate the average angular acceleration of the Earth during this time, we can use the following formula:

Average angular acceleration = (Δω) / Δt

where Δω is the change in angular velocity and Δt is the elapsed time.

The angular velocity of the Earth can be calculated using the formula:

ω = 2π / T

where T is the period of rotation. In this case, T is equal to 365.25 days, or approximately 31.54 × $10^6$ seconds.

Using the given information, we can calculate the change in angular velocity as follows:

Δω = (2π / 31.54 × $10^6$ seconds) - (2π / (31.54 × $10^6$ + 0.840 seconds))

Δω ≈ -2.34 × $10^{-15$ rad/s

The elapsed time is simply the difference between the two years, or 100 years. Therefore, the average angular acceleration is:

Average angular acceleration = (-2.34 × $10^{-15$ rad/s) / (100 years × 3.154 × $10^7$ seconds/year)

Average angular acceleration ≈ -7.42 × $10^{-22$ rad/s²

Therefore, the average angular acceleration of the Earth during this time was approximately -7.42 × $10^{-22$ rad/s². The negative sign indicates that the angular acceleration is causing the Earth's rotation to slow down.

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