All categories

2 years ago

Anyone good at physics, that can help me? Will give 100 point!!!

Physics

1 answer:

gayaneshka [121]2 years ago

6 0

Hi there!

Part 1:

Assuming this is from the work-energy unit, we can use the work-energy theorem to solve.

Since the block starts from rest:

Initial energy = GPE = mgh

Final energy = KE = 1/2mv²

We are only given the diagonal distance of the ramp, so we must solve for its height using trigonometry.

sin(30) = O/H

Hsin(30) = 11.9sin(30) = 5.95 m

Now, solve by setting the GPE and KE equal:

mgh = 1/2mv²

gh = 1/2v²

v = √2gh

v = √2(9.8)(5.95) = 10.8 m/s

Part 2:

Now, we can use the equation for work:

W = ΔKE

The final velocity of the block after sliding is 0 m/s, so:

W = 1/2mv²

Recall the equation for work:

W = Fdcosθ

Since friction works AGAINST motion, cos(180) = -1.

Thus, the work done by friction is:

W = -Fd

Recall the equation for kinetic friction:

F = μmg

Thus:

0 = 1/2mv² - Fd

0 = 1/2mv² - μmg

μmgd = 1/2mv²

Cancel out the mass and rearrange to solve for the coefficient of friction:

μgd = 1/2v²

μd = 0.5v²/gd

μ = 0.5(10.8²)/(9.8)(21) = 0.283

Part 3:

The energy lost due to friction is equivalent to the WORK DONE by friction, which is equivalent to the initial kinetic energy of the block at the bottom of the ramp.

Thus:

W = 1/2mv² = Fd

W = 1/2(10)(10.8²) = 583.1 J

You might be interested in

Help i am having a mental breakdown help! need this done...

guajiro [1.7K]

1. he traveled a total of 24 miles
2. peter is not moving between 10 and 30 minutes

4 0

3 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

4 years ago

20 Accelerated motion is represented by a line on a nonlinear distance time graph.

Varvara68 [4.7K]

Curved line

Explanation:

Acceleration of motion is represented by a curved line on a non-linear distance-time graph.

The acceleration of a non-linear motion is depicted using a parabola which is a curve. This implies that the velocity is constantly changing and the distance covered by the body is also changing with equal amount of time.

A plot of this will give a parabola. This can be further established using one of the equations of motion below:

x = u + $\frac{1}{2}$ at ²

This is a quadratic function where:

x is the distance

u is the initial velocity

t is the time

a is acceleration

A quadratic function gives a curved line which is a parabola.

Learn more:

Acceleration brainly.com/question/10932946

#learnwithBrainly

3 0

3 years ago

Glider one and glider two collided. The data table above shows the momentum of each before and after the collision. Perform an a

k0ka [10]

I think there was momentum conserved

Explanation: I took the test

7 0

3 years ago

A campground official wants to measure the distance across an irregularly-shaped lake, but can't do so directly. He picks a poin

sleet_krkn [62]

Answer:

149 m

Explanation:

The distances across the lake is forming a triangle.

let the distance between the point and the left side be 'x'

and the distance between the point and the right be 'y'

and the distance across the lake be 'z' and the angle opposite to 'z' be 'Z' given:

∠Z = 83°

x = 105 m

y = 119 m

Now, applying the Law of Cosines, we get

z² = x² + y² - 2xycos(Z)

Substituting the values in the above equation, we get

z² = 105² + 119² - 2×105×119×cos(83°)

z = √22140.48

z = 148.796 m ≈ 149 m

The point is 149 m across the lake

8 0

3 years ago