Two masses are being pulled up a 30.0-degree incline by a force F parallel to the incline. The acceleration up the incline is 1.00 m/s2 and the velocity is down the incline. The force is applied to a 200-kg mass and a string connects the 200-kg mass to a 150-kg mass. The coefficient of kinetic friction is 0.200. The force F is

Answers

Answer 1

Answer:

Explanation:

The acceleration up the incline is 1.00 m/s²

Net force acting on two masses = total mass x acceleration

= 350 x 1 = 350 N

weight acting down the plane = m g sinФ

= 350 x 9.8 x sin30 = 1715 N

Friction force acting down the plane = mg cosФ x μ where μ is coefficient of friction

= 350 x 9.8 x cos30 x .2 = 594N

Net force acting on total mass

= F - 1715 - 594 = 350 , where F is required force

F = 2659 N .


Related Questions

if the forces on an object are balanced the resultant force is equal to zero true false

Answers

Answer:

If the forces are balanced, the resultant force is zero. If the forces on an object are unbalanced, this is what happens: a stationary object starts to move in the direction of the resultant force. a moving object changes speed and/or direction in the direction of the resultant force.

Explanation:

A star's emission line of 400 nm appears shifted to 404 nm in the spectrum. What can you conclude from this shift?
A. The star is approaching you with the speed of 3000 km/s.
B. The star is approaching you with the speed of 30300 km/s.
C. The star is receding from you with the speed of 3000 km/s.
D. The star is receding from you with the speed of 30300 km/s.

Answers

Answer:

C. The star is receding from you with the speed of 3000 km/s

Explanation:

To get this answer we use the doppler effect equation . The formula for a receding emissor is given in the attachment.

We solve for V

V = 3x10⁶m/s

V = 3000km/s

We have the wavelength to be shifting towards red. Therefore we conclude that it is receding. We say the star is receding with speed of 3000km/s towards you.

Thank you!

Two motorcycles are traveling due east with different velocities. However, 5.68 seconds later, they have the same velocity. During this 5.68-second interval, motorcycle A has an average acceleration of 3.87 m/s2 due east, while motorcycle B has an average acceleration of 18.2 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 5.68-second interval, and (b) which motorcycle was moving faster?

Answers

Answer:

The answer is below

Explanation:

Let a be the initial velocity of motorcycle A and b be the initial velocity of motorcycle B.

After 5.68 seconds, both motorcycle had the same velocity (v), therefore for motorcycle A:

(a - v) / 5.68 = 3.87

a - v = 21.9816

v = a - 21.9816

For motorcycle B:

(b - v) / 5.68 = 18.2

b - v = 103.376

v = b - 103.376

Therefore:

a - 21.9816 = b - 103.376

b - a = -21.9816 + 103.376

b - a = 81.3944

a) The difference between their speeds at the beginning was 81.3944 m/s

b) Since b - a = 81.3944. This means that the initial velocity of motorcycle B is greater than that of motorcycle A by 81.3944 m/s.

Therefore motorcycle B was moving faster

Balance the following equation:
H3B03 →_B203 +_H20
a. 1, 3,2
b. 2,4,6
C. 4, 2, 6
d. 6, 4,2

Answers

It's c I think ( 4 , 2 , 6 , )

A skier pushes off the top of a hill with an initial speed of 3.30 m/s. How fast will she be moving after dropping 5.00 meters in elevation if friction is negligible?

Answers

Answer:

eeeeeeeeeeeeeeeeeeeee

Earth has seasons because _____.

it rotates on its axis as it moves around the sun

the temperature of the sun changes

its axis is tilted

the distance between Earth and the sun changes

Answers

Answer:

c, its axis is tilted

maybe

As it works its way around the sun, its tilted axis exposes different parts of earth.

C would be it because the roation of Earth on its axis doesn't have anything to do with the exposer of the revolution on its axis

It’s c .its axis is tilted

A long, straight wire has a uniform constant charge with linear charge density, - 3.60 nC/m. The wire is surrounded by a long nonconducting, thin-walled cylindrical shell that is charged on its outside surface, such that the electric field outside the shell is zero. The shell has a radius of 1.50 cm.

Required:
What uniform area charge density rho is needed on the shell for the electric field to be zero outside the shell?

Answers

Answer:

Uniform area charge density rho is needed is 3.82*10^-8 C.m^-2

Explanation:

See the attached files.

To find the rho, I used Gauss law for cylindrical shell which is equation 1 and Gauss law for the rod which is equation 4.

Note that in equation 4, Lamda is the charge per length while L is the length if the rod. Also R is the radius of the shell.

The final answer is 3.82*10^-8 C.m^-2 which is the uniform area charge density rho is needed.

Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.

Required:
a. Will the final temperature of the gas in A be greater, less than, or equal to the temperature in B?
b. Show both processes on a single PV diagram.
c. What are the initial pressures in containers A and B?
d. Suppose the heaters have 25 W of power and are turned on for 15s. What is the final volume of container B?

Answers

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial  pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial  pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) /  0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial  pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

what is the acceleration of a satellite moving in a circular orbit around the earth of radius 2r​

Answers

Explanation:

You do the radius times the circumference of the earth

A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has a 9.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.80 times that due to gravity

Answers

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration ([tex]a_c[/tex]) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration ([tex]a_c[/tex]) is given by:

[tex]a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s[/tex]

The velocity (v) is given by:

v = ωr;  where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

HELP ! ILL MARK BRAINLIEST HELP ASAP

Answers

Answer:

A

Explanation:

In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.

Hope this helps!

As the distance between the sun and earth decreases, the speed of the planet

a
increases
b
decreases
c
stays the same

Answers

Answer:

Explanation:

Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.

What is the shortest time that a jet pilot starting from rest can take to reach Mach-3.60 (3.60 times the speed of sound) without graying out? (Use 331 m/s for the speed of sound in cold air.)

Answers

Answer:

30.4 s

Explanation:

A pilot , with plane accelerated at 4 g starts greying out . In the problem , the acceleration of jet is 4 g

a = 4 x 9.8 = 39.2 m /s²

initial velocity u = 0

Final velocity = 3.60 times speed of sound

= 3.6 x 331 = 1191.6 m /s

v = u + at

Putting the values

1191.6 = 0 + 39.2 t

t = 30.4 s .

Which statement BEST explains why a bouncing basketball will not remain in motion forever?

Group of answer choices

The energy is transferred to sound and heat energy.

The energy is used up and destroyed.

The energy is transferred to light and potential energy.

The energy is transferred to chemical and heat energy.

Answers

Answer:

The energy is transferred to chemical and heat energy.

Explanation:

If you define "bouncing" as leaving the ground for any amount of time, the ball stops bouncing when the elastic energy stored in the compression phase of the bounce is not enough to overcome the weight of the ball. This is the proof of the answer i Hope this helps :)

If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is


364N

185N

173N

73N


ANSWER AND I WILL GIVE YOU BRAINILIEST

Answers

Answer:

73N

Explanation:Just multiply 1.2^2 by 50

a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time​

Answers

HERE IS YOUR ANSWER!

A constant electric field of 5.00 N/C points along the positive x-direction. An electron, initially at rest, moves a distance of 2.00 m in this space. How fast is the electron moving after its 2.00 m journey

Answers

Answer:

1.875 x 10⁶ m /s .

Explanation:

Force on electron = E e where E is electric field and e is charge on electron

acceleration generated = Ee / m where m is mass of the electron .

Putting the values

acceleration generated = 5 x 1.6 x 10⁻¹⁹ / 9.1 x 10⁻³¹

= .879 x 10¹² m /s²

v² = u² + 2 as , initial velocity u = 0 , displacement s = 2 m

v² = 0 + 2 x .879 x 10¹² x 2

v = 1.875 x 10⁶ m /s .

I need help will mark brainliest

Answers

Answer: ITS 1 TRUST ME MAN BYE K

Explanation: OK BYE TRUST YEAH

I’ll mark brainless please hurry

Answers

Answer:

Covalent bonds can form between similar atoms.

Explanation:

77. A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
a).004
s2
b).0056 m/s2 c).0079"
d).01 m/s2
M
m

Answers

Answer:

a

Explanation:

i just took the test

A spring has a spring constant of 25 Newtons per meter. The minimum force required to
stretch the spring 0.20 meter from its equilibrium position is approximately

Answers

Answer:

6.3N

Explanation:

Guessed it right on castle learning

Answer:

6.3 N

Explanation:

F=kx

F=(25N/m)(0.25m)

6.3 N

What are the two rules that light follows.​

Answers

ok so i dont know srry5

Calculate the magnitude and direction of the resultant of the following forces



Answer and I will give you brainiliest

Answers

Answer:

250N same direction

Explanation:

100+150 = 250N

same direction

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