What can be said about the sign of the work done by the force f⃗ 1??

Sophie [7]

Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

• The net force in the parallel direction is

∑ F (para) = -mg sin(21°) - f = ma

• The net force in the perpendicular direction is

∑ F (perp) = n - mg cos(21°) = 0

Solving the second equation for n gives

n = mg cos(21°)

n = (0.200 kg) (9.80 m/s²) cos(21°)

n ≈ 1.83 N

Then the magnitude of friction is

f = µn

f = 0.25 (1.83 N)

f ≈ 0.457 N

Solve for the acceleration a :

-mg sin(21°) - f = ma

a = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

a ≈ -5.80 m/s²

so the block is decelerating with magnitude

a = 5.80 m/s²

down the ramp.

5 0

2 years ago

STATE THE LAWS OF CONSERVATION OF MOMENTUM

dalvyx [7]

$\Huge \mid \underline {\mathcal {{{\color{purple}{Answer...}}}}} \mid$

When two bodies collide with each other in the absence of an external force, then the total final momentum of the bodies is equal to their total initial momentum.

6 0

3 years ago

a roller coaster is at the top of a hill and rolls to the top of a lower hill.If mechanical energy is conserved,on the top of wh

pychu [463]

Hello!!

Here we have a simple matter of conservation of energy. ME=PE+KE.

At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.

Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!

7 0

3 years ago

at location a, what are the directions of the electric fields contributed by the electron. calculate the magnitudes of the elect

Lisa [10]

We can use the equation E = k | Q | r 2 E = k | Q | r2 to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge,

<h3>What is electric and magnetic field ?</h3>

With the use of electricity and other types of artificial and natural illumination, invisible energy fields known as electric and magnetic fields (EMFs) and radiation are created.

While the magnetic field is discernible by the force it exerts on other magnetic particles and moving electric charges, the electric field is actually the force per unit charge experienced by a non-moving point charge at any given location inside the field.

Learn more about Electromagnetic field here:

brainly.com/question/14372859

#SPJ4

4 0

1 year ago

The acceleration due to gravity, g , is constant at sea level on the Earth's surface. However, the acceleration decreases as an

blsea [12.9K]

Answer:

g = g₀ [1- 2 h / Re + 3 (h / Re)²]

Explanation:

The law of universal gravitation is

F = G m Me / Re²

Where g is the universal gravitational constant, m and Me are the mass of the body and the Earth, respectively and R is the distance between them

F = G Me /Re² m

We call gravity acceleration a

g₀ = G Me / Re².

When the body is at a height h above the surface the distance is

R = Re + h

Therefore the attractive force is

F = G Me m / (Re + h)²

Let's take Re's common factor

F = G Me / Re² m / (1+ h / Re)²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

(1+ h / Re)⁻² = 1 -2 h / Re + 6/2 (h / Re) 2 + ...

Let's replace

F = G Me /Re² m [1- 2 h / Re + 3 (h / Re)²]

F = g₀ m [1- 2 h / Re + 3 (h / Re)²]

If we call the force of attraction at height

m g =g₀ m [1- 2 h / Re + 3 (h / Re)²]

g = g₀ [1- 2 h / Re + 3 (h / Re)²]

3 0

3 years ago