PHYSICS COLLEGE

. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force

Answers

Answer 1

Answer:

Answer:

elative magnitude of the two forces is the same and they are applied in a constant direction.

Explanation:

Newton's second law states that the sum of the forces is equal to the mass times the acceleration

∑ F = m a

in this case there are two forces on the x axis

F_applied - fr = 0

since they indicate that the velocity is constant, consequently

F_applied = fr

the relative magnitude of the two forces is the same and they are applied in a constant direction.

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In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes were sliding toward each other and crashed. The crash caused both boxes to change speed. Based on the information in the diagram, which statement is correct? In your answer, explain what the forces were like and why the boxes changed speed.Box 1 has more mass than Box 2.Box 1 and Box 2 are the same mass.Box 1 has less mass than Box 2.**YOU MUST BE DESCRIPTIVE! Any short answers not explaining it wont get brainliest!**

Define the term energy density of a body under strain

Answers

Answer:

Please mark as Brainliest!!

Explanation:

Strain energy is defined as the energy stored in a body due to deformation. The strain energy per unit volume is known as strain energy density and the area under the stress-strain curve towards the point of deformation. When the applied force is released, the whole system returns to its original shape.

An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ.Find an expression for the angular velocity ω in terms of g, L and angle θ.

Answers

Answer:

The expression would be ω = $\sqrt{(g)/(L sin 0 ) }$

Explanation:

Given that ω is the angular velocity

g is the acceleration due to gravity

L is the length

θ is the angle of downward tilt

For an object we compare the horizontal and vertical component of the forces acting on the body;

For vertical component

T sinθ = mg............1

For the horizontal component

T cos θ = $(mv^(2) )/(R)$ .............2

R is our radius and is = L cos θ

v = ωR

substituting into equation 2 we have

T cos θ = m(ωR $)^(2)$ /R

T cos θ=m(ω $)^(2)$ R ..................3

Now comparing the vertical and the horizontal component we have;

equation 1 divided by equation 3 we have

T sin θ /T cos θ = mg / m(ω $)^(2)$ R

Tan θ = g / (ω $)^(2)$ R............4

Making ω the subject formula we have;

(ω $)^(2)$ = g/ R Tan θ

But R = L cos θ and Tan θ = sin θ/ cosθ

putting into equation 4 we have;

(ω $)^(2)$ = g /[( L cos θ) x( sin θ/ cosθ)]

(ω $)^(2)$ = g/ L sinθ

ω = $\sqrt{(g)/(L sin 0 ) }$

Therefor the expression for the angular velocity ω in terms of g, L and angle θ would be ω = $\sqrt{(g)/(L sin 0 ) }$

Let g, r, L and T are gravity, radius, length, and angle of string w/r/t vertical, respectively.
Then
ω²r = rg/Lcos T
ω² = g/L cos T
ω = √(g / L cos T)

We can calculate the force that the atmospheric pressure produces on a surface. Consider a living room that has a 4.0m×5.0m floor and a ceiling 3.0m high. What is the total force on the floor due to the air above the surface if the air pressure is 1.00 atm?

Answers

Answer:

Force, $F=2.02* 10^6\ N$

Explanation:

It is given that,

Length of the room, l = 4 m

breadth of the room, b = 5 m

Height of the room, h = 3 m

Atmospheric pressure, $P=1\ atm=101325\ Pa$

We know that the force acting per unit area is called pressure exerted. Its formula is given by :

$P=(F)/(A)$

$F=P* l* b$

$F=101325* 4* 5$

$F=2.02* 10^6\ N$

So, the total force on the floor due to the air above the surface is $2.02* 10^6\ N$ . Hence, this is the required solution.

why does the value of capacitance of a capacitor increases in parallel combination but not in series??

Answers

It is easiest to consider problems like this by thinking exclusively about parallel plate capacitors for which $C \equiv (Q)/(V) =\kappa \epsilon_0 (A)/(d)$ where Q is the charge separated (+Q on one plate, -Q on the other), V is the voltage difference between the plates, A is the area of each plate, and d is the separation between the plates.

When capacitors are connected in parallel, the voltage across each capacitor is the same. But with two capacitors, it will require more charge to reach the voltage V than it would with just one capacitor. In fact, if capacitor 1 requires charge

A bowling ball of mass 3 kg is dropped from the top of a tall building. It safely lands on the ground 3.5 seconds later. Neglecting air friction, what is the height of the building in meters? (Give the answer without a unit and round it to the nearest whole number)

Answers

The height of the building is 60 m.

calculation of building height:

The velocity of the ball should be provided by

v = u + gt

here,

u is the initial velocity of the ball = 0

v = 0 + 9.8 x 3.5

v = 34.3 m/s

Now

When the ball hits the ground, energy is conserved;

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (0.5 v²) / g

h = (0.5 x 34.3²) / (9.8)

h = 60.025 m

h = 60 m

Learn more about friction here: brainly.com/question/14455351

Answer:

The height of the building is 60 m.

Explanation:

Given;

mass of the mass of the ball, m = 3 kg

time of motion, t = 3.5 s

The velocity of the ball is given by;

v = u + gt

where;

u is the initial velocity of the ball = 0

v = 0 + 9.8 x 3.5

v = 34.3 m/s

When the ball hits the ground, energy is conserved;

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (0.5 v²) / g

h = (0.5 x 34.3²) / (9.8)

h = 60.025 m

h = 60 m

Therefore, the height of the building is 60 m.

When a 5.0 kg box is hung from a spring, the spring stretches to 50 mm beyond its relaxed length. (a) In an elevator accelerating upward at 2.0 m/s2 , how far does the spring stretch with the same box attached? (b) How fast and in which direction should the elevator accelerate for the spring stretch to be zero (that is, the spring returns to its relaxed length)?

Answers

The extension of the spring in the elevator is 60 mm.

For the extension of the spring to be zero, the elevator must be moving downwards under free fall.

The given parameters;

mass of the box, m = 5 kg
extension of the spring, x = 50 mm = 0.05 m

The spring constant is calculated as follows;

F = kx

mg = kx

$k = (mg)/(x) \n\nk = (5 * 9.8)/(0.05) \n\nk = 980 \ N/m$

The tension on the spring in an elevator accelerating upwards is calculated as follows;

T = mg + ma

T = m(g + a)

T = 5(9.8 + 2)

T = 59 N

The extension of the spring is calculated as follows;

$T = kx\n\nx = (T)/(k) \n\nx = (59)/(980) \n\nx = 0.06 \ m\n\nx = 60 \ mm$

For the extension of the spring to be zero, the elevator must be under free fall, such that the tension on the spring is zero.

For free fall, a = g

T = m(g - a) = 0

Learn more here:brainly.com/question/4404276

Answer:

a) the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Explanation:

Given that;

Gravitational acceleration g = 9.81 m/s²

Mass m = 5 kg

Extension of the spring X = 50 mm = 0.05 m

Spring constant k = ?

we know that;

mg = kX

5 × 9.81 = k(0.05)

k = 981 N/m

Given that; Acceleration of the elevator a = 2 m/s² upwards

Extension of the spring in this situation = X1

Force exerted by the spring = F

we know that;

ma = F - mg

ma = kX1 - mg

we substitute

5 × 2 = 981 × X1 - (5 ×9.81 )

X1 = 0.06019 m

X1 = 60.19 mm

Therefore the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

Acceleration of the elevator = a

The spring is relaxed i.e, it is not exerting any force on the box.

Only the weight force of the box is exerted on the box.

ma = mg

a = g

a = 9.81 m/s² downwards.

Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

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