A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 n at an angle of 32.0° above the horizontal. the coefficient of kinetic friction between the box and the surface is 0.350. what is the acceleration of the box?

Answers

Answer 1

Answer:

The acceleration of the box is 0.81 m/sec².

What is acceleration?

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration

According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:

$\sum F= ma$ (1)

we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:

- the horizontal component of the force exerted by the rope, which is equal to

$F_x = F cos\theta = 250*cos 32 = 212 N$

the frictional force, acting in the opposite direction, which is equal to

$F_f = \mu *mg = 171.7 N$

By applying Newton's law (1), we can calculate the acceleration of the

box,

$F_x - F_f = ma\na = 0.81 m/sec^2$

The acceleration of the box is 0.81 m/sec².

To learn more about acceleration refer to the link:

brainly.com/question/12550364

#SPJ2

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An eighteen gauge copper wire has a nominal diameter of 1.02mm. This wire carries a constant current of 1.67A to a 200w lamp. The density of free electrons is 8.5 x 1028 electrons per cubic metre. Find the magnitude of:i. The current density ii. The drift velocity

Answers

Answer:

The current density is $J = 2.04 * 10^(6) A /m^2$

The drift velocity is $v_d = 1.5 * 10^(-4) m/s$

Explanation:

From the question we are told that

The nominal diameter of the wire is $d = 1.02 mm= (1.02)/(1000) = 0.00102 \ m$

The current carried by the wire is $I = 1.67 A$

The power rating of the lamp is $P = 200 W$

The density of electron is $n = 8.5 * 10^(28) \ e/m^3$

The current density is mathematically represented as

$J = (I)/(A)$

Where A is the area which is mathematically evaluated as

$A = \pi (d^2)/(4)$

Substituting values

$A = 3.142 * ((1.02 * 10^(-3))^2 )/(4)$

$A = 8.0*10^(-4)m^2$

$J = (1.67)/(8.0*10^(-4))$

$J = 2.04 * 10^(6) A /m^2$

The drift velocity is mathematically represented as

$v_d = (J)/(ne)$

Where e is the charge on one electron which has a value $e = 1.602 *10^(-19) C$

$v_d =(2.04 * 10^6 )/(8.5 *10^(28) * 1.6 * 10^(-19))$

$v_d = 1.5 * 10^(-4) m/s$

A disk of mass 5 kg and radius 1m is rotating about its center. A lump of clay of mass 3kg is dropped onto the disk at a radius of 0.5m , sticking to the disk. If the system is rotating with an angular velocity of 11 rad/s, what is the final angular momentum of the disk h the clay lump?wit? ( Idisk = MR^2/2)

Answers

Answer:

$27.5 kgm^2/s$

Explanation:

We can solve for the final angular velocity of the system using the law of momentum conservation

$I_1\omega_1 = I_2\omega_2 = M_2$

Where $I_1 = MR^2/2 = 5*1^2/2 = 2.5 kgm^2$ is the moments of inertia of the disk before. $I_2 = I_1 + mr^2 = 2.5 + 3*0.5^2 = 2.5 + 0.75 = 3.25 kgm^2$ is the moments of inertia of the disk after (if we treat the clay as a point particle). $\omega_1 = 11rad/s$ is the angular speed before.

$2.5*11 = M_2$

$M_2 = 27.5 kgm^2/s$

So the final momentum of the system is 27.5 kgm2/s

Answer:

The final angular momentum is 35.75 kg.m²/s

Explanation:

Given;

mass of disk, M = 5 kg

radius of disk, R = 1 m

mass of clay, M = 3 kg

radius of clay, R = 0.5 m

final angular momentum, $\omega _f$ = 11 rad/s

Final angular momentum angular momentum of the disk that the clay lumped with;

$P = I_f\omega_f$

where;

$I_f$ is the final moment of inertia

$I_f = I_(disk) + I _(sand)\n\nI_f = (M_DR^2)/(2) + M_SR^2\n\nI_f = (5*1^2)/(2)+ 3*0.5^2\n\nI_f = 2.5 + 0.75=3.25 \ kg.m^2$

Final angular momentum of the disk;

= $I_f \omega_f$

= 3.25 x 11 = 35.75 kg.m²/s

Therefore, the final angular momentum is 35.75 kg.m²/s

A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, `A D 0:28 m, and `B D 0:36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.

Answers

Answer:

The acceleration of the mower will be "4.7 m/s²".

Explanation:

Balance of vertical force will be:

⇒ $Ra + Rb = mg$

For wheel to take off at A,

⇒ $Ra = 0$

Hence,

$Rb=mg$

Balancing moments about G will be:

⇒ $F* h = Rb* LB$

As we know,

Force, F = $(Rb* LB )/(h)$

On putting the values, we get

⇒ = $(38* 9.81* 0.36)/(0.75)$

⇒ = $178.9 \ N$

Now,

Acceleration, a = $(F)/(m)$

⇒ = $(178.9)/(38)$

⇒ = $4.7 \ m/s^2$

You are measuring the volume of a chemical beaker how would u take the measurment? a.lift the beaker to eye level
b.you look down at the liquid above
c.use a ruler to measure
d.you squat to be eye level to the beaker

Answers

I think its d because lifting it would make the chemical swish around and that will make it so you cant get the right measurement. hope this helps :)

Circuit A in a house has a voltage of 218 V and is limited by a 45-A circuit breaker. Circuit B is at 120.0 V and has a 25-A circuit breaker.What is the ratio of the maximum power delivered by circuit A to that delivered by circuit B?

Answers

Answer:

3.27

Explanation:

Electric Power: This can be defined as the rate at which electric energy is consumed. The unit of power is Watt (W).

Mathematically, electric power is represented as

P = VI ..................................... Equation 1.

Where P = power, V = voltage, I = Current.

For Circuit A,

P₁ = V₁I₁ ................................... Equation 2

Where P₁ = maximum power delivered by circuit A, V₁ = Voltage of circuit A, I₁ = circuit breaker rating of circuit A.

Given: V₁ = 218 V, I₁ = 45 A.

Substituting into equation 2

P₁ = 218×45

P₁ = 9810 W.

For Circuit B,

P₂ = V₂I₂............................. Equation 3

Where P₂ = maximum power delivered by the circuit B, V₂ = voltage of circuit B, I₂ = circuit breaker rating of circuit B

Given: V₂ = 120 V, I₂ = 25 A.

Substitute into equation 3

P₂ = 120(25)

P₂ = 3000 W.

Ratio of maximum power delivered by circuit A to that delivered by circuit B = 9810/3000

= 3.27.

Thus the ratio of maximum power delivered by circuit A to circuit B = 3.27

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).1. For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.Express your answer in terms of g and the variables m, l, x, and θ.(U^G=?)2. Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)Express your answer in terms of g and the variables m, l, and θ.

Answers

Answer:

UG (x) = m*g*x*sin(Q)

Vx,f (x)= sqrt (2*g*x*sin(Q))

Explanation:

Given:

- The length of the friction less surface L

- The angle Q is made with horizontal

- UG ( x = L ) = 0

- UK ( x = 0) = 0

Find:

derive an expression for the potential energy of the block-Earth system as a function of x.

determine the speed of the block at the bottom of the incline.

Solution:

- We know that the gravitational potential of an object relative to datum is given by:

UG = m*g*y

Where,

m is the mass of the object

g is the gravitational acceleration constant

y is the vertical distance from datum to the current position.

- We will consider a right angle triangle with hypotenuse x and angle Q with the base and y as the height. The relation between each variable can be given according to Pythagoras theorem as follows:

y = x*sin(Q)

- Substitute the above relationship in the expression for UG as follows:

UG = m*g*x*sin(Q)

- To formulate an expression of velocity at the bottom we can use an energy balance or law of conservation of energy on the block:

UG = UK

- Where UK is kinetic energy given by:

UK = 0.5*m*Vx,f^2

Where Vx,f is the final velocity of the object @ x:

m*g*x*sin(Q) = 0.5*m*Vx,f^2

-Simplify and solve for Vx,f:

Vx,f^2 = 2*g*x*sin(Q)

Hence, Velocity is given by: