A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the horizontal and the sand is moved without slipping at the rate of 2 m/s. The sand is collected in a big drum 5 m below the end of the conveyor belt. Determine the horizontal distance between the end of the conveyor belt and the middle of the collecting drum.

Answers

Answer 1

Answer:

The motion of sand is due to the movement of conveyor belt. The horizontal distance between the end of the conveyor belt and the middle of the collecting drum is 2.044 meters.

What is equation of motion?

The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.

The second equation of the motion for distance can be given as,

$y=ut+(2)/(2)gt^2$

Here, $u$ is the initial body, $g$ is the acceleration of the body due to gravity and $t$ is the time taken by it.

Given information-

The conveyor is tilted at an angle of 18° above the horizontal.

The Sand is moved without slipping at the rate of 2 m/s.

The sand is collected in a big drum 5 m below the end of the conveyor belt.

The horizontal component of the velocity is given as,

$v_y=2\cos 18$

The vertical component of the velocity is given as,

$v_y=2\sin18$

Put the value in the above equation as,

$y-y_0=v_yt+(1)/(2)gt^2$

$0-5=2\sin18 (t)+(1)/(2)*9.8\tiems t^2\nt=1.075\rm sec$

The horizontal distance between the end of the conveyor belt and the middle of the collecting drum is,

$d=v_xt\nd=2\cos18*1.075\nd=2.044\rm m$

Thus, the horizontal distance between the end of the conveyor belt and the middle of the collecting drum is 2.044 meters.

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Answer 2

Answer:

Answer:

x = 2.044 m

Explanation:

given data

initial vertical component of velocity = Vy = 2sin18

initial horizontal component of velocity = Vx = 2cos18

distance from the ground yo = 5m

ground distance y = 0

from equation of motion

$y = yo+ V_y t +(1)/(2)gt^2$

$0 = 5 + 2sin18+ (1)/(2)*9.8t^2$

solving for t

t = 1.075 sec

for horizontal motion

$x = V_x t$

x = 2cos18*1.075

x = 2.044 m

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What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the braketrail braking
controlled braking
threshold braking
coasting

Answers

Answer:

Controlled braking

Explanation:

CONTROLLED BRAKING occur in a situation where a person or an individual driving a vehicle releases the brake and slowly apply smooth as well as firmly pressure on the brake without the wheels been locked which is why CONTROLLED BRAKING are often used for emergency stops by drivers reason been that it helps to reduce speed when driving as fast as possible while the driver maintain the steering control of the vehicle.

Therefore the form of braking which is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake is called CONTROLLED BRAKING.

Final answer:

The method of braking that involves applying smooth, steady pressure to the brake to bring the vehicle to a smooth stop is called controlled braking. It helps prevent skidding and provides the driver with more control over the vehicle.

Explanation:

The form of braking used to bring a vehicle to a smooth stop by applying smooth, steady pressure to the brake is known as controlled braking. This method of braking involves applying consistent, even pressure to the brake pedal, which allows the car to slow down gently and gradually. It helps prevent uncontrolled skidding and provides the driver with more control over the vehicle's direction and speed during the stop. Unlike other methods like trail braking, threshold braking, or coasting, controlled braking is typically the safest and most effective method for daily driving conditions.

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Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.

Answers

Answer:

Point motion will eventually stops due to action of g exactly perpendicular...

Explanation:

If ignoring the air resistance, the magnitude of gravitational acceleration is already strong enough to stops the acceleration. As we know that, the spring constant of a bungee spring cord will be F = -k/x, where x is the stretched length and k is the spring constant of bungee cord. If F = ma = w = mg, the g = -m k/x. Now we can clearly see that the value of g remains constant due to the fluctuating length of the cord as the motion progresses back and forth in SHM say from x1 to x2 and x2 to x1.

Consider position [x] = L, time [t] = T, velocity [v] = L/T and acceleration [a] = L/T 2 . Find the exponent A in the equation v = a^2 t^ A /x

Answers

Answer:

The exponent A in the equation is 3.

Explanation:

v = a^2 t^ A /x

$v = (a^2t^A)/(x) \n\nvx = a^2t^A\n\n((L)/(T))(L) = ((L)/(T^2))^2(T)^A\n\n (L^2)/(T)= ((L^2)/(T^4))(T)^A\n\n (L^2)/(T) *(T^4)/(L^2) = (T)^A\n\nT^3 = T^A\n\n(T^3)/(T^3) = (T^A)/(T^3)\n\nT^(3-3) = T^(A-3)\n\n3-3 = A-3\n\n0 = A-3\n\nA = 3$

Therefore, the exponent A in the equation is 3.

As shown in the figure below, Greta walks 30m toward her truck. She notices she forgot hercoffee and returns back to the house. Her total travel time is 240 seconds.
30 m
30 in
What is Greta's average velocity over the 240s period?
m/s
What is Greta's average speed over the 240s period?
m/s

Answers

Average velocity: 0 m/s. Average speed: 0.25 m/s. Greta returns to her starting point, so her displacement is 0m.

Greta's average velocity is 0 m/s because she ends up at the same point where she started. Her displacement is 0 meters, and since velocity is displacement divided by time, her average velocity is 0 / 240 = 0 m/s.

Her average speed, on the other hand, is calculated using the formula: Average Speed = TotalDistance / Total Time.

Initially, Greta walks 30 meters away from her truck, and then she returns 30 meters back to her starting point. So, the total distance she covers is 30 + 30 = 60 meters. Her total travel time is 240 seconds.

AverageSpeed = 60 meters / 240 seconds = 0.25 m/s.

In summary, Greta's average velocity is 0 m/s because her net displacement is 0 meters. Her average speed is 0.25 m/s because she covers a total distance of 60 meters in 240 seconds.

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Answer:

0 | for Velocity

.25 | for speed

Explanation:

A popular physics lab involves a hand generator and an assortment of wires with different values of resistance. In the lab, the leads of the generator are connected across each wire in turn. For each wire, students attempt to turn the generator handle at the same constant rate. Students must push harder on the handle when the leads of the generator are connected__________. This is because turning the handle at a given constant rate produces__________ , regardless of what is connected to the leads. So, when turning the handle at a constant rate, lab students must push harder in cases where there is________

Answers

Answer:

Explanation:

Students must push harder on the handle when the leads of the generator are connected across the wire with the lowest resistance.

This is because turning the handle at a given constant rate produces a constant voltage across the leads, regardless of what is connected to the leads.

So, when turning the handle at a constant rate, lab students must push harder in case where there is a greater current through the connected wire.

A mass of 0.14 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.28 m)cos[(8 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass .