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Sloan [31]
1 year ago
14

A conveyer belt moves a 40 kg box at a velocity of 2 m/s. What is the kinetic energy of the box while it is on the conveyor belt

?
Physics
2 answers:
andrew-mc [135]1 year ago
4 0
Kinetic energy =  \frac{mass*velocity ^{2} }{2} =  \frac{40*2 ^{2} }{2} = 80J. Therefore kinetic energy of the box while on the conveyor belt is 80J. 
dlinn [17]1 year ago
4 0

Answer : The kinetic energy of the box is, 80 J

Explanation :

The formula used for kinetic energy is:

K.E=\frac{1}{2}mv^2

where,

K.E = kinetic energy = ?

m = mass = 40 kg

v = velocity = 2 m/s

Now put all the given values in the above formula, we get the kinetic energy of the box while it is on the conveyor belt.

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}\times (40kg)\times (2m/s)^2

K.E=80kg.m^2/s^2=80J

Thus, the kinetic energy of the box is, 80 J

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Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.
aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

 

6 = (20 / 32.2 + 50 / 32.2)a

 

2.173a = 6

 

A = 2.76 ft/s^2

 

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

 

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0
1 year ago
An experiment is conducted in which red light is diffracted through a single slit. Listed below are alterations made, one at a t
Xelga [282]

Answer:

B. The distance between the slits and the screen is halved.

C. The slit width is doubled.

D. A green, rather than red, light source is used.

E. The experiment is conducted in a water-filled tank.

Explanation:

As we know that the position of first minimum is given as

a sin\theta = N\lambda

so we have

\theta = sin^{-1}(\fracN\lambda}{a})

so width of minimum is given as

w = L\times sin^{-1}(\fracN\lambda}{a})

now if we need to decrease the angular position of minimum

1). so we can decrease the distance of screen from the slit

2). we can decrease the wavelength

3). We can increase the width of the slit

So correct answer will be

B. The distance between the slits and the screen is halved.

C. The slit width is doubled.

D. A green, rather than red, light source is used.

E. The experiment is conducted in a water-filled tank.

6 0
2 years ago
Your eye is designed to work in air. Surrounding it with water impairs its ability to form images. Consequently, scuba divers we
grin007 [14]

Answer:

Check the explanation

Explanation:

A) There are two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface) and the angle immediately before the second refraction (the plastic/air interface).

To find out how they relate, draw a picture with the path the light follows in the plastic and the normal to both surfaces.

Once you have labeled both angles, keep in mind that the surfaces are parallel, and thus their normal are parallel lines. An important theorem from geometry will give you the relationship between the angles.

Using Snell's Law,   θa = asin[(nw/na)*sin(θw)]

B) D = l/tan(θw)

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8 0
2 years ago
A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet
Gnom [1K]

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

6 0
2 years ago
A couch is pushed with a horizontal force of 80 N and moves the couch a
Lapatulllka [165]

Answer:

400 J

Explanation:

Work = force × distance

W = (80 N) (5 m)

W = 400 J

5 0
1 year ago
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