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trapecia [35]
3 years ago
13

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.2 m. The mug s

lides off the counter and strikes the floor 1.35 m from the base of the counter. Assume the mug moves in the positive x-direction. Give your answer in three significant figures.
a) With what velocity did the mug leave the counter?
b) What was the direction (in degrees and with respect to the horizontal) of the mug's velocity just before it hit the floor?
Physics
1 answer:
ZanzabumX [31]3 years ago
6 0

Part a can be solve using the equation of trajectory:

Y = x tana + (g*x^2)/ [2(V0^2)*(cos a)^2]

Where y is the height

X is the length

G is the acceleration due to gravity

Vo Is the initail velocity

a is the angle of trajectory

1.2 = 1.35  tan(0) + (9.81*1.35^2)/ [2(V0^2)*(cos 0)^2]

Solve for V0 = 2.729 m/s

b. can be solve using the formula

v = sqrt(2gy)

= sqrt ( 2*1.2*9.81)

= 4.852 m/s  going down ( 0 degree from the horizontal)

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A block of mass 10 kg and measuring 250 mm on each edge is pulled up an inclined surface on which there is a film of SAE 10W-30
Ivan

Answer:

a) 2.53 * 10^-2 m/s

b) -4.78 * 10^-2 m/s

c)  1.21 * 10^-1 m/s

Explanation:

Given data :

Mass of block = 10 kg

Measuring 250mm on each side

a) calculate the speed  when a force of 75N is applied to pull block upwards

F = f + W sin∅ ( equation for applying the force of equilibrium condition in the x axis )  ----- ( 1 )

f ( friction force )=  ( 16400v * 6.25 *10^-2) =  1025 v

F ( force applied ) = 75

W ( weight of  block ) = 10 * 9.81 = 98.1 N

∅ = 30°

input values into equation 1

V = \frac{75- (98.1*sin30^{0}) }{1025} = 2.53 * 10^-2 m/s

b) Speed when no force is applied on the block

F = f + W sin∅

F = 0

f = 1025 V

W = 98.1 N

∅ = 30°

hence V = \frac{0 - (98.1*sin30^{0}) }{1025} =  - 4.78 * 10^-2 m/s

c) when a force is applied to push block down the incline

F = f + W sin∅ ----- ( 3 )

F = 75 N

f = 1025 V

W = 98.1 N

∅ = 30°

input values into equation 3 considering the fact that the weight of the block is acting in the opposite direction

75 = 1025 V - 98.1 ( sin 30° )

V = \frac{75+( 98.1*sin30^{0})  }{1025} = 1.21 * 10^-1 m/s

5 0
3 years ago
How much work does and elephant do pulling a circus wagon 1000m with<br> an applied force of 200 N?
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Answer:

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NEED ASAP, WILL GIVE BRAINLIEST ANSWER
Brrunno [24]
12kN is equal to 12000 N
and the perpendicular component will be
[email protected]
mg = 12000 N
cos10°= 0.984
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sin10° = 0.1736
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