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leva [86]
3 years ago
11

How much work did the movers do (horizontally) pushing a 41.0-kg crate 10.6 m across a rough floor without acceleration, if the

effective coefficient of friction was 0.60?

Physics
1 answer:
VLD [36.1K]3 years ago
5 0

Answer:

The required work done is 2555.448~J

Explanation:

Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '\mu_{k}' be the coefficient of friction, then

f = \mu_{k} \times N = \mu_{k} \times Mg

where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.

Since the application of force by the movers does not create any acceleration to the block, we can write

F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N

So the work done (W) in moving the crate by a distance s = 10.6 m is

W = F \times s = 241.08~N \times 10.6~m = 2555.448 J

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You hang a heavy ball with a mass of 30 kg from a tungsten rod 2.8 m long by 1.5 mm by 2.6 mm. You measure the stretch of the ro
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Answer:

Young's modulus (Y) = 3.56×10^11 N/m^2

The speed of sound in tungsten = 6166.4 m/s

Explanation:

Young's modulus (Y) = stress/strain

Stress = force/area

Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

From the equations of motion

v^2 = u^2 + 2gh =

Initial speed (u) = 0 m/s

v = sqrt (2×9.8×1.94×10^6)

v = 6166.4 m/s

7 0
3 years ago
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Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and
Liono4ka [1.6K]

a) 5 \Omega, 1.6 A

b) 6 \Omega, 1.33 A

Explanation:

a)

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

R=R_1+R_2

where

R_1=4\Omega

R_2=1 \Omega

Therefore, the equivalent resistance is

R=4+1=5 \Omega

Now we can use Ohm's Law to find the current flowing through the circuit:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=5\Omega is the equivalent resistance of the circuit

Substituting,

I=\frac{8}{5}=1.6 A

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

b)

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

R=R_1+R_2+R_3

where:

R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega

Substituting, we find the equivalent resistance:

R=4+1+1=6 \Omega

Now we can find the current through the circuit by using again Ohm's Law:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=6\Omega is the equivalent resistance

Substituting,

I=\frac{8}{6}=1.33 A

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

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3 years ago
An inventive child named Nick wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rop
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Because their blood cells could be transfused to any blood type
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Answer:

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