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elena55 [62]
3 years ago
10

Raul dug a hole in his yard to repair a water pipe. It took him 2 seconds to apply a force of 50 Newtons to push the shovel 0.25

m into the ground.
How much power was used?
A 0.01 watts
B. 6.25 J/S
C. 25.00 N-m-s
D.100.00 watts
Physics
1 answer:
Sergeu [11.5K]3 years ago
3 0

Answer:

Option B. 6.25 J/S

Explanation:

Data obtained from the question include:

t (time) = 2secs

F (force) = 50N

d (distance) = 0.25m

P (power) =?

The power can be obtained by using the formula P = workdone/time.

P = workdone / time

P = (50 x 0.25)/ 2

P = 6.25J/s

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In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an
natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa

σ=5517.25 Pa

Strain in x direction

ε=σ/E

=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0
3 years ago
CAN SOMEONE PLEASE TELL ME WHAT THIS WHEEL IS CALLED.WILL GIVE BRAINLIEST
kolbaska11 [484]

Answer:

its a pie chart

Explanation:

.........................

3 0
3 years ago
Read 2 more answers
Which scenario is an application of Newtons Second Law of Motion?
svlad2 [7]
D would be the answer because The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
4 0
3 years ago
A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop s
skelet666 [1.2K]

Answer:

frictional force = 0.52 N

Explanation:

diameter of turn table (D1) = 30 cm = 0.3 m

mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I =  0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

frictional force = 0.52 N

6 0
4 years ago
Why does the cord of an electric heater not glow while the heating element does​
Strike441 [17]

Answer: Why does the cord of an electric heater not glow while the heating element does? The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminum is very low so it does not glow.

Explanation:

7 0
3 years ago
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