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QveST [7]
4 years ago
7

An aluminium bar 600mm long1 with diameter 40mm, has a hole drilled in the centre of the bar. The hole is 30mm in diameter and i

s 100mm long. If the modulus of elasticity for the aluminium is 85GN/M^2 Calculate the total contraction on the bar due to a compressive load of 180KN

Physics
1 answer:
nexus9112 [7]4 years ago
6 0

Answer:

Total contraction = 1.2277 mm

Explanation:

the solution is given in the picture below and it is more explanatory

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What is the first step in cell division?
ohaa [14]

Answer:

d

Explanation:

4 0
3 years ago
a transformer has 1,400 turns in the primary coil and 100 turns in the secondary coil. if the primary coil is connected to a 120
harkovskaia [24]

0.56 A is the current of the secondary coil.

transformer is the device that transfer electric energy from one AC circuit to another.

we know that ;

I ∝ 1/n;

where I IS CURRENT and n is NUMBER OF TURNS.

(I) secondary = X A

(n) secondary = 100

(I)  primary  = 0.04 A

(n)  primary  = 1400

USING THIS-----> I ∝ 1/n;

=>

0.04 A/X A =100/1400

=>X=0.56 A

as the current in the transformer is inversely proportional to the number of turns so 0.56 A is the current in the secondary coil.

learn more about transformers-

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4 0
1 year ago
Yo, help will give brainliest
kaheart [24]

Answer:

30

Explanation:

30

6 0
3 years ago
What is the mass of a rock lifted 2 meters off the ground that has 196 J of potential energy?
eimsori [14]

Answer:

10kg

Explanation:

Let PE=potential energy

PE=196J

g(gravitational force)=9.8m/s^2

h(change in height)=2m

m=?

PE=m*g*(change in h)

196=m*9.8*2

m=10kg

4 0
3 years ago
When you skid to a stop on your bike, you can significantly heat the small patch of tire that rubs against the road surface. Sup
Wittaler [7]

Answer:

W_f = 148.17J

Explanation:

During the exchange of applied force, thermal energy is generated by the friction that exists between the ground and the tire.

Said force according to the statement is the reaction of half the force on the rear tire. In this way the normal force acted is,

N=\frac{mg}{2} = \frac{90*9.8}{2} = 441N

The work done is given by the friction force and the distance traveled,

W_f = fd = \mu_k Nd

Where \mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)

The thermal energy released through the work done is,

W_f = 148.17J

3 0
4 years ago
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