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antoniya [11.8K]
2 years ago
15

An angle grinder is best suited for use with which material?.

Engineering
1 answer:
Kaylis [27]2 years ago
4 0

Answer:

Angle grinders are used mostly for copper, iron, steel, lead, and other metals.

Explanation:

I hope it helps! Have a great day!

Lilac~

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A 4-kW electric heater runs for 2 hours to raise the room temperature to the desired level. Determine the amount of electric ene
Anna35 [415]

Answer:

Q' = 8 KW.h

Q'=28800 KJ

Explanation:

Given that

Heat Q= 4 KW

time ,t = 2 hours

The amount of energy used in KWh given as

Q ' = Q x t

Q' = 4 x  2 KW.h

Q' = 8 KW.h

We know that

1 h = 60 min = 60 x 60 s  = 3600 s

We know that W  = 1 J/s

The amount of energy used in KJ given as

Q' = 8 x 3600 = 28800 KJ

Therefore

Q' = 8 KW.h

Q'=28800 KJ

6 0
3 years ago
A 75-kg piano is hosted on a crane and delivered throughout the window of a 6th story apartment (20 meters above ground). What i
Sonja [21]

Answer:

  14,700 J

Explanation:

PE = Mgh = (75 kg)(9.8 m/s²)(20 m) = 14,700 J

8 0
2 years ago
Luợng hơi nước có lưu lượng 100kg/s ở áp suất đầu p;= 20bar, nhiệt độ đầu t; =400°C giãn nở đoạn nhiệt trong tuabin đến độ khô X
Valentin [98]

Answer:

Explanation:

.

3 0
3 years ago
The power supply converts the wall outlet AC power into DC power. T or F
Alika [10]

Answer:

True

Explanation:

All computer parts require DC power to operate, and wall outlets provide AC Power.

7 0
3 years ago
2. The block is released from rest at the position shown, figure 1. The coefficient of
denis23 [38]

Answer:

Velocity = 4.73 m/s.

Explanation:

Work done by friction is;

W_f = frictional force × displacement

So; W_f = Ff * Δs = (μF_n)*Δs

where; magnitude of the normal force F_n is equal to the component of the weight perpendicular to the ramp i.e; F_n = mg*cos 24

Over the distance ab, Potential Energy change mgΔh transforms into a change in Kinetic energy and the work of friction, so;

mg(3 sin 24) = ΔKE1 + (0.22)*(mg cos 24) *(3).

Similarly, Over the distance bc, potential energy mg(2 sin 24) transforms to;

ΔKE2 + (0.16)(mg cos 24)(2).

Plugging in the relevant values, we have;

1.22mg = ΔKE1 + 0.603mg

ΔKE1 = 1.22mg - 0.603mg

ΔKE1 = 0.617mg

Also,

0.813mg = ΔKE2 + 0.292mg

ΔKE2 = 0.813mg - 0.292mg

ΔKE2 = 0.521mg

Now total increase in Kinetic Energy is ΔKE1 + ΔKE2

Thus,

Total increase in kinetic energy = 0.617mg + 0.521m = 1.138mg

Putting 9.81 for g to give;

Total increase in kinetic energy = 11.164m

Finally, if v = 0 m/s at point a, then at point c, KE = ½mv² = 11.164m

m cancels out to give; ½v² = 11.164

v² = 2 × 11.164

v² = 22.328

v = √22.328

v = 4.73 m/s.

5 0
3 years ago
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