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OverLord2011 [107]
3 years ago
13

4. 2075 Q.No. 1 Name any two physical quantities which have same dimension. ​

Physics
1 answer:
vivado [14]3 years ago
4 0

Answer:

acceleration,joule,newton,watt

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If two parallel conductors are both free to move and are carrying current in the same direction, what would happen if the curren
pishuonlain [190]

The wires would remain attracted to each other.

Option D.

Explanation:

It is known that magnetic flux will be generated in conductors with varying emf. So when current is flowing in two parallel conductors, the magnetic flux will be generated in those wires. If the current is flowing in same direction in both the wires, then the magnetic flux will be generated towards inside and outside the wires. Thus, the wire will get attracted to each other till the time the current is flowing in the same direction in both the wires. So if the current flow in each wire was reversed at the same time, then the wire would remain attracted to each other.

8 0
3 years ago
A motorbike accelerates from 15m/s to 25m/s in 15 seconds.
RSB [31]

distance traveled by a uniformly accelerated bike is given as

d = \frac{v_f + v_i}{2} (t)

here we know that

v_f = 25 m/s

v_i = 15 m/s

t = 15 s

now we will have from above equation

d = \frac{15 + 25}{2} (15)

d = 20 (15) = 300 m

so it will cover the total distance of 300 m

5 0
4 years ago
61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
Mademuasel [1]

Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

6 0
3 years ago
16. Cassandra notices that when she breathes on a cool window, the water vapor in her breath forms
choli [55]

Answer: A

Explanation:I studied .

7 0
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Find the temperature
NNADVOKAT [17]

Answer:

-------1

Explanation:

beacuse that is what i know

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