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guapka [62]
4 years ago
15

-What is the potential different drop across an electric hot plate that draws 5A when its hot resistance?​

Physics
1 answer:
Bumek [7]4 years ago
7 0

Answer:

Potential difference across the plate is 25 R

Explanation:

Let the resistance of he hot plate is R

Current flowing in the plate is 5 A

We have to find the potential difference across the hot plate.

Potential difference across the plate is given by V=i^2R, here i is current and R is resistance

Therefore potential difference across plate

V=5^2\times R=25R

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A student throws a 0.160-kg ball straight upwards to a height of 4.70 m. How much work did the student do?
IgorC [24]

Answer:

7.4 J

Explanation:

m = mass of the ball = 0.160 kg

h = height gained by the ball = 4.70 m

g = acceleration due to gravity = 9.8 m/s²

U = Potential energy gained by the ball

Potential energy gained by the ball is given as

U = mgh

W = work done by student

Using conservation of energy

W = U

W = mgh

W = (0.160) (9.8) (4.70)

W = 7.4 J

7 0
3 years ago
Can someone plz check if what I typed is good :)
pochemuha

Answer:

Explanation:

6 0
3 years ago
What is the Doppler Effect?
pishuonlain [190]
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8 0
3 years ago
1. A force of 60 Newtons is applied to the end of a wrench 0.12 m long. How much torque is produced?
Advocard [28]

Answer:

Explanation:

Torque is given as = r × F

Then,

τ = rF sin θ

Force(F)=60N

r=0.12m

Then,

τ = rF sin θ

If it is applied at the end of the wrench it is applied at 90°

Then,

τ = rF sin θ

τ = 0.12×60×sin90

τ = 7.2Nm

6 0
4 years ago
Read 2 more answers
A 0.300 kg oscillator has a speed of 98.4 cm/s when its displacement is 4.00 cm and 81.4 cm/s when its displacement is 6.00 cm.
vovangra [49]

Answer:

v_{max}=1.101\ m.s^{-1}

Explanation:

Given:

  • mass of the oscillator, m=0.3\ kg
  • first case of displacement, x_1=0.04\ m
  • velocity in the first case, v_1=0.984\ m.s^{-1}
  • second case of displacement, x_2=0.06\ m
  • velocity in the second case, v_2=0.814\ m.s^{-1}

<u>Now as we know that the total energy in both the cases will remain conserved:</u>

TE_1=T_2

KE_1+PE_1=KE_2+PE_2

\frac{1}{2} \times m.v_1^2+\frac{1}{2} \times k.x_1^2=\frac{1}{2} \times m.v_2^2+\frac{1}{2} \times k.x_2^2

\frac{1}{2} \times 0.3\times 0.984^2+\frac{1}{2} \times k\times 0.04^2=\frac{1}{2} \times 0.3\times 0.814^2+\frac{1}{2} \times k\times 0.06^2

0.1452+k\times 0.0008=0.0994+k\times 0.0018

k=45.8\ N.m^{-1}

Now the total energy:

TE_1=\frac{1}{2} \times 0.3\times 0.984^2+\frac{1}{2} \times k\times 0.04^2

TE_1=\frac{1}{2} \times 0.3\times 0.984^2+\frac{1}{2} \times 45.8\times 0.04^2

TE_1=0.1819\ J

When the whole of the spring potential converts into kinetic energy:

TE_1=\frac{1}{2} m.v_{max}^2

0.1819=\frac{1}{2} \times 0.3\times v_{max}^2

v_{max}=1.101\ m.s^{-1}

7 0
3 years ago
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