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Stolb23 [73]
2 years ago
7

30. The length of mercury thread when it is at 0°C, 100°C and at an unknown temperature 0 is 25mm, 225mm and 175mm respectively.

What is the value of 02 A. 85.0°C B. 80.0°C C. 75.0°C D. 70.0°C ​
Physics
1 answer:
yulyashka [42]2 years ago
3 0

Answer:

75°C

Explanation:

175−25/225−25=x−0/100−0

150/200=x/100

x=150×100/200

= 75°C

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A 1.5m wire carries a 6 A current when a potential difference of 68 V is applied. What is the resistance of the wire?
ANTONII [103]

Answer:

11.3 \Omega

Explanation:

We can find the resistance of the wire by using Ohm's law:

V=RI

where

V is the voltage applied

R is the resistance

I is the current

In this problem, we know I = 6 A and V = 68 V, so we can re-arrange the equation to find the resistance of the wire:

R=\frac{V}{I}=\frac{68 V}{6 A}=11.3 \Omega

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5 0
2 years ago
A girl drops a stone from the top a tower 45m tall. At the same time, a boy standing at the base of the tower, projects another
Advocard [28]

Answer:

(i) The stones meet at 1.8 second

(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

Explanation:

(i)

First we consider the stone dropped by the girl. We have data:

Vi = Initial Velocity of Stone = 0 m/s   (Since the stone was initially at rest)

t = Time Period

g = 10 m/s²

s₁ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₁ = Vi t + (0.5)gt²

s₁ = (0)(t) + (0.5)(10)t²

s₁ = 5t²   ----- equation (1)

Now, we consider the stone throne vertically upward by the boy. We have data:

Vi = Initial Velocity of Stone = 25 m/s

t = Time Period

g = - 10 m/s²   (negative sign due to upward motion)

s₂ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₂ = Vi t + (0.5)gt²

s₂ = (25)(t) + (0.5)(-10)t²

s₂ = 25t - 5t²   ----- equation (2)

At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).

s₁ + s₂ = 45

using values from equation (1) and equation (2)

5t² + 25t - 5t² = 45

25t = 45

t = 45/25

<u>t =  1.8 sec</u>

(ii)

using this value of of t in equation (2)

s₂ = (25)(1.8) - (5)(1.8)²

<u>s₂ = 28.8 m</u>

using this value of of t in equation (1)

s₁ = (5)(1.8)²

<u>s₁ = 16.2 m</u>

<u>Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.</u>

6 0
3 years ago
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