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pickupchik [31]
1 year ago
8

An aluminum block of mass 12.00 kg is heated from 20 c to 118 c. if the specific heat of aluminum is 913 j-1 kg k-1 then how muc

h energy is required?
Physics
1 answer:
Jobisdone [24]1 year ago
6 0

Answer:

107 kJ

Explanation:

12.00 kg * 913 J /kg-C * (118-20)°C        ( I corrected spec heat units)

                    ( a  C degree change is the same as a kelvin change)

                   = 1 073 688 J   = 107 kJ   ( three sig digits)

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A child (approximately 4 years old) takes her metal “slinky Toy” (a flexible coiled metal spring) and does various tests to dete
anzhelika [568]

Answer:

248

Explanation:

L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H

l = length of the slinky = 3 m

N = number of turns in the slinky

r = radius of slinky = 4 cm = 0.04 m

Area of slinky is given as

A = πr²

A = (3.14) (0.04)²

A = 0.005024 m²

Inductance is given as

L = \frac{\mu _{o}N^{2}A}{l}

130\times 10^{-6} = \frac{(12.56\times 10^{-7})N^{2}(0.005024)}{3}

N = 248

8 0
3 years ago
Describe how you could measure the thickness of paper with an ordinary ruler
Svetllana [295]
Measure a whole stack (one in which you know the number of sheets), then divide your measurement by the number of sheets in that stack
5 0
4 years ago
When motion is measured a starting point is called ?
AleksAgata [21]
The starting point for measuring motion is called : The Reference Point.


5 0
3 years ago
In 2014, about how far in meters would you have to travel on the surface of the Earth from the North Magnetic Pole to the Geogra
pogonyaev

Answer:

267.07 km

Explanation:

We have given the radius of the earth = 6378.1 km

In 2014 the difference between the magnetic north pole and geographical north pole is 2.40°

2.40°=2.40\times \frac{\pi }{180}=0.041866 radian

We know that linear distance is given by S=R\Theta =6378.1\times 0.041866=267.07km

So we have to travel 267.07 km in going from magnetic north pole to geographic north pole

3 0
3 years ago
A coin dropped in the lift it takes time 0.5 s to reach the floor when lift is staionary it takes time t when lift is moving up
BARSIC [14]

Answer:

t₁ > t₂

Explanation:

A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂,  t₁ = t₂,  t₁ >> t₂ ,  t₂ > t₁

Solution:

Newton's law of motion is given by:

s = ut + (1/2)gt²;

where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.

u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.

hence:

When stationary:

s=\frac{1}{2}gt_1^2\\\\t_1^2=\frac{2s}{g}  \\\\When\ moving\ with\ acceleration(a):\\\\s=\frac{1}{2}(g+a)t_2^2\\\\t_2^2=\frac{2s}{g+a}

Hence t₂ < t₁, this means that t₁ > t₂.

4 0
3 years ago
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