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oksian1 [2.3K]
3 years ago
7

ametal of mass 0.6kg is heated by an electric heater connected to 15v batter when the ammeter reading is 3A its tempeeature rise

s feom 20c to 85c in 10 minutes calculate the s.h.c of metal cylinder​
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
6 0

Answer:

692 J/kg/°C

Explanation:

Electric energy added = amount of heat

Power × time = mass × SHC × increase in temperature

Pt = mCΔT

(15 V × 3 A) (10 min × 60 s/min) = (0.6 kg) C (85°C − 20°C)

C = 692 J/kg/°C

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mel-nik [20]

Answer:

option C is correct

Explanation:

8 0
3 years ago
How much work would a child do while puling a 12-kg wagon a distance of 3m with a 22 N force directed 30 degrees with respect to
Alika [10]

Answer:

The work done will be 57.15 J

Explanation:

Given that,

Mass = 12 kg

Distance = 3 m

Force = 22 N

Angle = 30°

We need to calculate the work done  

The work done is defined as,

W = Fd\cos\theta

Where, F = force

d = displacement

Put the value into the formula

W=22\times3\times\cos30^{\circ}

W=22\times3\times\dfrac{\sqrt{3}}{2}

W = 57.15\ J

Hence, The work done will be 57.15 J

6 0
3 years ago
two balls are dropped to the ground from the different heights. one is dropped 1.5s after the other, but they reach the ground a
SCORPION-xisa [38]

Answer:

h = 120 m

h = 60 m

when rounded to two significant digits.

Explanation:

height of the higher ball drop point

h = ½gt² = ½(9.8)5.0² = 122.5 m

lower ball drop point

h = ½(9.8)(5.0 - 1.5)² = 60.025

8 0
2 years ago
Blaine steps onto a ski slope with an angle of 25°. There is a coefficient of kinetic friction of 0.15 between him and the groun
Ymorist [56]

θ = angle of the incline surface from the horizontal surface = 25⁰

μ = Coefficient of friction = 0.15

m = mass of the person = 65 kg

f_{k} = kinetic frictional force acting on the person as he slides down

mg = weight of the person acting in down direction

F_{n} = normal force by the incline surface on the person

the free body diagram showing the forces acting on the person is given as


3 0
2 years ago
1. Potential energy and a roller coaster. A 1000-kg roller coaster car moves from point A to B then C.
zalisa [80]

Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,

          U

B

​

=0

When the car is at point A, the potential energy of the car-Earth system is given by

          U

A

​

=mgy

where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:

y=(41.1m)sin40.0

0

=26.4m

Thus,

U

A

​

=(1000kg)(9.80m/s

2

)(26.4m)=2.59∗10

5

J

The change in potential energy of the car-Earth system as the car moves from A to B is

U

B

​

−U

A

​

=0−2.59∗10

5

J=−2.59∗10

5

J

(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U

A

​

=0. The potential energy of the system when the car is at point B is given by U

B

​

=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.

Thus,

8 0
2 years ago
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