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Novosadov [1.4K]
4 years ago
8

A bag of ice cubes absorbs 149,000 J of heat, which causes its temperature to increase by 5.23 degrees celsius. what is the mass

of the ice in the bag?
Ice=c=2000J/(kg*c) (unit=kg)
Physics
1 answer:
Dima020 [189]4 years ago
6 0

Answer:

14.3kg

Explanation:

Given parameters:

Quantity of heat = 149000J

Change in temperature = 5.23°C

specific heat of the ice = 2000J/kg°C

Unknown:

Mass of the ice in the bag = ?

Solution:

The heat capacity of a substance is given as:

            H = m c Ф

H is the heat capacity

m is the mass

c is the specific heat

Ф is the temperature change;

  since m is the unknown, we make it the subject of the expression;

                    m = H/ mФ

                  m = \frac{149000}{2000 x 5.23}  = 14.3kg

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Answer:

313.6 m

Explanation:

From the question given above, the following data were obtained:

Time (t) = 8 s

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The height at which the package was dropped can be obtained as follow:

h = ½gt²

h = ½ × 9.8 × 8²

h = 4.9 × 64

h = 313.6 m

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Answer:

The magnetic field is doubled.

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When the current doubles, the magnetic field produced by the wire is also  doubled.

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3 years ago
A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0
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x(t)=v_0 \cos \alpha t
y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2
where
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The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
0=h-v_0 \sin \alpha t -  \frac{1}{2}gt^2
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c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that 
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