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Shalnov [3]
3 years ago
10

If a 42kg rolling object slows from 11.5m/s to 3.33m/s how much work did friction do

Physics
1 answer:
geniusboy [140]3 years ago
8 0

Answer: 1608.39 J

Explanation: Given that the

mass M = 42kg

U = 11.5m/s

V = 3.33m/s

how much work did friction do

Work done = Force × distance

Work done = Ma × distance

But acceleration a = V/t

Work done = M × V/t × d

Work done = M × V × d/t

Where d/t = velocity

Therefore,

Work done = M × U × V

Work done = 42 × 11.5 × 3.33

Work done = 1608.39 J

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A girl is riding her bicycle. She rides 4 miles in 2.5 hours. What was her average speed?
xz_007 [3.2K]

Explanation:

4/2.5 = 1.6 miles/hour

4 0
3 years ago
If you had started with a larger mass, how would the half-life change?
iogann1982 [59]

Answer:

There is no change, unless your mass is somehow at the quantum level, at which the concept of half-life breaks down.

Half life is a property of the specific radioactive isotope...NOT of the initial sample's mass.

3 0
3 years ago
Jessie and Jaime complete a 5.0 km race. Each has a mass of 68 kg. Jessie runs the race at 15 km/h; Jaime walks it at 5 km/h. Ho
Leto [7]

The total metabolic energy used by each to complete the course is determined as 656.91 J.

<h3>Kinetic energy of Jessie and Jaime</h3>

The kinetic energy of Jessie and Jaime is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of Jaime
  • v is speed

15 km/h = 4.17 m/s

5 km/h = 1.39 m/s

K.E = ¹/₂(68)(4.17)² + ¹/₂(68)(1.39)²

K.E = 656.91 J

Thus, the total metabolic energy used by each to complete the course is determined as 656.91 J.

Learn more about kinetic energy here: brainly.com/question/25959744

5 0
2 years ago
The density for gold is 19.3 g/cm3. What would be the mass of a 45 cm3 piece of gold?
lyudmila [28]

Answer:

868.5 g

Explanation:

Mass= Density x Volume

Mass= 19.3 x 45

=868.5

6 0
3 years ago
Read 2 more answers
An engineer is investigating energy loss through windows. The windowpane of interest is 0.650 cm thick, has dimensions of 1.19 m
krok68 [10]

Answer:

7908.92307 W

683330953.248 J

Explanation:

k = Heat conduction coefficient = 0.8 W/(m·°C)

A = Area = 1.19\times 2.25\ m^2

l = Thickness = 0.65 cm

T_2 = 24°C

T_1 = 0°C

Rate of heat transfer is given by

Q=\frac{kA(T_2-T_1)}{l}\\\Rightarrow Q=\frac{0.8\times 1.19\times 2.25(24-0)}{0.65\times 10^{-2}}\\\Rightarrow Q=7908.92307\ W

The rate of heat transfer is 7908.92307 W

Amount of energy is given by

E=Qt\\\Rightarrow E=7908.92307\times 24\times 3600\\\Rightarrow E=683330953.248\ J

The energy transferred through the window in one day is 683330953.248 J

8 0
3 years ago
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