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gtnhenbr [62]
3 years ago
11

ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements.

Physics
1 answer:
vekshin13 years ago
6 0

Answer:

The answer is 500

Explanation:

You might be interested in
A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co
Genrish500 [490]

Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be= 147.1∘C

Explanation:

H = mcΘ

heat lost by block = heat gained by water

m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.

0.5*900*(200-20) = m₁*4186*(20-0)

m₁ = 450*180/83270

<em>m₁ = 0.973kg</em>

<em>when 1.0kg of aluminium block is used, the final temperature of the mixture will be </em><em>T</em>

heat lost by block = heat gained by water

1.0*900*(200-T) = 0.973*4186*(T-0)

180000 - 900T = 4073T

4973T = 180000

T = 180000/4973 = 36.2∘C

<em>If 1.0kg copper block is used, T of the mixture will be</em>

heat lost by block = heat gained by water

1.0*387*(200-T) = 0.973*4186*(T-0)

77400 - 387T = 4073T

4460T = 77400

T = 77400/4460 = 17.4∘C

<em>If 100g (0.1kg) of ice at 0∘C is used, T will be</em>

<em>heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C</em>

heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg

0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg

90000 - 450T =  418.6T + 33600

418.6T + 450T = 90000 - 33600

868.6T = 56400

T = 56400/868.6 = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be

0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg

90000 - 450T =  104.65T + 8400

104.65T + 450T = 90000 - 8400

554.65T = 81600

T = 81600/554.65 = 147.1∘C

7 0
3 years ago
What is the average speed of an object with a total distance traveled of 45 meters in 50 seconds
Anettt [7]

Answer: 0.9m/s

Explanation:

Use equation for speed:

V=S/t

S-distance; S=45m

t-time; t=50 s

V=S/t

V=45m/50s

V=0.9 m/s

7 0
3 years ago
1)Calculate the mass of a girl that has a weight of 113 Newtons.
Blababa [14]

Answer:

Explanation:We should know that weight = mass * gravity.

That is weight equals mass times gravity.

Gravity is a force of attraction between any two bodies in the universe. It is directly proportional to product of their masses and inversely proportional to the square of the distance between them.

Gravity is generally measured in terms of acceleration due to gravity, denoted as g. For Earth it is, 9.8 m/s². And for moon, it is about 1.62 m/s².

On Earth, your weight is 70 kg = W

W = mass x 9.8

70 = mass x 9.8

Your mass is 70/ 9.8

i.e approximately 7.14

Weight at the Moon, W' = 7.14 x 1.62

Hence, your weight on the surface of the moon is just 11.56 kg.

Congratulations, you've lost about 58.14 kilograms without any hard exercise. And you're as light as a Sweedish Vallhund! Cheers!

3 0
3 years ago
A mountain climber weighs 42.0 N. If he climbs a hill 100m high. Calculate the work done in joules​
quester [9]

Answer:

The answer is 4200 J.

Explanation:

The formula of work done is, W = F×D where F is the force of an object and D is the distance. Then you just substitute the values into the equation :

W = F×D

F = 42N

D = 100m

W = 42 × 100

= 4200 J

5 0
3 years ago
A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl
Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

4 0
3 years ago
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