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2 years ago

A force of 150N is applied to a sphere which rolls 0 m. How much work has been done? Include the unit.

Physics

1 answer:

Serggg [28]2 years ago

3 0

Answer:

0 J

Explanation:

Because W=F*D, and the sphere's total distance is 0m, the work done is automatically 0 J.

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Which meteorites are the most useful for defining the age of the solar system? Why?

scoundrel [369]

Answer:

meteorite is a piece of interplanetary debris that lives its fiery drops during a through the earth's atmosphere and strikes the surface of the earth.

Explanation:

the meteorites which are most useful for the determination of the age of the solar system are the primitive meteorites. they consist light of colored or grey silicates mixed with metallic grains. the parent bodies of these meteorites are also mostly believed to be pieces asteriods left after they formed in the solar system.

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3 years ago

1. What is evaporation and how does it affect weather?

alina1380 [7]

Answer:

Explanation:don’t know

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3 years ago

A child pulls a sled up a snow covered hill. If the child does 504J of work on the sled while pulling the sled 23m up the hill t

zhenek [66]

Explanation:

ans is equal to 504j* 23 m* 10 ms

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2 years ago

If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m

algol [13]

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

$H=ut+\dfrac{1}{2}at^2$

Here, a = g

$H=\dfrac{1}{2}gt^2$

$H=\dfrac{1}{2}\times 9.8\times (2.261)^2$

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

7 0

2 years ago

Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter

stellarik [79]

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

$m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a$

where,

$m_c$ = mass of copper = 227 g

$m_w$ = mass of water = 844 g

$m_a$ = mass of aluminum = 155 g

$C_c$ = specific heat capacity of calorimeter = 385 J/kg.°C

$C_w$ = specific heat capacity of water = 4200 J/kg.°C

$C_a$ = specific heat capacity of aluminum = 890 J/kg.°C

$\Delta T_c$ = change in temperature of copper = 283°C - T

$\Delta T_w$ = change in temperature of water = T - 14.6°C

$\Delta T_a$ = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

$(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}$

8 0

2 years ago