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2 years ago

Which of the following is required for work to be done on an object?

Physics

1 answer:

ad-work [718]2 years ago

7 0

I think that in order for work to be done, the object must move in the direction of the force and move over a distance.

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A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–d

Ad libitum [116K]

Answer:

Option (D) is correct.

Explanation:

The balloon lands horizontally at a distance of 420 m from a point where it as released.

Velocity of air balloon along +X axis =10 m/s

velocity of ball=4 m/s along + X axis

the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s

time taken= 30 s

The distance traveled is given by d= v t

d= 14 (30)

d= 420 m

Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.

6 0

3 years ago

A cheetah, which is the fastest land mammal, can accelerate at 9 m/s from rest to in 31 m/s. How long does

Anna007 [38]

31 m/s ÷ 9 m/s² = 3.44 s

Time = Change in velocity divided (÷) by acceleration.

8 0

3 years ago

The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. Determine the amount o

kotykmax [81]

Answer:

time taken is equal to 14,156 years

Explanation:

we know,

$Y=Ae^{-kt}$

at t = 0

Y(0) = A

given that half life of plutonium 239 = 24,200

$\dfrac{A}{2}=Ae^{-kt}\\0.5=e^{-kt}\\k\times 24200 = ln(2)\\k = \dfrac{ ln(2)}{24200}$

$Y=Ae^{-kt}$

$\frac{3}{2} = e^{-kt}\\ln(1.5)=-\dfrac{ ln(2)}{24200}\times t\\t=-\dfrac{ln(1.5)\times 24200}{ ln(2)}\\t=14,156 \ years$

hence time taken is equal to 14,156 years

5 0

3 years ago

A uniform metre rod of 5 metre length is suspended horizontally by two strings P and Q.Pis attached 0.8 metre from one end and Q

denpristay [2]

Answer:

The tension in string P is 25 N, while that of Q is 85 N.

Explanation:

Considering the conditions for equilibrium,

i. Total upward force = Total downward force

$T_{P}$ + $T_{Q}$ = 110 N

ii. Taking moment about P,

clockwise moment = anticlockwise moment

110 × (2.5 - 0.8) = $T_{Q}$ × (3 - 0.8)

110 × 1.7 = $T_{Q}$ × 2.2

187 = 2.2 $T_{Q}$

$T_{Q}$ = $\frac{187}{2.2}$

$T_{Q}$ = 85 N

From the first condition,

$T_{P}$ + $T_{Q}$ = 110 N

$T_{P}$ + 85 N = 110 N

$T_{P}$ = 110 - 85

$T_{P}$ 25 N

Therefore, the tension in string P is 25 N while that of Q is 85 N.

5 0

3 years ago

Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w

ioda

Answer:

$Q_{week} = 458884.6\, BTU$

Explanation:

The weekly water consumption of Kimonoski is:

$m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)$

$m_{bath.week} = 948.205\,lbm$

$m_{others, week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (23\,gal)\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)$

$m_{others, week} = 1346.218\,lbm$

$m_{week} = m_{bath,week} + m_{others, week}$

$m_{week} = 2294.423\,lbm$

The total energy required per week for hot water is:

$Q_{week} = m_{week}\cdot c_{p,water}\cdot \Delta T$

$Q_{week} =(2294.423\,lbm)\cdot (1\,\frac{BTU}{lbm\cdot ^{\textdegree}F} )\cdot (50^{\textdegree}F)$

$Q_{week} = 458884.6\, BTU$

3 0

2 years ago