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3 years ago

If a truck travels 100 miles in 2 hours, what is its speed?

Physics

2 answers:

densk [106]3 years ago

6 0

Answer:

50 miles/hr

Explanation:

$Distance = 100\: miles\\Time = 2\:hours\\\\Speed = \frac{Distance}{Time} \\\\Speed = \frac{100\:miles}{2\:hours}\\\\ Speed = 50\:miles/hour$

miskamm [114]3 years ago

5 0

Answer:50 miles per hour 50/1hr

Explanation:100 divided by 2 is 50, divide 2 by 2 thats 1

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What requirement must a force acting on a object satisfy in order for the object to undergo simple harmonic motion?

viktelen [127]

Answer:

Simple harmonic motion is the movement of a body or an object to and from an equilibrium position. In a simple harmonic motion, the maximum displacement (also called the amplitude) on one side of the equilibrium position is equal to the maximum displacement.

The force acting on an object must satisfy Hooke's law for the object to undergo simple harmonic motion. The law states that the force must be directed always towards the equilibrium position and also directly proportional to the distance from this position.

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3 years ago

What is the speed of an object that travels 60 metres in 4 seconds

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What happens when you boil seawater(no copying)

geniusboy [140]

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3 years ago

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In doing a load of clothes, a clothes dryer uses 16 A of current at 240 V for 45 min. A personal computer, in contrast, uses 2.7

EastWind [94]

Answer:

(a) Total time will be 8.05 hour

(B) Total cost will be $417.6

Explanation:

We have given dryer uses 16 A of current so current i = 16 A

Voltage V = 240 volt

Time t = 45 minutes =45×60 = 2700 sec

So power P = VI = 240×16 = 3840 W

So energy E = power×time = 3840×45×60 = 9396 KJ

(A) Now current in the computer i = 2.7 A

Voltage V = 120 V

So power P = 120×2.7 = 324 W

Now computer is using energy of dryer

So time by which it will use energy is $t=\frac{9396000}{324}=29000sec=8.05hour$

So time will be 8.05 hour

(b) Cost per kilowatt is $0.12

So total cost =3480×0.12 = $417.6

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3 years ago

A horizontal force of magnitude 46.3 n pushes a block of mass 4.14 kg across a floor where the coefficient of kinetic friction i

IrinaVladis [17]

A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
$W=Fd = (46.3 N)(4.25 m)=196.8 J$

b) The work done by the frictional force against the motion of the block is equal to:
$W_f = -F_f d =- (\mu mg) d =-(0.609)(4.14 kg)(9.81 m/s^2)(4.25 m)=$
$=-105.1 J$
Part of these 105.1 Joules of work becomes increase of thermal energy of the block ( $\Delta E_B$ ), and part of it becomes increase of thermal energy of the floor ( $\Delta E_F$ ). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
$\Delta E_F = 105.1 J - \Delta E_B = 105.1 J-38.2 J=66.9 J$

c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
$W_{net}=W-W_f=196.8 J-105.1 J=91.7 J$
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
$W_{net} = \Delta K$
So, we have $\Delta K=+91.7 J$

5 0

3 years ago