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artcher [175]
3 years ago
12

If an object has a mass of 1417g and it is moved 47 meters in 90 seconds, how much power was used?

Physics
1 answer:
astraxan [27]3 years ago
3 0

Mass of the object is given as

m = 1417 g = 1.417 kg

now the speed of object is given as

v = \frac{d}{t}

here we know that

d = 47 m

t = 90 s

now we will have

v = \frac{47}{90} = 0.52 m/s

now we will have kinetic energy of the object as

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(1.417)(0.52)^2

KE = 0.19 J

now the power is defined as rate of energy

so here we can find power as

P = \frac{KE}{t}

P = \frac{0.19}{90} = 2.14\times 10^{-3} W

so above is the power used for the object

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Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc
Lilit [14]

Answer:

Explanation:

Image of distant object will be made at far point or at 52.5 so

object distance u = infinity

image distance v = - 52.5 cm

focal length required = f

Lens formula

1 / v - 1 / u = 1 / f

1 /  - 52.5 - 0 = 1 / f

f =  -52.5 cm

= -.525 m

Power P = 1 / f = -  1 / .525

= -  1.90

now , for eye with glass we shall find new near point .

v = ?

u = - 17.2 cm

f = -  52.5 cm

1 / v - 1 / u = 1 / f

  1 / v + 1 / 17.2 = -  1 / 52.5

1 / v  = - 1 / 17.2 -    1 / 52.5

= - .05813 -  .019

= - .07713

u = - 12.96 cm

so new near point will be 12.96 cm

5 0
3 years ago
A 3.00 kg object is fastened to a light spring, with the intervening cord passing over a pulley. The pulley is frictionless, and
finlep [7]

Answer:

Part a)

k = 588.6 N/m

Part b)

v = 0.7 m/s

Explanation:

As we know that initially block is at rest

now if block is released from rest then it will go down by 10 cm and again comes to rest

so here we have

Part a)

Work done by gravity + work done by spring force = change in kinetic energy

W_g + W_{spring} = 0 - 0

mg(0.10) + \frac{1}{2}k(0^2 - 0.10^2) = 0

3(9.81)(0.10) - \frac{1}{2}k(0.10)^2 = 0

k = 588.6 N/m

Part b)

Now when spring is stretch by x = 5 cm then the speed of the block is given as

mgx' + \frac{1}{2}k(0^2 - x'^2) = \frac{1}{2}mv^2 - 0

here we have

x' = 0.05 m

3(9.81)(0.05) - \frac{1}{2}(588.6)(0 - 0.05^2) = \frac{1}{2}(3) v^2

1.4715 - 0.736 = 1.5 v^2

v = 0.7 m/s

3 0
3 years ago
Select the correct answer.
Natali5045456 [20]

(NH4)2CO3(s) ⇄ 2NH3(g) + CO2(g) + H2O (g)

The  effect  that adding  water (H2O)   have in reaction above is  that The  concentration of the product  would  decrease, and the concentration of reactant  would increase. (answer  C)


   <u><em>Explanation</em></u>

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  • H2O   in reaction above  is  a product.

  • Increase of water (H2O)  concentration will led to the shift of equilibrium  to the left ( reactant  side)  .The concentration of  reactant  increase  as that of product decrease.
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Answer:

C

Explanation:

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shtirl [24]
They all end with suffix "-ide"

In short, Your Answer would be Option C

Hope this helps!
5 0
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