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Tpy6a [65]
2 years ago
8

A 4.0-kg object has 72 J of kinetic energy. Determine its speed.

Physics
1 answer:
Svetlanka [38]2 years ago
8 0

Answer:

6m/s

Explanation:

v = sqrt of 2KE/m

Where:

KE = kinetic energy

m = mass of a body

v = velocity of a body

= sqrt of 2(72)/4

= sqrt of 144/4

= sqrt of 36

= 6m/s

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Compare the light gathering power of a 1 meter diameter telescope to that of the human eye ,which has a diameter of roughly 2.5
lesantik [10]

Answer:

The telescope can gather light 1600 times more than the human eyes can!

Explanation:

The light gathering ability of an optical element is directly proportional to its area of opening.

So, in comparing the light gathering abilities for two objects, it is just the ratio of their area of opening.

Let the diameter of the telescope be D = 1 m

And the diameter of the human eyes be d = 2.5 cm = 0.025 m

Light gathering ability of the telescope compared to the eyes = D² ÷ d²

= (D²/d²) = (1²/0.025²) = 1600 times.

The telescope can gather light 1600 times more than the human eyes can!

Hope this Helps!!!

7 0
4 years ago
G (where g=9.8 m/s2). If an object’s mass is m=10. kg, what is its weight?
Semenov [28]
F_{w} =m*g \\ F_{w} =10*9.8=98N
7 0
4 years ago
A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float
storchak [24]

Answer:

 t ’= \frac{1450}{0.6499 + 2 v_r},  v_r = 1 m/s       t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

        v_b - v_r = d / t

By the time the boat goes down the river

        v_b + v_r = d '/ t'

let's subtract the equations

       2 v_r = d ’/ t’ - d / t

       d ’/ t’ = 2v_r + d / t

       t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }

In the exercise they tell us

         d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

         d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

      t ’= \frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \  + 2 \ v_r}

      t ’= \frac{1450}{0.6499 + 2 v_r}

In order to complete the calculation, we must assume a river speed

          v_r = 1 m / s

       

let's calculate

      t ’= \frac{ 1450}{ 0.6499 + 2 \ 1}

      t ’= 547.19 s

8 0
3 years ago
How long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘C? Note that t
QveST [7]

Answer : The time required is, 16.1 minutes.

Explanation :

First we have to calculate the amount of heat required to increase the temperature is:

Q=mC\Delta T\\\\Q=\rho VC\Delta T

(m=\rho V)

where,

Q = amount of heat required = ?

m = mass

\rho = density of air = 1.20kg/m^3

V = volume of air

C = specific heat of air = 1006J/kg^oC

\Delta T = change in temperature = 10.0^oC

Now put all the given values in above formula, we get:

Q=\rho VC\Delta T

Q=(1.20kg/m^3)\times (3.00m\times 5.00m\times 8.00m)\times (1006J/kg^oC)\times (10.0^oC)

Q=1.449\times 10^6J

Now we have to calculate the time required.

Formula used :

t=\frac{Q}{P}

where,

t = time required = ?

Q = amount of heat required = 1.449\times 10^6J

P = power = 1500 W

Now put all the given values in above formula, we get:

t=\frac{1.449\times 10^6J}{1500W}

t=966s\times \frac{1min}{60s}=16.1min

Thus, the time required is, 16.1 minutes.

3 0
3 years ago
While painting the top of an antenna 225 m in height, a worker accidentally lets a 1.00‐L water bottle fall from his lunchbox. T
yaroslaw [1]

Answer:

Increase in temperature of water = 0.53 °C

Explanation:

Change in mechanical energy = Potential energy

Potential energy = mgh

Mass, m = Mass of 1 L water = 1 kg

Acceleration due to gravity, g = 9.81 m/s²

Height, h = 225 m

Potential energy = 1 x 9.81 x 225 = 2207.25 J

Because of this 2207.25 J water gets heated.

Heat energy, E = mcΔT

Mass, m = Mass of 1 L water = 1 kg

Specific heat of water, c = 4200 J/kg/C

Energy, E = 2207.25 J

Change in temperature, ΔT = ?

Substituting

         2207.25 = 1 x 4200 x ΔT

           ΔT = 0.53 °C

          Increase in temperature of water = 0.53 °C

4 0
3 years ago
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