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3 years ago

5

A 5 kilogram block is dropped from a height of 3 meters and falls straight to the ground. What is the work done by the force of

gravity?

1 answer:

vazorg [7]3 years ago

5 0

Answer:

Workdone = 147Nm

Explanation:

Given the following data;

Mass = 5kg

Height = 3m

We know that acceleration due to gravity is 9.8m/s²

First of all, we would find the force applied;

$Force = mass * acceleration$

Force = 5*9.8

Force = 49N

Now, to find the workdone;

$Workdone = force * distance$

Substituting into the equation, we have;

Workdone = 49 * 3

Workdone = 147Nm

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For a specific volume of 0.2 m3/kg, find the quality of steam if the absolute pressure is (a) 40 kPa and (b) 630 kPa. What is th

ICE Princess25 [194]

Answer:

$x=0.0498$

$x'=0.659$

Explanation:

Specific Volume $V=0.2m_3/kg$

Absolute Pressure (a) $P_a= 40kpa$

Giving

$T_a=75.87$

$v_f=1.265*10^{-3}m^3/kg$

$v_g=3.993m^3/kg$

(b) $P_a= 630kpa$

Giving

$T_b=160.13C$

$v_f'=1.10282*10^{-3} m^3/kg$

$v_g'=0.30286 m^3/kg$

(a)

Generally the equation for quality of Steam X is mathematically given by

$x=\frac{v-v_f}{v_g-v_f}$

$x=\frac{0.2-1.0265*10^{-3}}{3.993-1.0265*10^{-3}}$

$x=0.0498$

(b)

Generally the equation for quality of Steam X is mathematically given by

$x'=\frac{v-v_f'}{v_g'-v_f'}$

$x'=\frac{0.2-1.10*10^{-3}}{3.30-1.1*10^{-3}}$

$x'=0.659$

4 0

3 years ago

Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?

Stolb23 [73]

Answer:

$200 cal/^{\circ}C$

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

$Q=C\Delta T$

where

Q is the heat supplied

C is the heat capacity of the object

$\Delta T$ is the change in temperature

In this problem we have:

$Q=10,000 cal$ is the energy supplied

$\Delta T=75C-25C=50C$ is the change in temperature of the object

Therefore, the heat capacity of the object is:

$C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C$

6 0

3 years ago

A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

8 0

3 years ago

The intensity of light from a star varies inversely as the square of the distance. If you lived on a planet ten times farther aw

frosja888 [35]

Answer:

the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.

That is,

Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100

Explanation:

Let the intensity of light be represented by I

Let the distance of the star be d

I ∝ (1/d²)

I = k/d²

For the earth,

Iₑ = k/dₑ²

k = Iₑdₑ²

For the other planet, let intensity be Iₒ and distance be dₒ

Iₒ = k/dₒ²

But dₒ = 10dₑ

Iₒ = k/(10dₑ)²

Iₒ = k/100dₑ²

But k = Iₑdₑ²

Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100

Iₒ = Iₑ/100

Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.

3 0

4 years ago

The coefficient of performance of a residential heat pump is 1.6. Calculate the heating effect in kJ/s this heat pump will produ

grin007 [14]

Answer:

8 KJ/ s

Explanation:

Heat pumps Transfer thermal energy through absorbing of heat that comes from cold region and then release to warmer area by utilizing external power.

The coefficient of performance known as COP provide the ratio of both heating and cooling that are supplied to required work.

✓QH=The rate at which heat is produced = ?

✓COP= Coefficient of performance of a residential heat pump = 1.6

✓ W(in)= power consumption= 5KW

QH=The rate at which heat is produced=[Coefficient of performance of a residential heat pump] × [power consumption]

= 1.6 × 5KW

=8 KJ/ s

7 0

3 years ago