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3 years ago

7

What is force? Do systems have inputs, outputs, do they have both, do they have one, do they have the other, what are the system

s.

1 answer:

Irina18 [472]3 years ago

7 0

a force is a push or pull. hope i helped!

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A child in a tree house uses a rope attached to a basket to lift a 24 N dog upward through a distance of 4.9 m into the house.

jenyasd209 [6]

<h3>Answer:</h3>

117.6 Joules

<h3>Explanation:</h3>

<u>We are given;</u>

Force of the dog is 24 N
Distance upward is 4.9 m

We are required to calculate the work done

Work done is the product of force and distance
That is; Work done = Force × distance
It is measured in Joules.

In this case;

Force applied is equivalent to the weight of the dog.

Work done = 24 N × 4.9 m

= 117.6 Joules

Hence, the work done in lifting the dog is 117.6 Joules

3 0

3 years ago

a gas has a volume of 8 liters at a temperature of 300 K the volume is then increased to 12 Liters what is the new temperature (

natali 33 [55]

300/8 = 37.5
37.5 x 12 = 450
New temp. = 450 K
Hope this helps!

6 0

3 years ago

How old is our universe

Alja [10]

Answer:

13.8 billion years

Explanation:

In physical cosmology, the age of the universe is the time elapsed since the Big Bang. The current measurement of the age of the universe is around 13.8 billion.

6 0

3 years ago

Read 2 more answers

A U-shaped tube with both arms open to the air has a 43.0 cm column of liquid of unknown density in its right arm. Beneath this

puteri [66]

Answer:

$\delta_x=732.5581395 \approx 737.56kh/m^3$

Explanation:

From the question we are told that

Height of unknown liquid $h_1= 43.0cm =0.43m$

Height of glycerin $h_2= 18.0cm =0.18m$

Generally at equal levels of liquid pressure

$P_1=P_2$

Therefore

$\delta_gh_1=\delta_gh_2$

$\delta g=Density\ of \ glycerin=>1260kg/m^3$

Therefore

$\delta_g=\frac{\delta_gh_2}{h_1}$

$\delta_x=\frac{1260*0.43-0.18}{0.43}$

The Density of the unknown liquid is given by

$\delta_x=732.5581395 \approx 737.56kh/m^3$

8 0

3 years ago

Degger [83]

Explanation:

The initial kinetic energy $KE_0$ is

$KE_0 = \frac{1}{2}mv_0^2 = \frac{1}{2}(400\:\text{kg})(10\:\text{m/s})^2$

$\:\:\:\:\:\;\:= 2×10^4\:\text{J} = 20\:\text{kJ}$

The final kinetic energy $KE$ is

$KE = \frac{1}{2}mv^2 = \frac{1}{2}(400\:\text{kg})(30\:\text{m/s})^2$

$\:\:\:\:\:\:\:= 1.8×10^5\:\text{J} = 180\:\text{kJ}$

The work done W on the car is

$W = \Delta{KE} = KE - KE_0$

$\:\:\:\:\:\:\:= 180\:\text{kJ} - 20\:\text{kJ} = 1.6×10^5\:\text{J}$

The power expended P is

$P = \dfrac{W}{t} = \dfrac{1.6×10^5\:\text{J}}{15\:\text{s}} = 10667\:\text{Watts}$

$\:\:\:\:\:= 10.7\:\text{kW}$

8 0

2 years ago