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3 years ago

How is work calculated?

Physics

1 answer:

BartSMP [9]3 years ago

6 0

Work is force times distance. If there's no distance, there's no work being done.

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675 km equals how much cm ?

irga5000 [103]

KHDMDCM.
Now go from Kilometer to Centimeter: 5.
Move the decimal 5 places to the right: 67,500,000 centimeters.
Hope this helps :)

7 0

3 years ago

The chart shows the percentage of different elements in the human body.

GenaCL600 [577]

I believe that answer is nitrogen.

8 0

3 years ago

An airplane starts from A and flies to B at a constant speed. After reaching B it returns to A at the same speed. There was no w

Dafna1 [17]

Answer:

When there is wind it takes longer

Explanation:

With no wind, the round trip time is

$t_1=\frac{d}{v}+\frac{d}{v}=\frac{2d}{v}$

When we have a constant wind speed w

$t_2=\frac{d}{v-w} +\frac{d}{v+w} =\frac{2vd}{v^{2} -w^{2}}$

comparing the reciprocal times;

$\frac{1}{t_2}=\frac{v^{2}-w^{2} }{2vd}=\frac{v}{2d}-\frac{w^{2}}{2vd} \leq \frac{v}{2d}=\frac{1}{t_1}$

This means that t1 is smaller than t2, ergo, it takes longer with wind

4 0

2 years ago

What happens to thermal energy when Kinetic energy is Increased/Decreased?

Elis [28]

Answer:If kinetic energy increases, so does the thermal energy, and vice versa.

Please brainliest!

3 0

3 years ago

An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f

IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

$v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}$

Similarly, The equation of mption:

$x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})$

replacing our assumed values:

where $v_=0 \ and \ g= 9.81$

$x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})$

$x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m$

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

$1000= 0.981t-0.981(1-e^{-(10)t}) \ m$

By diregarding $e^{-(10)t} \ m$

$1000= 0.981t-0.981$

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0

3 years ago