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2 years ago

How does area affect the pressure?

1 answer:

5 0

The smaller the area the greater the pressure, while the bigger the smaller the pressure. So they are inversely proportional

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In a certain chemical process, a lab technician supplies 292 J of heat to a system. At the same time, 68.0 J of work are done on

vekshin1

Answer:

The increase in the internal energy of the system is 360 Joules.

Explanation:

Given that,

Heat supplied to a system, Q = 292 J

Work done on the system by its surroundings, W = 68 J

We need to find the increase in the internal energy of the system. It can be given by first law of thermodynamics. It is given by :

$dE=dQ+dW\\\\dE=292\ J+68\ J\\\\dE=360\ J$

So, the increase in the internal energy of the system is 360 Joules. Hence, this is the required solution.

3 0

2 years ago

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Sup

Tema [17]

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

$15.8 min*\frac{60s}{1min}=948s$

The equation of motion you need is:

$v_f = a*t+v_0$

where $v_f$ is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

$v_f = a*t = 20.2*948 = 19,149.6 m/s$

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

$x=\frac{v_0+v_f}{2}t$

solving for first and last interval:

Since the spaceship starts and finish with 0 velocity:

$x=\frac{v}{2}t=\frac{19,149.6}{2}948=9,076,910.4m=9,076.9104km$

Then the ship traveled $384,000-9,076.9104*2 = 361,846.1792km$ at constant speed, which means that it traveled:

$f_{constant_speed} =\frac{ x_{constant_speed}}{x_total} =\frac{361,846.1792}{380,000} =0.95$

Which in percentage is 95% of the trip.

to calculate total time you need to calculate the time used during constant speed:

$t = \frac{361,846,179.2}{19,149.6} = 18,895.75s = 314min$

That added to the other interval times:

$t_{total} = t_1+t_2+t_3=15.8+314.93+15.8=346.5min$

5 0

2 years ago

A 5.0 cm object is 12.0 cm from a concave mirror that has a focal length of 24.0 cm. The distance between the image and the mirr

MA_775_DIABLO [31]

Answer:

The answer is 24cm

Explanation:

This problem bothers on the curved mirrors, a concave type

Given data

Object height h= 5cm

Object distance = 12cm

Focal length f=24cm

Let the image distance be v=?

Applying the formula we have

1/v +1/u= 1/f

Substituting our given data

1/v+1/12=1/24

1/v=1/24-1/12

1/v=1-2/24

1/v=-1/24

v= - 24cm

This implies that the image is on the same side as the object and it is real

7 0

3 years ago

Read 2 more answers

After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg

GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses $m_1$ and $m_2$ whose respective velocities before collision are $u_1$ and $u_2$ ;