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Vesna [10]
3 years ago
12

Hi please may someone help me especially on the sketch part.

Physics
1 answer:
vaieri [72.5K]3 years ago
7 0

Ignoring the air resistance it will take about 3 seconds for the object to reach the ground.We know that the acceleration due to gravity is 10m/s2.

We also know that the final velocity is 30 m/s while the initial velocity is 0 m/s

we can use the formulae for acceleration to calculate the time taken/

(final - initial velocity)/timetaken=10

(30-0)/timetaken=10

timetaken =30/10=3 seconds

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Does wind have atoms and molecules in it?
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When you rub a plastic rod with fur, the plastic rod becomes negatively charged and the fur becomes positively charged. As a con
Yuri [45]

Answer: C. the rod gains mass and the fur loses mass.

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4 0
3 years ago
An object whose weight is 10kg is placed on smooth plane inclined at 30° to the horizontal. find the acceleration of the object
velikii [3]

Answer:

4.9 m/s²

Explanation:

Draw a free body diagram.  There are two forces on the object:

Weight force mg pulling straight down,

and normal force N pushing perpendicular to the plane.

Sum the forces in the parallel direction.

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 30°)

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8 0
3 years ago
0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a
klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}

(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }

where h_{fg is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

V_{fg is specific volume ( 1.5 ft³ )

we substitute

(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5}) / ( (15+460)\frac{1.5}{0.5})  

(\frac{dP}{dT} )_{sat } = 272.98 Ibf-ft²/R

Now,

(\frac{dP}{dT} )_{sat } = (\frac{P_2 - P_1}{T_2 - T_1})_{sat

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

T_2 = ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}

T_2 = 475 + 5.2751\\

T_2 = 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

3 0
3 years ago
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