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umka2103 [35]
3 years ago
14

As a space shuttle climbs, _______

Physics
2 answers:
n200080 [17]3 years ago
4 0
Rocket fuel will and smoke will emit from the thrusters.
lesya692 [45]3 years ago
3 0

ITS MASS  DECREASES

Fuel is released and the rocket takes off into space.

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A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s g
Sliva [168]
To solve for force, you need to get the product of mass and acceleration. 
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration. 
Acceleration is the change in velocity over a period of time.

a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

a =  \frac{vf-vi}{t}

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s. 

So we can put it into our formula now:

a = \frac{0m/s-50m/s}{20ms}

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s. 

20ms x \frac{1s}{1000ms} = \frac{20s}{1000} = 0.02s

So your new time is 0.02s. Now we put this time into the formula:


a = \frac{0m/s-50m/s}{0.02s}
a =  \frac{-50m/s}{0.02s}  = -2,500 m/ s^{2}

As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop. 

So now we have our acceleration. Now using this, we can get our force. 

F= ma

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is kg.m/ s^{2} but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first. 

150g =  \frac{1kg}{1,000g}  =  \frac{150kg}{1,000}  = 0.150kg

Our new mass is 0.150kg. 

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 m/ s^{2}

Now that all our units match, we can put that into our formula:

F= ma
F= (0.150kg)(-2,500m/s^{2})
F = -375kg.m/ s^{2}  or -375N

The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign. 



4 0
3 years ago
Calculate the wave length of a water wave with a speed of 20 m/s and a frequency of 2.5 Hz
12345 [234]

Wavelength of the water wave is 8 m

Explanation:

  • Wavelength measures the distance between two successive crests or troughs of the wave. It is given by the following equation

λ = v/f, where f is the frequency, v is the velocity of the wave

Here, v = 20 m/s and f = 2.5 Hz

⇒ λ = 20/2.5

      = 8 m

5 0
3 years ago
If you need 40.0 Nm of torque in order to loosen a nut on a wn
KonstantinChe [14]

Answer:

0.301 m

Explanation:

Torque = Force × Radius

τ = Fr

40.0 Nm = 133 N × r

r = 0.301 m

The mechanic must apply the force 0.301 m from the nut.

6 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
Suppose a capacitor is fully charged by a battery and then disconnected from the battery. The positive plate has a charge +q and
dimaraw [331]

Answer:-q

Explanation:

Given

Capacitor is charged to a battery and capacitor acquired a charge of q i.e.

+q on Positive Plate and -q on negative Plate.

If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by

c=\epsilon_0 \cdot \frac{A}{d}

where A=area of capacitor plate

d=Separation between plates

This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q

8 0
3 years ago
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