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kupik [55]
3 years ago
7

What effect does the vertical acceleration have on the horizontal velocity of the projectile?

Physics
1 answer:
KengaRu [80]3 years ago
3 0
Answer:
None, if air resistance is ignored.

Explanation:
At any instant, the projectile has vertical and horizontal components of velocity.
Vertical acceleration due to gravity affects the vertical velocity by accelerating the object toward the center of the earth, and by decreasing the upward vertical velocity.. 
The horizontal component of velocity makes the object travel horizontally as long as the projectile is airborne.
Thsi discussion assumes that air resistance is ignored.
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A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west
ivolga24 [154]

Answer:

Magnitude of avg velocity, |v_{avg}| = 18.9 km/h

\theta' = 56.85^{\circ}

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = \frac{27}{60} h

Angle, \theta = 50^{\circ}

Time taken in 50^{\circ}east of due North direction, t' = 29 min =  \frac{29}{60} h

Time taken in west direction, t'' = 37 min =  \frac{27}{60} h

Solution:

Now, the displacement, 's' in east direction is given by:

\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km

Displacement in  50^{\circ} east of due North for 29.0 min is given by:

\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}

\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}

\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km

Now, displacement in the west direction for 37 min:

\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km

Now, the overall displacement,

\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}

\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}

\vec{s_{net}} =  16.03\hat{i} + 24.54\hat{j} km

(a) Now, average velocity, v_{avg} is given:

v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}

v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}

v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h

Magnitude of avg velocity is given by:

|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h

(b) angle can be calculated as:

tan\theta' = \frac{15.83}{10.34}

\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}

6 0
3 years ago
You have gas in a container with a movable piston. The walls of the container are thin enough so that its temperature stays the
bearhunter [10]

Answer:

New pressure of the gas increases by 26.5% with respect to initial pressure, new volume decreases 27% with respect to initial volume and new temperature decreases 8% with respect to initial volume.

Explanation:

If we assume the gas is a perfect gas we can use the perfect gas equation:

PV=nRT

  • For Isothermal process:

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}(1)

Where subscripts 1 shows before the isothermal process and 2 after it, because isothermal means constant temperature T1=T2, and pressure increases by 10% means P2=1,1*P1, using these facts on (1) we have:

V_{2}=\frac{V_{1}}{1.1} (2)

  • For Isobaric process:

\frac{P_{2}V_{2}}{T_{2}}=\frac{P_{3}V_{3}}{T_{3}} (3)

Where subscripts 2 shows before the isobaric process and 3 after it, because isobaric means constant pressure P2=P3, and volume decreases by 20% means V3=0.8*V2, using these facts on (3) we have:

T_{3}=0.8T_{2} (4)

  • For Isochoric process:

\frac{P_{3}V_{3}}{T_{3}}=\frac{P_{4}V_{4}}{T_{4}} (5)

Where subscripts 3 shows before the isochoric process and 4 after it, because isochoric means constant volume V3=V4, and temperature increases by 15% means T4=1.15*T3, using these facts on (5) we have:

P_{4}=1.15P_{3} (6)

So now because P4=1.15*P3, P2=P3 and P2=1.1*P1:

P_{4}=1.15*1.1P_{1}=1.265P1

This is, the new pressure of the gas increases by 26.5%  with respect to initial pressure.

Similarly, we have V3=V4, V3=0.8*V2 and V1=1,1*V2:

V_{4}=\frac{0.8}{1.1}V_{1}=0.72V1

so the final volume decreases 27% with respect to initial volume.

T4=1,15*T3, T3=0.8*T2 and T1=T2:

T_{4}=1.15*0.8T_{1}=0.92T1

The new temperature decreases 8% with respect to initial volume.

3 0
3 years ago
Plz help urgent <br> i will give brainliest
In-s [12.5K]

Answer:

The answer is D. Balanced forces

Hope it helps.......... pls mark as brainliest

5 0
2 years ago
She left the cubes in the water for three hours which of the following describes a heat flow that took place during those three
quester [9]

Answer:

veffvevfevve

Explanation:

3 0
2 years ago
Read 2 more answers
A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?
rosijanka [135]

Answer:

a

     l_o  =52 \  m

b

      l = 37.13 \ LY

Explanation:

From the question we are told that

    The  speed of the spaceship is  v  =  0.800c

    Here  c is the speed of light with value  c =  3.0*10^{8} \ m/s

    The  length is  l = 31.2 \  m

     The  distance of the star for earth is d = 145 \  light \  years

     The  speed is v_s = 2.90 *10^{8}

     

Generally the from the length contraction equation we have that

       l  =  l_o  \sqrt{1 -[\frac{v}{c } ]}

Now the when at rest the length is  l_o

So  

      l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }

      l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }

      l_o=52 \  m

Considering b  

  Applying above equation

            l  =l_o \sqrt{1 -  [\frac{v}{c } ]}

Here l_o  =145 \  LY(light \ years )

So

           l=145 *  \sqrt{1 -  \frac{v_s^2}{c^2 } }

            l =145 *  \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }

            l = 37.13 \ LY

4 0
3 years ago
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