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3 years ago

give an example of motion in which displacement of the particle is zero but acceleration is not zero in journey?

1 answer:

4 0

Motion of a ball thrown by a person upwards and caught after some time is an example of motion in which displacement of the particle is zero but acceleration is not zero in journey.

The displacement of the ball is zero because the starting and end point of the motion are same, i.e, the person's hands.During its motion, the acceleration of ball is constant and non zero called as acceleration due to gravity, g= -9.8 m/s². The velocity of ball is continuously changing. It first decreases during the upward motion of the ball and then increases during the downward journey.The acceleration remains constant and non zero all the time.

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A rocket moves upward, starting from rest with an acceleration of +30.0 m/s2 for 5.00 s. It runs out of fuel at the end of this

olchik [2.2K]

Answer:

The distance covered by the rocket after fuel ran out is $3442.04 m$

Explanation:

Given that the rocket moves with an acceleration $a=30m/s^2$

time $t=5 s$

Since the rocket starts from rest initial velocity $u=0 s$

The distance it travelled within this time is given by $s=ut+ \frac{1}{2} at^2$ $=0 \times 5+ \frac{1}{2} (30\times25)=375 m$

Velocity at this point is given by $v=u+at$

$v=0+30\times5=150m/s$

Given that at this height it runs out of fuel but travels further. Here final velocity $v=0$ (maximum height), initial velocity $u=150 m/s$ and time to zero velocity $t=\frac{v}{g} = \frac{150}{9.8} =15.3 s.$

Thus it travels $15.3 seconds$ more after fuel running out. The distance covered during this period is given

$s= ut+\frac{1}{2} gt^2=150 \times 15.3+1/2 \times9.8 \times 15.3^2=3442.04 m$

7 0

3 years ago

the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance

Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

$F = G\frac{Mm}{r^{2} }$

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = $\frac{1}{4}$ G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = $\frac{1}{4}$ F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.

6 0

3 years ago

The metric unit to measure work which equals one newton meter<br> is called?

Solnce55 [7]

The metric unit to measure work which equals one newton meter is called One Joule.

8 0

3 years ago

Kon'nichiwa~<br>please help me with this question!!

andrezito [222]

Let's check the relationship

$\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}$

$\\ \rm\hookrightarrow g\propto G$

Raindrops will fall faster . .
Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

$\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}$

So time period of simple pendulum would decrease.

4 0

2 years ago

Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const

kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is L= 20cm = 0.2m

Thermal conductivity of the wall is K = 2.79 W/m·K

Temperature at the left side surface is T₁ = 50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

$Q_{conduction} = Q_{convection}$

Expression for the heat conduction process is

$Q_{conduction} = \frac{K(T_1 - T)}{L}$

Expression for the heat convection process is

$Q_{convection} = h(T_2 - T)$

Substitute the expressions of conduction and convection in equation above

$Q_{conduction} = Q_{convection}$

$\frac{K(T_1 - T_2)}{L} = h(T_2 - T)$

Substitute the values in above equation

$\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC$

Now heat flux through the wall can be calculated as

$q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2$

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0

3 years ago