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2 years ago

Ratio of velocity ratio

Physics

1 answer:

Zolol [24]2 years ago

4 0

Answer:

The ratio of a distance through which any part of a machine moves to that which driving part moves during the same time.

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A car increased its velocity to 62m/s in 10s starting from rest. Calculate the distance it covers during this time?

Roman55 [17]

Answer:

distance= velocity ×time

distance= 62×10

distance=620m

hope it helps you mate please mark me as brainliast

6 0

2 years ago

Read 2 more answers

Which represents the net force

ipn [44]

Answer:

The net force is 500N downwards

Explanation:

When Haley is trying to pull an object upward. The below forces are acting on the object.

Fp = 5500N

Fg = 6000N

because the force of gravity is more than the force of the pull.

Fnet = Fg - Fp = 6000N - 5500N = 500N

And, the direction of the resultant force is the direction of the larger force.

4 0

2 years ago

Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds

solong [7]

Answer:

1.1 m/(s)^2

Explanation:

u=11 m/s

v=33 m/s

t=10s

v=u+at

=> 33=22+(a)(10)

=> 33-22=10a

=> 10a=11

=> a=11/10=1.1 m/(s)^2

7 0

2 years ago

Water is boiled at 1 atm pressure in a 25-cm-internal- diameter stainless steel pan on an electric range. if it is observed that

patriot [66]

3933 watts At 100 C (boiling point of water), it's density is 0.9584 g/cm^3. The volume of water lost is pi * 12.5^2 * 10 = 4908.738521 cm^3 The mass of water boiled off is 4908.738521 * 0.9584 = 4704.534999 grams. Rounding to 4 significant figures gives me 4705 grams of water. The heat of vaporization for water is 2257 J/g. So the total energy applied is 2257 J/g * 4705 g = 10619185 J Now we need to divide that by how many seconds we've spent boiling water. That would be 45 * 60 = 2700 seconds. Finally, the rate of heat transfer in Joules per second will be the total number of joules divided by the total number of seconds. So 10619185 J / 2700 s = 3933 J/s = 3933 (kg m^2/s^2)/s = 3933 (kg m^2/s^3) = 3933 watts

3 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

Ratio of velocity ratio​

Ratio of velocity ratio