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KonstantinChe [14]

2 years ago

12

The stopwatch used by a student to measure velocity of a pulse in a slinky was of least count 0.1 second. He stops the stopwatch

when a pulse has made 3 journeys from one end to the other of the slinky. He finds the seconds hand to be at 52nd division.
Calculate the correct time noted by him.

1 answer:

sweet-ann [11.9K]2 years ago

4 0

Least count of the pulse stopwatch is given by

$\Delta t = 0.1 s$

this means each division of the stopwatch will measure 0.1 s of time

After 3 journeys from one end to other we can see that total time that is measured here is shown by the clock as 52nd division

So here total time is given as

Time = (Number of division) (Least count)

now we will have

$T = 52 \times 0.1s$

$T = 5.2 s$

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1 year ago

A box has a mass of 150kg and the surface area of the bottom of the box is :

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2 years ago

IBM has a fast computer that it calls the Blue Gene/L that can do '136.8

dmitriy555 [2]

Answer:

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Kilo and Mega means the numbers are raised to the power of 3 and 6 respectively. There also exists the ones that indicates how small, such as milli and micro, which are to the powers of -3 & -6.

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136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

1 micro second = 1*10^-6.

If the IBM does

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Then it does

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When we cross multiply, we have

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6 0

2 years ago

Two people stand facing each other at roller skating rink then push off each other

9966 [12]

a) 0 kg m/s

b) 0 kg m/s

c) +3 m/s

d) 60 N

Explanation:

a)

The momentum of an object is a vector quantity given by:

$p=mv$

where

m is the mass of the object

v is the velocity of the object

In this problem, we have a system of two people, so the total momentum will be the sum of the individual momenta of the two people:

$p=p_1 + p_2$

Which can be rewritten as

$p=m_1 u_1 + m_2 u_2$

where $m_1,m_2$ are the masses of the two people and $u_1,u_2$ their initial velocities.

However, the two people are initially at rest, so

$u_1 = 0\\u_2 = 0$

Therefore the total momentum is

$p=0+0=0$

b)

The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

$p_i = p_f$

where

$p_i$ is the total momentum of the system before

$p_f$ is the total momentum of the system after

In this problem,

$p_i = 0$

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

Therefore, according to the law of conservation of momentum,

$p_f = p_i = 0$

So the total momentum is zero also after they push off each other.

c)

The total momentum of the girl and the boy after they push off each other can be written as:

$p_f = m_1 v_1 + m_2 v_2$ (1)

where:

$m_1 = 30 kg$ is the mass of the girl

$v_1 = -5 m/s$ is her velocity (she moves backward, so the negative sign)

$m_2 = 50 kg$ is the mass of the boy

$v_2$ is the velocity of the boy

As calculated in part b), we also know that the total momentum of the girl and the boy is

$p_f = 0$ (2)

By combining eq(1) and eq(2) we get

$0=m_1 v_1 + m_2 v_2$

And solving for v2 we find the velocity of the boy:

$v_2=-\frac{m_1 v_1}{m_2}=-\frac{(30)(-5)}{50}=+3 m/s$

and the positive sign means he is moving forward.

d)

We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

$\Delta p = F\Delta t$

where

$\Delta p$ is the change in momentum

F is the force

$\Delta t$ is the time during which the force is applied

In this problem:

$\Delta t = 2.5 s$

For the boy, the change in momentum is:

$\Delta p = m_2 (v_2 - u_2)$

And since

$m_2 = 50 kg\\u_2 = 0 m/s\\v_2 = 3 m/s$

We have

$\Delta p = (50)(3-0)=150 kg m/s$

So, the force exerted between the boy and the girl is:

$F=\frac{\Delta p}{\Delta t}=\frac{150}{2.5}=60 N$

8 0

2 years ago