PHYSICS COLLEGE

A 50 kg woman and an 80 kg man stand 12.0 m apart on frictionless ice.(a) How far from the woman is their CM?
m

(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 1.3 m, how far from the woman will he be now?
m

(c) How far will the man have moved when he collides with the woman?
m

Answers

Answer 1

Answer:

Answer:

Given that

m₁ = 50 kg

m₂=80 kg

d= 12 m

We know that center of mass given as

X = (x₁m₁+x₂m₂)/(m₁+m₂)

Lets take distance of CM from woman is X

So now by putting the value

X = (0 x 50+12 x 80)/(50+80)

x=7.38 m

There is no any external force so the CM will not move.

So we can say that

x₁m₁+x₂m₂ = 0

50(x) - 80(1.3)=0

x=2.08

So the distance move by woman d=12-2.08-1.3=8.62 m

d=8.62 m

c) lets take distance move by man is x

50 (x) - 80 (12-x) =0

x=7.38

So the distance move by woman d=12-7.38

d=4.62 m

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Which two types of simple machines can be found in a bicycle?

Answers

Answer:

The correct answer is A) lever and wheel and axle

Explanation:

I took the quiz

hope this helps :)

A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?

Answers

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

$\mu mg = (mv^2)/(R)$

here we have

$\mu g = (v^2)/(R)$

now we know that

$v = √(\mu Rg)$

here we have

$\mu = 0.600$

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

$v = √((0.600)(35)(9.81))$

$v = 14.35 m/s$

The mass of a baseball is 0.145 kg and its acceleration as it falls to the ground is 9.81 m/s2. How much force is acting on the baseball

Answers

Answer:

The answer is 1.42 N

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

Force = 0.145 × 9.81 = 1.42245

We have the final answer as

1.42 N

Hope this helps you

A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes. If instead of a single wire we use a coil of thin Nichrome wire containing 23 turns, what does the ammeter read?

Answers

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

$\epsilon=NA(dB)/(dt)$

For N = 1

$\epsilon=A(dB)/(dt)$

We need to calculate the current

Using formula of current

$i=(\epsilon)/(R)$

Put the value of emf

$i=(A(dB)/(dt))/(R)$

Now, if the number of turn is 22 , then induced emf would be

$\epsilon'=NA(dB)/(dt)$

Then the current would be

$i'=(\epsilon')/(NR)$

$i'=(NA(dB)/(dt))/(NR)$

$i'=(A(dB)/(dt))/(R)$

$i'=i$

Hence, The current would be same in both situation.

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?