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OLEGan [10]
3 years ago
10

A 0.05kg dart is thrown at and sticks into a 0.4 kg block hanging on a string. After the collision the block and dart swing in a

circular arc rising 0.1m vertically. What must the speed have been just after the collision
Physics
1 answer:
zloy xaker [14]3 years ago
6 0

Answer:

v = 1.4  m /s

Explanation:

We shall apply law of conservation of mechanical energy

The kinetic energy of dart and block   is converted into potential energy of both dart and block .

1 /2 (m+M) v² = ( m +M) gH

.5  x v² =  9.8 x .1

=  v² = 1.96

v = 1.4

v = 1.4  m /s

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If the value of static friction for a couch is 400N, what could be a possible value of its kinetic friction? *
Dafna11 [192]

Answer:

kinetic friction may be greater than 400 N or smaller than 400 N

Explanation:

As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

F_s = \mu_s N

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

F_k = \mu_k N

now we know that

\mu_k < \mu_s

so here value of limiting static friction force is always more than kinetic friction

also we know that

initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction

and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction

so kinetic friction may be greater than 400 N or smaller than 400 N

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Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is λA = 622nm. The other wavelength is λ
Sergio039 [100]

Answer:

\lambda_{B}=414.67 nm

Explanation:

In this question we have given

\lambda_{A}=622nm

we have to find

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We know that

optical path difference for bright fringe is given as=n\lambda

Here,

n is order of fringe

and optical path difference for dark fringe is given as=(n+.5)\lambda

since the light with wavelength \lambda_{A} produces its third-order bright fringe at the same place where the light with wavelength \lambda_{B} produces its fourth dark fringe  

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Therefore,

3\lambda_{A}=(4+.5)\lambda_{B}...............(1)

Put value of \lambda_{A} in equation (1)

3 \times 622=(4+.5)\lambda_{B}

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\lambda_{B}=414.67 nm

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