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3 years ago

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An 80-kg man is skating northward and happens to suddenly collide with a 20-kg boy who is ice skating toward the east. Immediate

ly after the collision, the man and boy are seen to be moving together at 2.5 m/s in a direction 60° north of east. How fast was the boy moving just before the collision

1 answer:

Fantom [35]3 years ago

4 0

Answer:

6.25 m/s

Explanation:

mass of man (m1) = 80 kg

mass of boy (m2) = 20 kg

mass of man and boy after collision (m12)= 20 + 80 = 100 kg

velocity of man and boy after collision (v) = 2.5 m/s

angle θ = 60 °

How fast was the boy moving just before the collision ?

From the diagram attached, the first image shows the man and the boys motion while the second diagram shows their motion rearranged to form a triangle. With the momentum of the man and the boy forming the sides of the triangle.
M₁₂ = total momentum after collision = m12 x v = 100 x 2.5 = 250
Mboy = momentum of the boy before collision = m2 x Velocity of boy
Mman = momentum of the man before collision = m1 x velocity of man
from the triangle, cos θ = $\frac{Mboy}{M₁₂}$

cos 60 = $\frac{Mboy}{250}$

Mboy = 250 x cos 60 = 125

recall that momentum of the boy (Mboy) also = m2 x Velocity of boy

therefore

125 = 20 x velocity of boy

velocity of boy = 125 / 20 = 6.25 m/s

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Answer:

a) $v = 6.25\times 10^{5} m/s$

b) $v = 1.73\times 10^{4} m/s$

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is $4.55\times 10^5 m/s$

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

$a =\frac{q E}{m}$

$a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}$

$a = 2.58\times 10^{11} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355$

$v = 6.25\times 10^{5} m/s$

b) for proton

$a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}$

$a = -1.41\times 10^{8} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355$

$v = 1.73\times 10^{4} m/s$

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3 years ago

As part of his job, Warren attends seminars where he must listen to presentations given by experts in his field. When he is list

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Answer:

Option (B)

Explanation:

In terms of communication, a receiver is usually referred to as a person who listens, reads as well as observes. In simple words, a receiver is an individual or it can be a group of individuals, to whom any type of message is being diverted. The other name for the receiver is 'audience'.

In the given condition, Warren is attending a seminar in which he is listening to the speaker, as a part of the audience. So, it can be concluded that Warren is a receiver who is receiving information or hearing the speaker.

Thus, the correct answer is option (B).

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When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

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Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

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3 years ago

A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a

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Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

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3 years ago

A badger is running at a speed of 1 m/s. If the badger moves that was for 2600 seconds, how far will the badger travel?

katen-ka-za [31]

Answer:

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Explanation:

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