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shusha [124]
3 years ago
10

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
Physics
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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An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e
zimovet [89]

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

3 0
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
jeka57 [31]

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).

7 0
3 years ago
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

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,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
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
Yanka [14]

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]

4 0
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:

It would be 2600

Explanation:

M/S stands for meters per second. If it moved 1 meter for 2600 seconds, than it would be 2600. You just multiply 2600 by 1! I hope this helps :D

8 0
3 years ago
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