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nata0808 [166]
3 years ago
12

A boy is standing motionless on a skateboard. He throws a basketball forward. Describe his motion.

Physics
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

Random motion

Explanation:

If the boy throws the basketball forward while at a position on the skateboard, the motion of the ball will be a random motion since we are not told if the ball is moving on a straight line when thrown forward.

In this case, the boy will tend to move in the direction of the ball. Since the ball is moving in a random manner, the motion of the boy will also be a random motion.

A random motion is a motion of a body in a zig zag manner. It is also known as Brownian motion e.g motion of a buzzing mosquito, motion of a smoke coming out of a chimney etc.

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Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li
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Answer:

(a) 10 s

(b) 236.5 m

(c) Kathy's speed = 47.3 m/s

    Stan's speed = 42.9 m/s

Explanation:

<u>Given:</u>

  • u_k = initial speed of Kathy = 0 m/s
  • u_s = initial speed of Stan = 0 m/s
  • a_k = acceleration of Kathy = 4.73\ m/s^2
  • a_s = acceleration of Stan = 3.9\ m/s^2

<u>Assumptions:</u>

  • v_k = final speed of Kathy when see catches Stan
  • v_s = final speed of Stan when Kathy catches him
  • s_k = distance traveled by Kathy to catch Stan
  • s_s = distance traveled by Stan when Kathy catches him
  • t_k = time taken by Kathy to catch Stan = t
  • t_s = time interval in which Kathy catches Stan = t+1

Part (a):

 Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.

\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\

Hence, Kathy catches Stan after 11 s from the Stan's starting times.

Part (b):

Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.

s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5

Hence, Kathy traveled a distance of 236.5 m to overtake Stan.

Part (c):

v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3

The speed of Kathy at the instant she catches Stan is 47.3 m/s.

v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9

The speed of Stan at the instant Kathy catches him is 42.9 m/s.

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Our knowledge cell is a Scienctific_______.<br><br>A guess<br>B law<br>C fact<br>D theory
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The protons initially are located where the electric potential has a value of 7.60 MV and then they travel through a vacuum to a
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Answer:

(a) 3.82 x 10⁷ m/s

(b) 4.5 MV/m

Explanation:

(a)

ΔV = change in the electric potential as the proton moves = 7.60 x 10⁶ Volts

q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

v = speed gained by the proton

m = mass of proton = 1.67 x 10⁻²⁷ kg

Using conservation of energy

Kinetic energy gained by proton = Electric potential energy

(0.5) m v² = q ΔV

inserting the values

(0.5) (1.67 x 10⁻²⁷) v² = (1.6 x 10⁻¹⁹) (7.60 x 10⁶)

v = 3.82 x 10⁷ m/s

(b)

d = distance over which the potential change = 1.70 m

Electric field is given as

E = ΔV/d

E = 7.60 x 10⁶/1.70

E = 4.5 x 10⁶ V/m

E = 4.5 MV/m

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