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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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A bird is about 6.26.2 in.​ long, with a​ thin, dark bill and a​ wide, white wing stripe. If the bird can fly 9292 mi with the w
Trava [24]

Answer:

209 mph

Explanation:

V = Speed of bird in still air

v = Speed of wind = 44 mph

Consider the motion of the bird with the wind

D_{1} = distance traveled with the wind = 9292 mi

t_{1} = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

t_{1} = \frac{D_{1}}{V + v}

t_{1} = \frac{9292}{V + 44}                              eq-1

Consider the motion of the bird with the wind

D_{2} = distance traveled against the wind = 6060 mi

t_{2} = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

t_{2} = \frac{D_{2}}{V + v}

t_{2} = \frac{6060}{V - 44}                              eq-2

As per the question,

Time taken with the wind = Time taken against the wind

t_{1} = t_{2}

\frac{9292}{V + 44} = \frac{6060}{V - 44}

(9292) (V - 44) = (6060) (V + 44)

9292V - 408848 = 6060V + 266640

3232V = 675488

V = 209 mph

5 0
2 years ago
Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill
laila [671]

Answer:

a)   m =1  θ = sin⁻¹  λ  / d,  m = 2        θ = sin⁻¹ ( λ  / 2d) ,   c)     m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

         d sin θ = m λ

the maximum for m = 1 is at the angle

          θ = sin⁻¹  λ  / d

the second maximum m = 2

          θ = sin⁻¹ ( λ  / 2d)

the third maximum m = 3

        θ = sin⁻¹ ( λ  / 3d)

the fourth maximum m = 4

       θ = sin⁻¹ ( λ  / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

       I = I₀ cos² (Ф) (sin x / x)²

       Ф = π d sin θ /λ

       x = pi a sin θ /λ

where a is the width of the slits

with the values ​​of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference       d sin θ = m λ

first diffraction minimum    a sin θ = λ

we divide the two expressions

                       d / a = m

In our case

                   3a / a = m

                    m = 3

order three is no longer visible

7 0
2 years ago
How does a sound wave's amplitude affect the sound we hear? How does a sound
damaskus [11]

Answer:

cause we hear differently

Explanation:

7 0
3 years ago
Read 2 more answers
A capacitance C and an inductance L are operated at the same angular frequency.
Levart [38]

A) \omega = \frac{1}{\sqrt{LC}}

The magnitude of the capacitive reactance is given by

X_C = \frac{1}{\omega C}

where

\omega is the angular frequency

C is the capacitance

While the magnitude of the inductive capacitance is given by

X_L = \omega L

where L is the inductance.

Since we want the two reactances to be equal, we have

X_C = X_L

So we find

\frac{1}{\omega C}= \omega L\\\omega^2 = \frac{1}{LC}\\\omega = \frac{1}{\sqrt{LC}}

B) 7449 rad/s

In this case, we have

L=5.30 mH = 5.3\cdot 10^{-3}H is the inductance

C= 3.40 \mu F= 3.40 \cdot 10^{-6}F is the capacitance

Therefore, substituting in the formula for the angular frequency, we find

\omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(5.30\cdot 10^{-3}H)(3.40\cdot 10^{-6} F)}}=7449 rad/s

C) 39.5 \Omega

Now we can us the formulas of the reactances written in part A). We have:

- Capacitive reactance:

X_C = \frac{1}{\omega C}=\frac{1}{(7449 rad/s)(3.40\cdot 10^{-6}F)}=39.5 \Omega

- Inductive reactance:

X_L = \omega L=(7449 rad/s)(5.30\cdot 10^{-3}H)=39.5 \Omega

7 0
3 years ago
Why the efficiency of normal light is smaller than solar light ??
stellarik [79]

Answer:

part of energy is wasted in heat because of resistance in the filament (and that's how it glows)

8 0
3 years ago
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