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2 years ago

12

John see Hua running towards him at 11m/s. While running, Hua throws a ball to john at 5m/s. What is the speed of the ball as ob

served by John?

1 answer:

Sav [38]2 years ago

7 0

6 m/s HUUUUUUUUUUUAAAAAAAAAAAAAAAAAAAAAAAH

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A bug on the surface of a pond is observed to move up and down a total vertical distance of 6.5 cm , from the lowest to the high

m_a_m_a [10]

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to = 4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω² .................1

and kinetic energy is directly proportional to square of the amplitude.

so

$\frac{KE2}{KE1} = \frac{A2^2}{A1^2}$ .............2

$\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}$

$\frac{KE2}{KE1}$ = 0.52284

so factor that bug maximum KE change is 0.52284

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3 years ago

Two satellites, A and B are in different circular orbits

jek_recluse [69]

Answer:

The ratio of the orbital time periods of A and B is $\frac{1}{2}$

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

$v_{oA} = 2v_{oB}$ (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

$v_{o} = \farc{2\piR}{T}$

where

R = Radius of the orbit

Now,

For satellite A:

$v_{oA} = \farc{2\piR}{T_{a}}$

Using eqn (1):

$2v_{oB} = \farc{2\piR}{T_{a}}$ (2)

For satellite B:

$v_{oB} = \farc{2\piR}{T_{b}}$ (3)

Now, comparing eqn (2) and eqn (3):

$\frac{T_{a}}{T_{b}} = \farc{1}{2}$

6 0

2 years ago

After a model rocket reached its maximum height, it then took 5.0 seconds to return to the launch site. what is the approximate

kifflom [539]

Air resistance is ignored.
g = 9.8 m/s².
At maximum height, the vertical velocity is zero.

Let h = the maximum height reached.
Let u = the vertical launch velocity.

Because ot takes 5.0 seconds to reach maximum height, therefore
(u m/s) - (9.8 m/s²)*(5 s) = 0
u = 49 m/s

The maximum height reached is
h = (49 m/s)*(5 s) - (1/2)*(9.8 m/s²)*(5 s)²
= 122.5 m

Answer: 122.5 m

3 0

3 years ago

A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec

kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, $\lambda=\dfrac{\lambda_ox_o}{x}$

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

$dE=\dfrac{k\ dq}{x^2}$ ..........(1)

The linear charge density is given by :

$\lambda=\dfrac{dq}{dx}$

$dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx$

Integrating equation (1) from x = x₀ to x = infinity

$E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx$

$E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}$

$E=\dfrac{k\lambda_o}{2x_o}$

Hence, this is the required solution.

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2.5 better than the first one more game to go hard but I 46 the bucks

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