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3 years ago

If Joe rides his bike for 15 minutes at 20 km/hour south, how far has he ridden? (Give the answer in km) Watch the units!

Physics

1 answer:

Virty [35]3 years ago

3 0

Answer:

Distance covered by Joe, d = 5 km

Explanation:

It is given that,

Speed of Joe, v = 20 km/hour

Time taken by Joe, t = 15 minutes = 0.25 h

We need to find the distance he is covering. It can be calculated using definition of speed as :

$v=\dfrac{d}{t}$

$d=v\times t$

$d=20\times 0.25$

d = 5 km

So, Joe covers a distance of 5 km. Hence, this is the required solution.

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What is the relationship between radio waves and the visible spectrum

evablogger [386]

Answer:

They both are part of electromagnetic radiation.

Radio waves have longer wavelength than visible waves.

Radio waves have lower frequency than visible waves.

Explanation:

8 0

2 years ago

Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth,

jeka57 [31]

Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = $-\frac{GMm}{R}$

Therefore: W = Energy at earth surface - $\frac{GMm}{R}$

Energy at earth surface (E) at an altitude of 5R = $-\frac{GMm}{5r} +\frac{1}{2}mV^2$

But $V=\sqrt{\frac{GM}{5R} }$

Therefore: $E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2= -\frac{GMm}{5R}+\frac{GMm}{10R} = -\frac{GMm}{10R}$

W = E - PE

$W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}$

7 0

3 years ago

Urgent! Which is not an example of work? Question 2 options: pushing a box across the floor picking up a box off the floor raisi

OLga [1]

Answer:

trying to push a rock that never moves

Explanation:

8 0

2 years ago

PLEASE HELP ASAP!!!

kirza4 [7]

Answer:

0.5mv^2=50, v=5, 25/2×m=50, m=50×2/25, So, the answer is 4

6 0

3 years ago

A star is moving away from an observer at 1% of the speed of light. At what wavelength would the observer find an emission line

Ivan

Answer:

λ = 5940 Angstroms

Explanation:

This is an exercise of the relativistic Doppler effect

f’= f √((1- v / c) / (1 + v / c))

Where the speed in between the strr and the observer is positive if they move away

Let's use the relationship

c = λ f

f = c /λ

We replace

c /λ’ = c /λ √ ((1- v / c) / (1 + v / c))

λ = λ’ √ ((1- v / c) / (1 + v / c))

Let's calculate

v = 0.01 c

v = 0.01 3 10⁸

v= 3 10⁶ m / s

λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]

λ = 6000 √ [0.99 / 1.01]

λ = 5940 Angstroms

6 0

3 years ago