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3 years ago

9

How does wavelength and frequency affect the speed?

1 answer:

kvasek [131]3 years ago

3 0

Rather, an alteration in wavelength affects the frequency in an inverse manner. A doubling of the wavelength results in a halving of the frequency; yet the wave speed is not changed.

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If there is a doubt on a boundary call in tennis, you should do what?

Ronch [10]

Answer: Each player is responsible for all decisions in her half of the Court. She should be completely honest on all “calls” but, if in doubt, she should give her opponent the benefit of the doubt and play the ball as good. You should not play a let.

Explanation:

3 0

3 years ago

What would be the speed of a car if it covers a distance of 400 km in 5 hours?

iragen [17]

Answer:

80km/hr

Explanation:

Hope this helps! :>

p.s. is this right? i'm not 100% sure...:I

8 0

3 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

3 years ago

Your physics TA has a far point of 0.759 m from her eyes and is able to see distant objects in focus when wearing glasses with a

romanna [79]

Answer:

$d=0.019m$

Explanation:

From the question we are told that:

Far point $x=0.759m$

Refractive power $P=-1.35 D.$

Generally, the equation for Focal length is mathematically given by

$F=\frac{1}{P}$

$F=\frac{1}{-1.35}$

$F=-0.74m$

Therefore

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Where

$u=o$

$\frac{1}{-0.74}=\frac{1}{0}+\frac{1}{v}$

$v=-0.74m$

Therefore,The between her glasses and eyes

$d=x-v$

$d=0.759-0.74m$

$d=0.019m$

4 0

3 years ago

A 3000 kg freight train traveling at a rate of 2 m/s strikes a stationery 3000 kg freight car and couples up. What is the speed

Contact [7]

Answer:

1 m/s

Explanation:

Momentum before the collision = momentum after the collision

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(3000 kg) (2 m/s) + (3000 kg) (0 m/s) = (3000 kg) v + (3000 kg) v

6000 kgm/s = 6000 kg v

v = 1 m/s

5 0

4 years ago