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SOVA2 [1]
3 years ago
14

What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75

Physics
1 answer:
likoan [24]3 years ago
6 0

Answer:

1800 m/s

Explanation:

The equation is v = fλ

λ= 0.75

f = 2400 Hz

V = 2400 × 0.75

V = 1800 m/s

[ you did not give units for wavelength, I assumed it would be m/s]

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A motorcycle is speeding along at a velocity of120kph.Then, it changes to a velocity of 150 kph for 2 minutes.What is the motorc
Taya2010 [7]

Answer:

Explanation:

a=v-u/t

a=acceleration

v=final velocity

u=initial velocity

t=tme taken

we need to convert from kph to ms⁻¹

v= 150*1000/60*60= 41.67ms⁻¹

u= 120*1000/60*60= 33.33ms⁻¹

t= 2*60= 120s

a=41.67-33.33/120

a=8.34/120

a=0.0694ms⁻²

8 0
2 years ago
Betty (mass 40 kg), standing on slippery ice, catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Show that the
MA_775_DIABLO [31]

Answer:

v = 2.18m/s

Explanation:

In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.

Then you have:

Mv_{1o}+mv_{2o}=(M+m)v        (1)

M: mass Betty = 40kg

m: mass of the dog = 15kg

v1o: initial speed of Betty = 3.0m/s

v2o: initial speed of the dog = 0 m/s

v: speed of both Betty and her dog = ?

You solve the equation (1) for v:

v=\frac{Mv_{1o}+mv_{2o}}{M+m}=\frac{(40kg)(3.0m/s)+(15kg)(0m/s)}{40kg+15kg}\\\\v=2.18m/s

The speed fo both Betty and her dog is 2.18m/s

7 0
2 years ago
What is the force of a object that has a mass of 7 kg and an acceleration of 6 m/s/s
UkoKoshka [18]

Answer:

<h2>42 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 7 kg

acceleration = 6 m/s²

We have

force = 7 × 6 = 42

We have the final answer as

<h3>42 N</h3>

Hope this helps you

7 0
2 years ago
Paola can flex her legs from a bent position through a distance of 20.1 cm. Paola leaves the ground when her legs are straight,
makkiz [27]

The third equation of free fall can be applied to determine the acceleration. So that Paola's acceleration during the flight is 39.80 m/s^{2}.

Acceleration is a quantity that has a direct relationship with velocity and also inversely proportional to the time taken. It is a vector quantity.

To determine Paola's acceleration, the third equation of free fall is appropriate.

i.e V^{2} = U^{2} ± 2as

where: V is the final velocity, U is the initial velocity, a is the acceleration, and s is the distance covered.

From the given question, s = 20.1 cm (0.201 m), U = 4.0 m/s, V = 0.

So that since Poala flies against gravity, then we have:

V^{2} = U^{2} - 2as

0 = (4)^{2} - 2(a x 0.201)

  = 16 - 0.402a

0.402a = 16

a = \frac{16}{0.402}

  = 39.801

a = 39.80 m/s^{2}

Therefore Paola's acceleration is 39.80 m/s^{2}.

Visit: brainly.com/question/17493533

7 0
2 years ago
Read 2 more answers
Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}}), where imax=\frac{E}{R}

Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

⇒\frac{L}{R}=\frac{2}{log2}

Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

5 0
3 years ago
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