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anastassius [24]

2 years ago

5

A fish is swimming around the 720-meter perimeter of her pond. If she swims 10 laps in 120 minutes, what is her average speed in

m/min

2 answers:

natulia [17]2 years ago

6 0

Her average speed is 60 m/min

zavuch27 [327]2 years ago

5 0

Answer:

60

Explanation:

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A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t

Mashutka [201]

As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
$f = (\frac{g + v_{r}}{g + v_{s}})f_{o}$ -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
$v_{r}$ = Velocity of the receiver/observer relative to the medium = ?.
$v_{s}$ = Velocity of the source with respect to medium = 0 m/s.
$f_{o}$ = Frequency emitted from source = 400 Hz.
$f$ = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

$408 = ( \frac{340 + v_{r}}{340 + 0})*400$

$\frac{408}{400} = \frac{340 + v_{r}}{340}$

Solving above would give you,
$v_{r}$ = 6.8 m/s

The correct answer = 6.8m/s

7 0

2 years ago

I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =

Helen [10]

Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)

7 0

2 years ago

A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, th

jek_recluse [69]

Answer:

$a)v_{f}=0.88m/s$

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

$\\ p_{1}=p_{2}\\m_{ball}*v_{o.ball}+m_{girl}*v_{o.girl} = m_{ball}*v_{f.ball} + m_{girl}*v_{f.girl}$ (1)

At the beginning the girl is stationary:

$v_{o.girl}=0m/s$ (2)

If the girl catch the ball, both have the same speed:

$v_{f.girl}=v_{f.ball}=v_{f}$ (3)

We replace (2) and (3) in (1):

$m_{ball}*v_{o.ball} = (m_{ball}+m_{girl})*v_{f} \\$

We can now solve the equation for v_{f}:

$v_{f}=\frac{m_{ball}*v_{o.ball}}{(m_{ball}+m_{girl})}=\frac{15*4.7}{15+65}=0.88m/s$

4 0

2 years ago

Floor lamps usually have a base with large inertia, while the long body and top have much less inertia. Part A If you want to sh

Novay_Z [31]

Answer:

When a an object is been rotated its resistance capacity to that rotational force is know as rotational inertia and this mathematically given as

$I = mr^2$

Where m is the mass

r is the rotation radius

For the spinning of the lamp as a baton to work the location of the center of mass of the floor lamp needs to be located

This is more likely to be located closer to base of the lamp as compared to the top, so success of spinning a floor lamp like a baton is highly likely if the lamp is grabbed closer to the base because that is where the position of its center of mass is likely to be.

Explanation:

5 0

2 years ago

A squirrel in a tree drops an acorn. how long does it take the acorn to fall 20 feet?

mart [117]

We use the equation of motion,

$S= ut+\frac{1}{2}at^{2}$

Here, S is the height, u is initial velocity and a is acceleration.

Given, $S = 20 \ ft$ $S = 20 \ ft = 20 \times\frac{1 \ m}{3.2808399 ft} = 6.096 \ m$

As acorn falls from tree, therefore we take the value of $a = 9.8 \ m/s^2$ and initial velocity $u = 0$ .

Substituting these values in equation of motion,

$6.096 \ m = 0 \times t +\frac{1}{2} \times 9.8 m/s^2 (t)^2 \\\\\ t = 1.12 \ s$

Thus, the time taken by the acorn to fall 20 feet ( 6.096 m ) is 1.12 s.

5 0

2 years ago