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3 years ago

A student practicing for a cross country meet runs 250 m in 30 s. What is her average speed?

Physics

2 answers:

Novay_Z [31]3 years ago

6 0

It's 8.3 m per second

a_sh-v [17]3 years ago

5 0

Just divide it... and you get the answer

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Pirates drag a treasure chest to the left across a sandy beach. In which direction does the treasure chest experience a friction

yawa3891 [41]

Answer:

Chest experiences the friction forces towards right

Explanation:

Friction forces can be defined as a force a body experiences when it is made to slide against the surface. Friction force always acts to stop the the body which is moving. To stop the body, it always acts in the opposite directions to the motion of the body. Therefore, if the treasure chest is dragged across a sandy beach to the left, the frictional forces will act in the right direction.

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2 years ago

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Sholpan [36]

Answer:

Second option: High frequency and low amplitude

Explanation:

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2 years ago

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andrey2020 [161]

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3 years ago

What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m

pychu [463]

Complete question:

A taut rope has a mass of 0.123 kg and a length of 3.54 m. What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m if the waves are to travel at 28.0 m/s ?

Answer:

The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.

Explanation:

Velocity = Frequency X wavelength

V = Fλ ⇒ F = V/λ

F = 28/0.6 = 46.67 Hz

Angular frequency (ω) = 2πF = 2π (46.67) = 93.34π rad/s

Average power supplied to the rope will be calculated as follows

$P_{avg} =\frac{1}{2} \mu \omega^2 A^2 V$

where;

ω is the angular frequency

A is the amplitude

V is the velocity

μ is mass per unit length = 0.123/3.54 = 0.0348 kg/m

$P_{avg} =\frac{1}{2} ( 0.0348)(93.34 \pi )^2 (0.2)^2 (28)$ = 1676.159 watts

The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.

6 0

3 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude

Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

$a=\frac{1.2^2}{0.400}=3.6 m/s^2$

7 0

3 years ago