In the laboratory you will find three bottles with the following labels:

gives the speed v versus time t for a 0.500 kg object of radius 6.00 cm that rolls smoothly down a 30° ramp. The scale on the ve

ddd [48]

Answer:

Rotational inertia of the object is given as

$I = 7.2 \times 10^{-4} kg m^2$

Explanation:

As we know that the acceleration of the object on inclined plane is given as

$a = \frac{gsin\theta}{1 + k^2/R^2}$

now we know that velocity at any instant of time is given as

$v = at$

now we know that if the graph between velocity and time is given then the slope of the graph will be same as acceleration

so here we have

$\frac{gsin\theta}{1 + k^2/R^2} = slope$

now from the graph slope of the graph is given as

$slope = \frac{3.5 - 0}{1}$

$\frac{gsin\theta}{1 + k^2/R^2} = 3.5$

$\frac{9.81 sin30}{1 + k^2/R^2} = 3.5$

$k^2 = 0.4 R^2$

now rotational inertia is given as

$I = mk^2$

$I = 0.5(0.4)(0.06)^2$

$I = 7.2 \times 10^{-4} kg m^2$

8 0

3 years ago

A guitar string vibrates 600 times in 2 seconds.

vazorg [7]

frequency is equal to number of oscillations or vibrations upon time

therefore,

check picture

4 0

2 years ago

A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin

astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

$v_{1x} = 406 mph\\v_{1y} = 0$

While the components of the velocity of the wind are

$v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph$

So the components of the resultant velocity of the jet are

$v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph$

And the new speed is the magnitude of the resultant velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph$

6 0

3 years ago

Three people pull simultaneously on a stubborn donkey. jack pulls eastward with a force of 92.5 n, jill pulls with 89.9 n in the

alekssr [168]

Jack------------ force of 92.5 n eastward-------Fjack(X)=92.5 n Fjack(Y)=0

<span>jill ------------------------------- force of 89.9 n northeast
Fjill(X)=cos45*89.9=63.57 n
</span>Fjill(Y)=sin45*89.9=63.57 n<span>

</span>jane -----------------------------force of 163 n southeast
Fjane(X)=cos45*163=115.26 n
Fjane(X)=-sin45*163=-115.26 n

Ftotal (X)=92.5+63.57+115.26=271.33 n
Ftotal (Y)=0+63.57-115.26=-51.69 n

Fotal=((271.33)^2+(-115.26)^2) ^0.5=294.80 n southeast

the magnitude of the net force the people exert on the donkey. is 294.80 n southeast

7 0

3 years ago

What is Potential Energy

grandymaker [24]

Answer:

energy that is ponteaila

Explanation:

8 0

3 years ago

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