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3 years ago

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What happens when an object is dropped? A. It decelerates at a rate of 9.8 m/s/s. B. It travels 9.8 meters before stopping. C. I

t accelerates at a rate of 9.8 m/s/s. D. It falls at a constant rate of 9.8m/s/s.

1 answer:

andreev551 [17]3 years ago

5 0

I think its c I really don't take physics but I try to help.

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A playground merry-go-round of radius R = 2.20 m has a moment of inertia I = 260 kg · m2 and is rotating at 12.0 rev/min about a

MrRa [10]

Answer:

The new angular speed of the merry-go-round is 8.31 rev/min.

Explanation:

Because the merry-go-round is rotating about a frictionless axis there’re not external torques if we consider the system merry-go-round and child. Due that we can apply conservation fo angular momentum that states initial angular momentum (Li) should be equal final angular momentum (Lf):

$L_f=L_i$ (1)

The initial angular momentum is just the angular momentum of the merry-go-round (Lmi) that because it's a rigid body is defined as:

$L_i=L_{mi}=I\omega_i$ (2)

with I the moment of inertia and ωi the initial angular speed of the merry-go-round

The final angular momentum is the sum of the final angular momentum of the merry-go-round plus the final angular momentum of the child (Lcf):

$L_f=L_{mf}+L{cf}=I\omega_f+L{cf}$ (3)

The angular momentum of the child should be modeled as the angular momentum of a punctual particle moving around an axis of rotation, this is:

$L{cf}=mRv_f$ (4)

with m the mass of the child, R the distance from the axis of rotation and vf is final tangential speed, tangential speed is:

$v_f=\omega_f R$ (5)

(note that the angular speed is the same as the merry-go-round)

using (5) on (4), and (4) on (3):

$L_f=I\omega_f+m\omega_f R^2$ (6)

By (5) and (2) on (1):

$I\omega_f+m\omega_f R^2=I\omega_i$

Solving for ωf (12.0 rev/min = 1.26 rad/s):

$\omega_f= \frac{I\omega_i}{]I+mR^2}=\frac{(260)(1.26)}{260+(24.0)(2.20)^2}$

$\omega_f=0.87\frac{rad}{s}=8.31 \frac{rev}{min}$

8 0

3 years ago

If a diver-in-training is put into a pressurized suit, by how much would the pressure have to be raised (in atmospheres) above a

Sever21 [200]

This is an incomplete question, here is a complete question.

The bulk modulus for bone is 15.0 (GPa). If a diver-in-training is put into a pressurized suit, by how much would the pressure have to be raised (in atmospheres) above atmospheric pressure to compress her bones by 0.130% of their original volume?

Answer : The change in pressure will be, $1.95\times 10^7Pa$

Explanation : Given,

Bulk modulus = $15.0GPa=15.0\times 10^9Pa$

Change in volume = 0.130 % of original volume

Let the original volume be, V

So, Change in volume = $\frac{0.130}{100}\times V$

Formula used for change in pressure is:

$\Delta P=\beta \frac{\Delta V}{V}$

Now put all the given values in this formula, we get:

$\Delta P=(15.0\times 10^9Pa)\times \frac{(\frac{0.130}{100}\times V)}{V}$

$\Delta P=1.95\times 10^7Pa$

Thus, the change in pressure will be, $1.95\times 10^7Pa$

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4 years ago

When the forces applied to an object act in the same direction, the net force is

love history [14]

Sum of the individual forces

5 0

3 years ago

The only force acting on a 1.9 kg canister that is moving in an xy plane has a magnitude of 3.9 N. The canister initially has a

Hunter-Best [27]

Answer:

The work done on the canister is 15.34 J.

Explanation:

Given;

mass of canister, m = 1.9 kg

magnitude of force acting on x-y plane, F = 3.9 N

initial velocity of canister in positive x direction, $v_i$ = 3.9 m/s

final velocity of the canister in positive y direction, $v_j = 5.6 \ m/s$

The change in the kinetic energy of the canister is equal to net work done on the canister by 3.9 N.

ΔK.E = $W_{net}$

ΔK.E $= K.E_f -K.E_i$

The initial kinetic energy of the canister;

$K.E_i = \frac{1}{2} mv_i^2\\\\K.E_i = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_i = \frac{1}{2} *1.9(\sqrt{3.9^2 +0^2 + 0^2}\ )^2 = 14.45 \ J$

The final kinetic energy of the canister;

$K.E_f =\frac{1}{2} mv_j^2 \\\\K.E_f = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_f = \frac{1}{2} *1.9(\sqrt{0^2 +5.6^2 + 0^2}\ )^2 = 29.79 \ J$

ΔK.E = 29.79 J - 14.45 J

ΔK.E = $W_{net}$ = 15.34 J

Therefore, the work done on the canister is 15.34 J.

5 0

3 years ago

Where do incident rays that are parallel to the principal axis converge to after reflecting off a concave mirror?

morpeh [17]

Answer:

At focus

Explanation:

A concave mirror is converging in nature. In a mirror, concave in nature, the rays which are parallel to the principal axis are supposed to be coming from very large distances or we assume the source to be placed at infinity for such rays which are parallel to the principal axis.

These rays, parallel to the principal axis, coming from infinity, converges at the focus of the mirror concave in nature after reflecting from the concave mirror

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3 years ago