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3 years ago

7

In which of these cases is a waiter doing work on the object

1 answer:

Gnom [1K]3 years ago

8 0

Section 2 is right,, i think. good luck

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Sharon the ant (Aaron’s sister) sits at the edge of a turntable of radius R that is spinning with period T. As she makes one-hal

Dmitry_Shevchenko [17]

Answer:

$a = \dfrac{4\pi^2R}{T^2}$

Explanation:

The acceleration of a circular motion is given by

$a = \omega^2 R$

where $\omega$ is the angular velocity and $R$ is the radius.

Angular velocity is related to the period, T, by

$\omega=\dfrac{2\pi}{T}$

Substitute into the previous formula.

$a = (\dfrac{2\pi}{T})^2 R$

$a = \dfrac{4\pi^2R}{T^2}$

This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.

6 0

3 years ago

High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g

tamaranim1 [39]

Explanation:

It is given that,

Mass of golf club, m₁ = 210 g = 0.21 kg

Initial velocity of golf club, u₁ = 56 m/s

Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg

After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.

Initial momentum of golf ball, $p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s$

After the collision, final momentum $p_f=0.21\ kg\times 42\ m/s+0.046v$

Using the conservation of momentum as :

$p_i=p_f$

$11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v$

v = 63.91 m/s

So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.

3 0

3 years ago

Here is a graph of speed vs time. If the object is moving to the east, which BEST describes the speed and velocity of the graph?

Artemon [7]

Answer:

Both speed and velocity are changing.

Explanation:

They are both going up so both are changing

5 0

3 years ago

Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23

Sloan [31]

Answer:

The ratio is $\frac{T_1}{T_2} = 3.965$

Explanation:

From the question we are told that

The radius of Phobos orbit is R_2 = 9380 km

The radius of Deimos orbit is $R_1 = 23500 \ km$

Generally from Kepler's third law

$T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }$

Here M is the mass of Mars which is constant

G is the gravitational constant

So we see that $\frac{ 4 * \pi^2 }{G * M } = constant$

$T^2 = R^3 * constant$

=> $[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3$

Here $T_1$ is the period of Deimos

and $T_1$ is the period of Phobos

So

$[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}$

=> $\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]$

=> $\frac{T_1}{T_2} = 3.965$

8 0

4 years ago

According to the inverse square law of light, how will the apparent brightness of an object change if its distance to us triples

aleksley [76]

According to the inverse square law of light, <span>apparent brightness will decrease by a factor of 9. Use the formula </span> $B=L/(4*pD^2)$ , to check it.

4 0

3 years ago