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slega [8]
3 years ago
11

A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad

ius 1.50 m at an angular speed of 6.00 rad/s. As the superhero swings the hammer, she then ascends vertically at a constant 2.00 m/s. What is the speed of the end of the hammer relative to the ground? What is the magnitude of acceleration of the end of the hammer? What is the direction of acceleration of the end of the hammer measured as an angle between a⃗and the direction toward the center of the circular path?
Physics
1 answer:
Korolek [52]3 years ago
3 0

Answer:

9.21954 m/s

54 m/s²

Angle is zero

Explanation:

r = Radius of arm = 1.5 m

\omega = Angular velocity = 6 rad/s

The horizontal component of speed is given by

v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s

The vertical component of speed is given by

v_v=2\ m/s

The resultant of the two components will give us the velocity of hammer with respect to the ground

v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s

The velocity of hammer relative to the ground is 9.21954 m/s

Acceleration in the vertical component is zero

Net acceleration is given by

a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2

Net acceleration is 54 m/s²

As the acceleration is towards the center the angle is zero.

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Explanation:

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3 0
3 years ago
Any help is appreciated please
stiv31 [10]

Answer:

C. It speeds up, and the angle increases

Explanation:

We can answer by using the Snell's law:

n_i sin \theta_i = n_r sin \theta_r

where

n_i, n_r are the refractive index of the first and second medium

\theta_i is the angle of incidence (measured between the incident ray and the normal to the surface)

\theta_r is the angle of refraction (measured between the refracted ray and the normal to the surface)

In this problem, light moves into a medium that has lower index of refraction, so

n_r < n_i

We can rewrite Snell's law as

sin \theta_r =\frac{n_i}{n_r}sin \theta_i

and since

\frac{n_i}{n_r}>1

this means that

sin \theta_r > sin \theta_i

which implies

\theta_r > \theta_i

so, the angle increases.

Also, the speed of light in a medium is given by

v=\frac{c}{n}

where c is the speed of light and v the refractive index: we see that the speed is inversely proportional to n, therefore the lower the index of refraction, the higher the speed. So, in this problem, the light will speed up, since it moves into a medium with lower index of refraction.

4 0
3 years ago
A light bulb connects to a battery. A second, identical light bulb connects in parallel to the first light bulb. The connecting
mr Goodwill [35]

Answer:d

Explanation:

Suppose V is the voltage of battery and R is the resistance of bulb

so Power drop for initial stage

P_1=\frac{V^2}{R}

When another identical bulb of same resistance is applied in parallel so voltage Drop across both the resistor will be same i.e. V

so Power consumed  P_2=\frac{V^2}{R}

so there is no change in power and hence no dip in brightness  

8 0
2 years ago
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