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3 years ago

Why do objects that are thrown or shot follow a curved path

1 answer:

8 0

The initial force of the throw overcomes gravity quite easily. Then, gravity begins to bring it back down to earth, making a curved path.

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What angle is formed by the sun, the earth, and the moon during an eclipse?.

Andrew [12]

Answer:

The Sun-Earth-Moon system happens to exhibit a striking geometric coincidence, which we examine in the first problem. PROBLEM 1. To an observer on Earth, the Sun and the Moon subtend almost the same angle in the sky. The average angle is 0.52 degrees for the Moon and 0.53 degrees for the Sun.

4 0

2 years ago

An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis.

zhenek [66]

Answer:

2 m/s and -2 m/s

Explanation:

The object travels with an angle of

60.0°

with the positive direction of the y-axis: this means that it lies either in the 1st quadrant (positive x) or in the 2nd quadrant (negative x).

If it lies in the 1st quadrant, the value of vx (component of v along x direction) is:

$v_x = v cos \theta = (4.0 m/s) cos 60.0^{\circ}=2 m/s$

If it lies in the 2nd quadrant, the value of vx (component of v along x direction) is:

$v_x = -v cos \theta = -(4.0 m/s) cos 60.0^{\circ}=-2 m/s$

5 0

3 years ago

What type of bulb is unaffected

emmasim [6.3K]

c) Parallel

Nothing happens to the brightness of the light bulbs in the parallel circuit if the power supply is capable of supplying the additional current.

3 0

2 years ago

By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d

Kipish [7]

Answer:

The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

Explanation:

Given;

rock concert sound intensity level, β₁ = 120 dB

whisper sound intensity level, β₂ = 20 dB

The sound intensity level is given as;

$\beta = 10Log(\frac{I}{I_o} )\\\\$

where;

I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²

I is the sound intensity

Intensity of sound at rock concert ;

$120 = 10Log(\frac{I}{10^{-12}} )\\\\12 = Log(\frac{I}{10^{-12}} )\\\\10^{12} = \frac{I}{10^{-12}}\\\\I = 10^{12} * 10^{-12}\\\\I = 10^0\\\\I = 1 \ W/m^2$

The intensity of sound of a whisper;

$20 = 10Log(\frac{I}{10^{-12}} )\\\\2 = Log(\frac{I}{10^{-12}} )\\\\10^{2} = \frac{I}{10^{-12}}\\\\I = 10^{2} * 10^{-12}\\\\I = 10^{-10} \ W/m^2\\\\$

Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper

$\frac{I_{Concert}}{I_{whisper}} = \frac{1}{10^{-10}} \\\\\frac{I_{Concert}}{I_{whisper}} = 1 * 10^{10}\\\\I_{Concert} = 1 * 10^{10}*I_{whisper}$

Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

3 0

3 years ago

A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval, the angular velocity d

Nikitich [7]

Answer:

The angular displacement of the wheel is 45 radians

Explanation:

Given;

initial angular velocity, ω₀ = 20 rad/s

final angular velocity, ωf = 10 rad/s

time interval, t = 5

Angular acceleration is calculated as;

$\alpha = \frac{\omega _f - \omega_0}{t} \\\\\alpha = \frac{10 -20}{5} \\\\\alpha = -2 \ rad/s^2$

|α| = 2 rad/s²

Angular displacement is calculated as;

$\theta = \omega_0 \ + \ \frac{1}{2} \alpha t^2\\\\\theta = 20 \ + \ \frac{1}{2} *(2)*5^2\\\\\theta = 20 \ + 25\\\\ \theta = 45 \ radians$

Therefore, the angular displacement of the wheel is 45 radians

8 0

3 years ago