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4 years ago

Section 1 - Question 2

Physics

2 answers:

Gemiola [76]4 years ago

5 0

Answer:

C. Electromagnetic waves from the sun have the same wavelength, but more energy.

Explanation:

Electromagnetic waves from the sun are far more dangerous due to the fact they have some much more energy. They're the same wavelength which is why they can be considered the same type of wave.

monitta4 years ago

3 0

D. Electromagnetic waves from the sun have the same wavelength, but less energy.

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An object has an angular velocity. It also has an angular acceleration due to torques that are present. Therefore, the angular v

shusha [124]

Answer:

α = 0 , w = w₀

Explanation:

Torque is related to angular acceleration by Newton's second law for rotational motion.

τ = I α

Where τ is the torque, I the moment of inertia and α the angular acceleration.

If we apply an external torque for the sum of all torques to be zero, the angular acceleration must fall to zero

α = 0

Since the acceleration is zero, the angular velocity you have at that time is constantly killed.

w = w₀ + α t

w = w₀ + 0

8 0

3 years ago

A playground ride requires 21 seconds to make one complete revolution

luda_lava [24]

The angular speed of the playground ride is determined as 0.3 rad/s.

<h3>What is angular speed?</h3>

Angular speed is the rate at which an object changes it angles which we measure in radians in a given time.

<h3>Angular speed of the ride</h3>

The angular speed of the ride if the ride makes one complete revolution is calculated as follows;

ω = θ/t

ω = 2π/t

where;

ω is angular speed of the ride
t is time of motion of the ride

one complete revolution = 2π radians

ω = 2π/21

ω = 0.3 rad/s

Thus, the angular speed of the playground ride is determined as 0.3 rad/s.

Learn more about angular speed here: brainly.com/question/24158647

#SPJ1

The complete question is below;

A playground ride requires 21 seconds to make one complete revolution, what is angular speed of the ride in radian per second.

3 0

2 years ago

If all the stars in an elliptical galaxy traveled random directions in their orbits, the elliptical galaxy would be type

Lemur [1.5K]

The answer would be E7. Galaxies categorized as E0 look to be nearly perfect, while those registered as E7 seem much extended than they are widespread. It is worth noting, though, that a galaxy's look is connected to how it lies on the sky when viewed from Earth. An E7 galaxy is very long and thin or the flattest of them all.

7 0

3 years ago

Help ASAP plssssssssss

garik1379 [7]

C. 32400 because 540 * 60 = 32400

5 0

3 years ago

A particle's position is given by x = 7.00 - 9.00t + 3t2, in which x is in meters and t is in seconds. (a) What is its velocity

spin [16.1K]

$x(t)=7.00\,\mathrm m-\left(9.00\dfrac{\rm m}{\rm s}\right)t+\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2$

a. The particle has velocity at time $t$ ,

$\dfrac{\mathrm dx(t)}{\mathrm dt}=-9.00\dfrac{\rm m}{\rm s}+\left(6\dfrac{\rm m}{\mathrm s^2}\right)t$

so that after $t=1\,\mathrm s$ it will have velocity $\boxed{-3.00\dfrac{\rm m}{\rm s}}$ .

b. The sign of the velocity is negative, so it's moving in the negative $x$ direction.

c. Its speed is 3.00 m/s.

d. The particle's velocity changes according to

$\dfrac{\mathrm d^2x(t)}{\mathrm dt^2}=6\dfrac{\rm m}{\mathrm s^2}$

which is positive and indicates the velocity/speed of the particle is increasing.

e. Yes. The velocity is increasing at a constant rate. Solving for $\dfrac{\mathrm dx(t)}{\mathrm dt}=0$ is trivial; this happens when $\boxed{t=1.50\,\mathrm s}$ .

f. No, the velocity is positive for all $t$ beyond 1.50 s.

5 0

3 years ago