Glenn shoots an arrow at a 30.0 degree angle. It has a velocity of 65.0 m/s How far will the arrow travel?

What are greenhouse gasses? How does carbon dioxide enter the atmosphere? (Site 1)

Alexxx [7]

Answer:

greenhouse gas is a gas that absorbs and emits radiant energy within the thermal infrared range

Explanation:

Animals and plants need to get rid of carbon dioxide gas through a process called respiration. Carbon moves from fossil fuels to the atmosphere when fuels are burned.

4 0

3 years ago

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Compare the geology of Callisto, Ganymede, and Titan.

V125BC [204]

Explanation: Ganymede, Callisto, Titan are the moons of outer planets or gaseous planets which are made up of ice and rock. Callisto is an ice-covered moon and has no inner or outer activity and is considered basically geologically dead. Ganymede has rocky core and shows signs of tectonic activity, including long cracks in the crust and regions of young surface terrain. Titan has active geology of liquid hydrocarbons on the surface, rain back onto the surface and evaporation into the atmosphere. It has similar size, composition and mass to Ganymede and Callisto.

8 0

3 years ago

I really need help with this!! Will give brainliest if you help T^T

slava [35]

Answer:

What inferences can you make about the melting points of the different substances and the motion of their particles, based on the data? (ignore that needed it here so i could see it better.)

Explanation:

Butter has a lower melting point than the cheese and the wax. The motion of the cheese were a little separated while the butter articles have more space in between. The wax had the closest particles.

I dont know if that makes sense?

5 0

3 years ago

An Olympic discus thrower (~100 kg) launches the 2.0 kg discus by spinning rapidly (~4 times per second) with arm outstretched (

vladimir1956 [14]

Answer:

F = 1263.03 N

Explanation:s

given,

mass of the disk thrower = 100 Kg

mass of the disk = 2 Kg

angular speed of the disk = 4 rev/s

arm outstretched = 1 m

centripetal force of the disk in the circular path

F = m ω² r

ω = 4 x 2 x π

ω = 25.13 rad/s

F = m ω² r

F = 2 x 25.13² x 1

F = 1263.03 N

hence, centripetal force equal to the F = 1263.03 N

6 0

3 years ago

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

atroni [7]

(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

4 0

3 years ago