Solving this using the time, we know that range = horizontal velocity x time of flight
since
there are no horizontal forces acting on the ball, there are no
horizontal accelerations and the initial horizontal velocity of 36 cos
28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have
range = 36 cos 28 x 3.44 s = 109.3 m
When the dust is too thick to penetrate with visible light, such as the Nebula, Radio Waves are used to penetrate the dust. Longer radio waves can completely penetrate the thick cloud cover, allowing scientists to beam radar waves.
Answer:
Mass = 18.0 kg
Explanation:
From Hooke's law,
F = ke
where: F is the force, k is the spring constant and e is the extension.
But, F = mg
So that,
mg = ke
On the Earth, let the gravitational force be 10 m/
.
3.0 x 10 = k x 5.0
30 = 5k
⇒ k =
................ 1
On the Moon, the gravitational force is
of that on the Earth.
m x
= k x 5.0
= 5k
⇒ k =
............. 2
Equating 1 and 2, we have;
= 
m = 
= 18.0
m = 18.0 kg
The mass required to produce the same extension on the Moon is 18 kg.
Answer:
The temperature is 
Explanation:
From the question ewe are told that
The rate of heat transferred is 
The surface area is 
The emissivity of its surface is 
Generally, the rate of heat transfer is mathematically represented as

=> ![T = \sqrt[4]{\frac{P}{e* \sigma } }](https://tex.z-dn.net/?f=T%20%20%3D%20%20%5Csqrt%5B4%5D%7B%5Cfrac%7BP%7D%7Be%2A%20%5Csigma%20%7D%20%7D)
where
is the Boltzmann constant with value 
substituting value
![T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }](https://tex.z-dn.net/?f=T%20%20%3D%20%20%5Csqrt%5B4%5D%7B%5Cfrac%7B13.1%7D%7B%200.287%2A%205.67%20%2A10%5E%7B-8%7D%20%7D%20%7D)
