This assumption is correct. The emissivity of a substance is the ratio of the energy radiated from a body to the energy radiated from a black body (perfect emitter). Because these stars reflect negligible radiation compared to the amount they emit.
"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.
The weight of the object is
(mass) x (gravity in the place where the object is) .
On the surface of the Earth,
Weight = (60 kg) x (9.8 m/s²)
= 588 Newtons.
Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is
(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.
The object's weight would also be 0.04 of its weight on the surface.
(0.04) x (588 Newtons) = 23.52 Newtons.
Again, the object's mass is still 60 kg out there.
___________________________________________
If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink.
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
Answer: 1.4mi/hr²
Explanation:
Blank 1: We know that acceleration is the change in velocity/change in time
Using this, we know that the acceleration will be: (50-15)/25 = 1.4
Blank 2: mi/hr²
Answer:
See the explanation below
Explanation:
To solve this problem we must decompose the initial speeds into x & y.
![v_{o}_{x}=25*cos(45)=17.67[m/s]\\v_{o}_{y}=25*sin(45)=17.67[m/s]\\](https://tex.z-dn.net/?f=v_%7Bo%7D_%7Bx%7D%3D25%2Acos%2845%29%3D17.67%5Bm%2Fs%5D%5C%5Cv_%7Bo%7D_%7By%7D%3D25%2Asin%2845%29%3D17.67%5Bm%2Fs%5D%5C%5C)
The acceleration of gravity is equal to g = 9.81[m/s^2] downward.
The maximum height is when the velocity of the projectile is zero in the component y, that is, it will not be able to go higher, by means of the following kinematic equation we can find that time, for that specific condition.
a)
![v_{y}=(v_{y})_{0}+a*t\\0 = 17.67 - 9.81*t\\17.67 = 9.81*t\\t=1.8 [s]](https://tex.z-dn.net/?f=v_%7By%7D%3D%28v_%7By%7D%29_%7B0%7D%2Ba%2At%5C%5C0%20%3D%2017.67%20-%209.81%2At%5C%5C17.67%20%3D%209.81%2At%5C%5Ct%3D1.8%20%5Bs%5D)
Note: Acceleration is taken as negative as it is directed downwards.
b)
The position in the x component can be found using the following kinematic equation
![x=(v_{x})_{o}*t\\x=17.67*1.8\\x=31.82[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bx%7D%29_%7Bo%7D%2At%5C%5Cx%3D17.67%2A1.8%5C%5Cx%3D31.82%5Bm%5D)
The position in the y component can be found using the following kinematic equation
![y =(v_{y})_{o}*t+\frac{1}{2} *g*t^{2} \\y=17.67*1.8-0.5*9.81*(1.8)^{2}\\ y=15.91[m]](https://tex.z-dn.net/?f=y%20%3D%28v_%7By%7D%29_%7Bo%7D%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%20%5C%5Cy%3D17.67%2A1.8-0.5%2A9.81%2A%281.8%29%5E%7B2%7D%5C%5C%20y%3D15.91%5Bm%5D)
c)
Since the motion on the X-axis is at constant speed, there is no acceleration, so the only acceleration that exists is due to gravity
d)
In the attached image we can see, the projectile with the vectors of acceleration and velocity.
