Answer:
a = 3.125 [m/s^2]
Explanation:
In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.


where:
Vf = final velocity = 25 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 8 [s]
The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.
25 = 0 + a*8
a = 3.125 [m/s^2]
Answer:
The correct answer to the question is
B. It always decreases
Explanation:
To solve the question, we note that the foce of gravity is given by
where
G= Gravitational constant
m₁ = mass of first object
m₂ = mass of second object
r = the distance between both objects
If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have
= 
Therefore the gravitational force is halved. That is it will always decrease
Answer:
Earth: 22.246 N
Moon: 3.71 N
Jupiter: 58.72 N
Explanation:
The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.
To calculate anything, we need to convert to standard measurements.
5.00 lbs = 2.27 kg
On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N
Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N
Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N
Answer:
The magnification of an astronomical telescope is -30.83.
Explanation:
The expression for the magnification of an astronomical telescope is as follows;

Here, M is the magnification of an astronomical telescope,
is the focal length of the eyepiece lens and
is the focal length of the objective lens.
It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.
Put
and
in the above expression.

M=-30.83
Therefore, the magnification of an astronomical telescope is -30.83.
We know that
g = LcosΘ
<span>where g, L and Θ are centripetal gravity length, and angle of object
</span><span>ω² = g/LcosΘ </span>
<span>ω = √(g / LcosΘ) </span>