Explanation:
We have,
The initial position of an object is zero.
The starting velocity is 3 m/s and the final velocity was 10 m/s.
The object moves with constant acceleration..
The area covered under the velocity-time graph gives displacement of the object. The correct option is "the area of the rectangle plus the area of the triangle under the line".
a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):
It is given:
v = 8.5 m/s
m = 91 kg
So:
b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h
In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J
m = 91 kg
g = 9.81 m/s² - gravitational acceleration
h = ? - height
Now, let's replace those:
3,287.4= 91 * 9.81 * h
⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m
Answer:
-30° C
Explanation:
Data provided in the problem:
The formula for conversion as:
F = (9/5)C + 32
Now,
for the values of F = -22 , C = ?
Substituting the value of F in the above formula, we get
-22 = (9/5)C + 32
or
-22 - 32 = (9/5)C
or
(9/5)C = - 54
or
C = - 54 × (5/9)
or
C = - 30 °
Hence, -22 Fahrenheit equals to -30°C
Explanation:
We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.
Condition 1: The stars rise and set perpendicular to the horizon
The observer is at the equator
Condition 2: The stars circle the sky parallel to the horizon
The observer is at the Pole of the Earth
Condition 3: The celestial equator passes through the zenith
The observer is at the equator
Condition 4: In the course of a year, all stars are visible
The observer is at the equator
Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)
The observer is at North Pole
Answer:
The surface gravity g of the planet is 1/4 of the surface gravity on earth.
Explanation:
Surface gravity is given by the following formula:
So the gravity of both the earth and the planet is written in terms of their own radius, so we get:
The problem tells us the radius of the planet is twice that of the radius on earth, so:
If we substituted that into the gravity of the planet equation we would end up with the following formula:
Which yields:
So we can now compare the two gravities:
When simplifying the ratio we end up with:
So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.
