A car is stopped at a red light. When the light changes, the car begins to accelerate at a rate of 10 m/s^2 and it continues to
1 answer:
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The gravitational force acting on the man is 800 N towards the Earth's centre
Explanation:
The weight of an object on the Earth is exactly the gravitational force exerted by the Earth on the object.
The gravitational force exerted by the Earth on an object located at the Earth's surface is given by:
where
G is the gravitational constant
M is the Earth's mass
m is the mass of the object
R is the radius of the Earth
And the direction of the force is towards the Earth's centre.
Since G, M and R are constant, they are grouped into a single constant called g, acceleration of gravity:
therefore the gravitational force can be rewritten as
And this is the usual equation that we use to calculate the weight of an object.
Therefore, weight and gravitational force acting on an object on Earth are the same thing: so, the gravitational force acting on the man is equal to his weight, 800 N, and it acts towards the Earth's center.
Learn more about weight and gravitational force:
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Answer:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.
where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing:
Answer: 31.6ft
Explanation:
Check the attachment for the diagram.
According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|
that is 20ft - 10ft = 10ft
According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.
|AE|^2 + |EC|^2 = |AC|^2
10^2 + 30^2 = |AC|^2
100 + 900 = |AC|^2
|AC| = √1000
|AC| = 31.6ft
Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.
