Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:
we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:
Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:
t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:
So once we got the two distances we can now find the difference between them:
so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
A gravitational force between objects depends on two things- their masses and the distance between them. So the greater the mass and the less distance there is, the more gravitational force and is the mass is less and the distance is great the gravitational force is weak
Answer:
<em>2.753*10^-11N</em>
Explanation:
According to Newton's law of gravitation, the force between the masses is expressed as;
F = GMm/d²
M and m are the distances
d is the distance between the masses
Given
M = 3.71 x 10 kg
m = 1.88 x 10^4 kg
d = 1300m
G = 6.67 x 10-11 Nm²/kg
Substitute into the formula
F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²
F = 46.52*10^(-6)/1.69 * 10^6
F = 27.53 * 10^{-6-6}
F = 27.53*10^{-12}
F = 2.753*10^-11
<em>Hence the gravitational force between the asteroid is 2.753*10^-11N</em>
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Answer:
66.26 m/s
Explanation:
Horizontal velocity, Vx = 55.3 m/s
Vertical velocity, Vy = 36.5 m/s
The value of the resultant velocity is given by the vector sum of the two velocities which are acting at 90°.
V = 66.26 m/s
Thus, the velocity of the vehicle is 66.26 m/s along its descent path.