Answer:
(a) Acceleration = 0.1063 m/s^2 (Second team wins)
(b) Tension in rope = 65.106 N
Explanation:
Total mass of first team = 68 * 9 = 612 kg
Total force of first team = 1350 * 9 = 12150 N
Total mass of second team = 73 * 9 = 657 kg
Total force of seconds team = 1365 * 9 = 12285 N
Difference in force = 12285 - 12150 = 135 N (towards the second team as it has more force)
(a) For acceleration we get:
F = m * a
135 = (mass of both teams) * a
a = 135 / (612 + 657)
acceleration = 0.1063 m/s^2 (Second team wins)
(b) Since we know the acceleration of the first team (pulling being pulled towards the second team at an acceleration of 0.1063 m/s^2) , we can find out the force required to move them:
Force required for first team = mass of first team * acceleration
Force required = 612 * 0.1063
Force required = 65.106 N
This is the force exerted on the first team through the rope, so the tension in the rope will also be 65.106 N.
The equation of the car is given by the equation,
x(t) = 2.31 + 4.90t² - 0.10t⁶
If we are going to differentiate the equation in terms of x, we get the value for velocity.
dx/dt = 9.8t - 0.6t⁵
Calculate for the value of t when dx/dt = 0.
dx/dt = 0 = (9.8 - 0.6t⁴)(t)
The values of t from the equation is approximately equal to 0 and 2.
If we substitute these values to the equation for displacement,
(0) , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31
(2) , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51
Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters.
Answer:
73.5 m/s
Explanation:
The position of the first ball is:
y = y₀ + v₀ t + ½ at²
y = h + (0)(18) + ½ (-9.8)(18)²
y = h − 1587.6
The position of the second ball is:
y = y₀ + v₀ t + ½ at²
y = h + (-v) (18−6) + ½ (-9.8)(18−6)²
y = h − 12v − 705.6
Setting the positions equal:
h − 1587.6 = h − 12v − 705.6
-1587.6 = -12v − 705.6
1587.6 = 12v + 705.6
882 = 12v
v = 73.5
The second ball is thrown downwards with a speed of 73.5 m/s
Answer:
The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
Explanation:
From the question given above, the following data were obtained:
Height to which the target is located = 50 m
Initial velocity (u) = 20 m/s
To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 10 m/s²
Maximum height (h) =?
v² = u² – 2gh (since the ball is going against gravity)
0² = 20² – (2 × 10 × h)
0 = 400 – 20h
Collect like terms
0 – 400 = – 20h
– 400 = – 20h
Divide both side by – 20
h = – 400 / – 20
h = 20 m
Thus, the the maximum height to which the cannon ball attained is 20 m.
From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
Assuming that you have a triangular prism, the ray of light will undergo refraction twice. The first time is the transition from air to flint glass on the entry face, and the second time is the transition from the flint glass to air from the exit face. With the available data, there are two possible solution since saying "20Âş from the normal" isn't enough information. Depending upon which side of the normal that 20 degrees is, the interior triangle will have the angles of 35, 90-r, and 55+r, or 35, 90+r, 55-r degrees where r is the angle from the normal after the 1st refraction. I will provide both possible solutions and you'll need to actually select the correct one based upon the actual geometry which I don't know because you didn't provide the figure or diagram that you were provided with.
The equation for refraction is:
(sin a1)/(sin a2) = n1/n2
where
a1,a2 = angles from the normal to the surface.
n1,n2 = index of refraction for the transmission mediums.
For this problem, we've been given an a1 of 20Âş and an n1 of 1.60. For n2, we will use air which at STP has an index of refraction of 1.00029. So
(sin a1)/(sin a2) = n1/n2
(sin 20)/(sin a2) = 1.00029/1.60
0.342020143/(sin a2) = 0.62518125
0.342020143 = 0.62518125(sin a2)
0.547073578 = sin a2
asin(0.547073578) = a2
33.16647891 = a2
So the angle from the normal INSIDE the prism is 33.2Âş. The resulting angle from the surface of the entry face will be either 90-33.2 or 90+33.2 depending upon the geometry. So the 2 possible triangles will be either 35Âş, 56.8Âş, 88.2Âş or 35Âş, 123.2Âş, 21.8Âş. with a resulting angle from the normal of either 1.8Âş or 68.2Âş. I can't tell you which one is correct since you didn't tell me which side of the normal the incoming ray came from. So let's calculate both possible exits.
1.8Âş
(sin a1)/(sin a2) = n1/n2
(sin 1.8)/(sin a2) = 1.6/1.00029
0.031410759/(sin a2) = 1.599536135
0.031410759= 1.599536135(sin a2)
0.019637418= sin(a2)
asin(0.019637418) = a2
1.125213477 = a2
68.2Âş
(sin a1)/(sin a2) = n1/n2
(sin 68.2)/(sin a2) = 1.6/1.00029
0.928485827/(sin a2) = 1.599536135
0.928485827 = 1.599536135(sin a2)
0.58047193 = sin a2
asin(0.58047193) = a2
35.48374252 = a2
So if the interior triangle is acute, the answer is 1.13Âş and if the interior triangle is obtuse, the answer is 35.48Âş
