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Verdich [7]
3 years ago
13

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

know v0 and R, determine the general expressions for the two distinct launch angles θ1 and θ2 that will allow the projectile to hit D. For v0 = 42 m/s and R = 70 m, determine numerical values for θ1 and θ2?

Physics
1 answer:
NikAS [45]3 years ago
7 0

Answer:

Theta1 = 12° and theta2 = 168°

The solution procedure can be found in the attachment below.

Explanation:

The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).

In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.

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What determines color in the visible light region of the electromagnetic spectrum?
anyanavicka [17]

Answer:

In the electromagnetic spectrum, "Blue" is the "highest" frequency color of visible light.

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Hope this helps!

7 0
4 years ago
A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 k
Reika [66]

Answer:

179.47m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

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u1 and u2 are the initial velocities

v is the final velocity

Substitute

7750(179)+72(230) = (7750+72)v

1,387,250+16560 = 7822v

1,403,810 = 7822v

v = 1,403,810/7822

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Hence the final velocity of the probe is 179.47m/s

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3 years ago
PHYSICS HELP PLS
MrRissso [65]

Answer:

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8 0
3 years ago
How much sweat (in ml ) would you have to evaporate per hour to remove the same amount of heat a 150 w light bulb produces in an
allsm [11]
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7 0
3 years ago
Problem: At the local swimming hole, a favorite trick is torun
denis23 [38]

Answer:

2 revolutions

Explanation:

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s = gt^2/2

where s = 8.3 m is the distance that she falls, t is the time it takes to fall, which is what we are looking for

t^2 = \frac{2s}{g} = \frac{2*8.3}{9.8} = 1.694

t = \sqrt{1.694} = 1.3 s

Since she rotates with an average angular speed of 1.6rev/s. The number of revolutions she would make within 1.3s is

rev = 1.3 * 1.6 = 2 revolution

8 0
4 years ago
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