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Yanka [14]
3 years ago
13

A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Negle

ct air resistance and use g=9.8m/sec2 as the acceleration of gravity.
Physics
1 answer:
Artist 52 [7]3 years ago
7 0

Answer:

θ₀ = 84.78° (OR) 5.22°

Explanation:

This situation can be treated as projectile motion. The parameters of this projectile motion are:

R = Range of Projectile = 150 m

V₀ = Launch Speed of Projectile = 90 m/s

g = 9.8 m/s²

θ₀ = Launch angle (OR) Angle of Elevation = ?

The formula for range of a projectile is given as:

R = V₀² Sin 2θ₀/g

Sin 2θ₀ = Rg/V₀²

Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²

2θ₀ = Sin⁻¹ (0.18)

θ₀ = 10.45°/2

<u>θ₀ = 5.22°</u>

Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:

θ₀ = 90° - 5.22°

<u>θ₀ = 84.78°</u>

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3 years ago
How much work is done when a 100 N force moves a block 59 m?
xxTIMURxx [149]

Answer:

5900J

Explanation:

Work=Forse*Distance

work = J, Jewls

100*59=5900

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3 0
3 years ago
A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses
stellarik [79]

Answer:

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

k = 7840 N/0.46 m

<u>k = 17043.5 N/m = 17.04 KN/m</u>

5 0
3 years ago
An apple from the top branch of the tree and an apple from the bottom branch of a tree fall at the same time. Which apple will h
Vedmedyk [2.9K]
B. The apple from the bottom will hit the ground earlier. This is because an increase in height causes an increase in the time that the object will fall, and therefore will affect the final velocity of the falling object. Moreover, the reduction in velocity due to friction from the air should also be considered.
3 0
3 years ago
A projectile is launched
Flauer [41]

The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

The given parameters;

  • initial horizontal velocity, vₓ = 16 m/s
  • initial vertical velocity, v_y =0
  • time interval 1 seconds

The components of the velocity can be horizontal or vertical velocity.

The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.

The vertical component of the velocity is calculated as;

v_y = v_0_y -gt\\\\v_y = 0 - (1\times 9.8)\\\\v_y = -9.8 \ m/s

The horizontal component of the velocity is constant since it is not affected by gravity.

The horizontal component of the velocity = 16 m/s

Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

Learn more here:brainly.com/question/20349275

4 0
3 years ago
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