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andrey2020 [161]
3 years ago
15

Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of

100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)
Physics
1 answer:
Molodets [167]3 years ago
8 0

Answer:

C is the correct answer

Explanation:

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You are on an airplane traveling 30° south of due west at 180 m/s with respect to the air. The air is moving with a speed 31 m/s
BlackZzzverrR [31]
1) 211m/s
2)240<span>°
3)759,600m or 759.6 km</span>
6 0
3 years ago
A parked car begins to roll down a hill, what can you conclude from that observation?
Fynjy0 [20]

Answer:

its The rolling friction is greater than the force of the car’s weight against the hill.

and A force was required to start the car rolling.

Explanation:

3 0
3 years ago
Given:A=6x-2y B:-4x-8y C:-3x+9y. Commute A+B-C
DedPeter [7]
<span>A+B-C
</span><span>A = 6x - 2y
B = -4x - 8y
C = -3x + 9y

(</span>6x - 2y) + (-4x - 8y) - (-3x + 9y)
(6x - 2y) + (-4x - 8y) + (3x - 9y)
2x -10y + (3x - 9y)

5x - 19y
8 0
3 years ago
What is the wavelength that corresponds to a frequency of 6.00x1014 hz?
kipiarov [429]
The basic relationship between wavelength \lambda, frequency f and speed c of an electromagnetic wave is 
\lambda=  \frac{c}{f}
where c is the speed of light. Substituting numbers, we find:
\lambda=  \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{6.0 \cdot 10^{14} Hz}=5\cdot 10^{-7} m
4 0
3 years ago
A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2. Compu
boyakko [2]

Answer:

0.42 m/s²

Explanation:

r = radius of the flywheel = 0.300 m

w₀ = initial angular speed = 0 rad/s

w = final angular speed = ?

θ = angular displacement = 60 deg = 1.05 rad

α = angular acceleration = 0.6 rad/s²

Using the equation

w² = w₀² + 2 α θ

w² = 0² + 2 (0.6) (1.05)

w = 1.12 rad/s

Tangential acceleration is given as

a_{t} = r α = (0.300) (0.6) = 0.18 m/s²

Radial acceleration is given as

a_{r} = r w² = (0.300) (1.12)² = 0.38 m/s²

Magnitude of resultant acceleration is given as

a = \sqrt{a_{t}^{2} + a_{r}^{2}}

a = \sqrt{0.18^{2} + 0.38^{2}}

a = 0.42 m/s²

8 0
3 years ago
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