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Charra [1.4K]
3 years ago
14

A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point. . . Whats the bul

let's flight time?. Whats the bullet's speed as it left the barrel?
Physics
1 answer:
Marina CMI [18]3 years ago
5 0
The problem ask to calculate the bullet's flight time and the bullet's speed as it left the barrel. So base on the problem, the answer would be that the flight time is 0.076 seconds and the speed of the bullet is 657.9 m/s. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications. 
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The correct answer is A. Installation of rigid metal conduit requires grounding and the grounding equipment used may weaken the structure.
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Can I have help with this please- thank you!!!
seropon [69]

Answer:

I would say there is friction against the floor, air resistance, and gravity.

Explanation:

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The train slows down at the railroad crossing. Is that acceleration, velocity, speed, of none of the above?​
Vlad [161]

Answer:

none of the above

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im pretty positive this is the answer tell me if i am wrong please

4 0
3 years ago
a car, which has a mass of 2000kg traveled a distance of 200 meters in 5 seconds. After 20 seconds the car was raveling at a spe
iragen [17]

Answer:

Explanation:

vi = 200/5 = 40 m/s

a = (vf - vi)/t = (60 - 40)/20 = 1 m/s²

F = ma = 2000(1) = 2000 N

4 0
2 years ago
An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 μT. The electron follows a helical p
sladkih [1.3K]

Answer:

3.4 x 10⁴ m/s

Explanation:

Consider the circular motion of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of electron = 9.1 x 10⁻³¹ kg

v  = radial speed

r = radius of circular path = 2 mm = 0.002 m

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

For the circular motion of electron

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Consider the linear motion of the electron :

v' = linear speed

x = horizontal distance traveled = 9 mm = 0.009 m

t = time taken = \frac{2\pi m}{qB} = \frac{2\pi (9.1\times 10^{-31})}{(1.6\times 10^{^{-19}})(80\times 10^{-6})} = 4.5 x 10⁻⁷ sec

using the equation

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

Speed is given as

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s

6 0
3 years ago
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