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LenaWriter [7]
3 years ago
10

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20×105m/s2 for 8.10×10

-4s. What is its muzzle velocity (that is, its final velocity)?
Physics
1 answer:
kati45 [8]3 years ago
7 0

Answer:

502.2 m/s

Explanation:

initial velocity, u = 0 m/s

Let the final speed is v.

acceleration, a = 6.2 x 10^5 m/s^2

time taken, t = 8.10 x 10^-4 s

The muzzle speed is the final speed.

Use first equation of motion

v = u + a t

v = 0 + 6.2 x 10^5 x 8.10 x 10^-4

v = 502.2 m/s

Thus, the muzzle speed of the bullet is 502.2 m/s.

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A wave with a wavelength of 125 meters is moving at a speed of 20 m/s. What is it’s frequency
IceJOKER [234]

Answer:

0.16Hz

Explanation:

wavelength (λ) = 125 meters

speed (V) = 20 m/s

frequency (F) = ?

Recall that frequency is the number of cycles the wave complete in one

second. And its value depends on the wavelength and speed of the wave.

So, apply the formula V = F λ

Make F the subject formula

F = V / λ

F = 20 m/s / 125 meters

F = 0.16 Hz

8 0
3 years ago
The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20 m/s. find th
hram777 [196]

maximum allowed value of the speed in roller coaster is given as

v = 20 m/s

now from kinematics we can say

v^2 - v_i^2 = 2 a s

here initial speed will be

v_i = 0

acceleration is due to gravity

a = 9.8 m/s^2

now we can use this to find the height

20^2 - 0^2 = 2 * 9.8* h

400 = 19.6 *h

h = 20.4 m

so maximum allowed height will be 20.4 m

4 0
3 years ago
A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.
mestny [16]

Answer:

a = 17.68 m/s²

Explanation:

given,

length of the string, L = 0.8 m

angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

radial acceleration = ?

first we have to calculate the radius of the circle

 R = L sin θ

 R = 0.8 x sin 61°

 R = 0.7 m

now, calculating at the angular velocity

\omega =\dfrac{2\pi}{T}

\omega =\dfrac{2\pi}{1.25}

  ω = 5.026 rad/s

now, radial acceleration

 a = r ω²

 a = 0.7 x 5.026²

a = 17.68 m/s²

hence, the radial acceleration of the ball is equal to 17.68 rad/s²

7 0
3 years ago
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

6 0
3 years ago
Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.
GalinKa [24]
Wave D has the same wavelength as wave A, but the amplitude is lower. The answer is Wave D.
7 0
3 years ago
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