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inn [45]
3 years ago
10

A high powered rifle can shoot a bullet at a speed of 1500 mi/hr. On the moon, with almost no atmosphere and an acceleration due

to gravity of 1.6 m/s2, a bullet shot straight into the air would take how long to return to the moon's surface? On Earth, a standard .30 caliber bullet fired from a rifle pointed straight up would rise to an altitude of 9,000 feet (2,743.2 meters) in 18 seconds, and then would return to Earth in another 31 seconds. How much influence does the atmosphere have on a bullet fired from the surface of the Earth?
Physics
1 answer:
Amiraneli [1.4K]3 years ago
6 0

Answer:

Explanation:

On the Moon :----

1500 x 1.6 = 2400 m /s is initial velocity of bullet .

g = 1.6 m /s²

v = u - gt

0 = 2400 - 1.6 t

t = 1500 s

This is time of ascent

Time of decent will also be the same

Total time of flight = 2 x 1500 = 3000 s

On the Earth : ---

v = u - a₁ t

0 = u - a₁ x 18

u = 18a₁

v² = u² - 2 x a₁ x 2743.2

0 = (18a₁ )² - 2 x a₁ x 2743.2

a₁ = 16.93

For downward return

s = ut + 1/2 a₂ x t²

2743.2 = 0 + .5 x a₂ x 31²

a₂ = 5.7 m /s²

If d be the deceleration produced by air

g + d = 16.93 ( during upward journey )

g - d = 5.7

g = (16.93 + 5.7) / 2  

= 11.315 m / s

d = 5.6 m /s²

So air is creating a deceleration of 5.6 m /s².

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Linear Velocity What is the linear velocity in cm>min for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10
daser333 [38]

3,89,988 cm/min is the linear velocity

Given,

Diameter of CD = 12 cm

So, Radius of CD = 6 cm

CD is spinning at 10350 rev/min

Firstly , convert rev/min into rad/min

1 rev = 2π radians

10350 rev/min = 10350 × 2π

                        = 64998 rad/min

Formula used,

v=rw where,

v is the Linear velocity

r is the radius

w is the angular velocity

v = 6 cm × 64998rad/min

  = 3,89,988 cm/min

Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.

Learn more about Angular speed here brainly.com/question/540174

#SPJ4

                 

8 0
2 years ago
A spaceship accelerates from 0 m/s to 60 m/s north in 3 seconds. What is the acceleration of the spaceship?
Vesna [10]
Acceleration = ( final velocity - intial velocity ) / time

= (60-0)m/s / 3s
= 20 m/s2
the answer is a
5 0
3 years ago
A person who is described as intersex is
labwork [276]
I believe it means they are born with both parts that a man and woman would be born with.
6 0
3 years ago
A train that has wheels with a diameter of 91.44 cm (36 inches used for 100 ton capacity cars) slows down from 82.5 km/h to 32.5
podryga [215]

Answer:

The answer is below

Explanation:

The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm

radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m

a) Initial angular velocity (\omega_o) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity  (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s

θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

angular acceleration (α) is:

\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\

b)

\omega=\omega_o+\alpha t\\\\19.67=50.13+(-3.89t)\\\\3.89t=50.13-19..67\\\\3.89t=30.46\\\\t=7.83\ s

c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

a) When it stops, the final angular velocity is 0. Hence:

\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)}  =112.4\ rev

θ = 323 rad

4 0
3 years ago
How much charge does a 9.0 v battery transfer from the negative to the positive terminal while doing 39 j of work?
sdas [7]
The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
W=q \Delta V
where q is the charge and \Delta V is the potential difference.

In our problem, the work done is W=39 J while the potential difference of the battery is \Delta V = 9.0 V, so we can find the charge transferred by the battery:
q= \frac{W}{\Delta V}= \frac{39 J}{9.0 V}=4.33 C
3 0
3 years ago
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