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1 year ago

3

A rifle can shoot a 4.00 g bullet at a speed of 998 m/s. Find the kinetic energy of the bullet. What work is done on the bullet

if it starts from rest?

1 answer:

Svet_ta [14]1 year ago

5 0

Answer:

1992.008J

Explanation:

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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by

aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

$x(t)=\alpha t^2 - \beta t^3$

where

$\alpha = 1.50 m/s^2$

$\beta = 0.05 m/s^3$

The average velocity is given by the ratio between the displacement and the time taken:

$v=\frac{\Delta x}{\Delta t}$

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

The position at t = 2.00 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

So the displacement is

$\Delta x = x(2)-x(0)=5.6-0=5.6 m$

The time interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

And so, the average velocity in this interval is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

The position at t = 0 is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

While the position at t = 4.00 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = x(4)-x(0)=20.8-0=20.8 m$

The time interval is

$\Delta t = 4.0 - 0 = 4.0 s$

So the average velocity here is

$v=\frac{20.8}{4.0}=5.2 m/s$

c)

The position at t = 2 s is:

$x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m$

While the position at t = 4 s is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

So the displacement is

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

$\Delta t = 4.0 - 2.0 = 2.0 s$

So the average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

6 0

2 years ago

Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.91 J1.91 J of work to set the ball s

Black_prince [1.1K]

Answer:

ω = 4.07 rad/s

Explanation:

By conservation of the energy:

W = ΔK

$1.91J = I/2*\omega^2$

where $I = 2/3*m*R^2=0.23kg.m^2$

Solving for ω:

$\omega = \sqrt{W*2/I} =4.07rad/s$

7 0

2 years ago

(a) A small object with mass 3.75 kg moves counterclockwise with constant speed 1.55 rad/s in a circle of radius 2.55 m centered

Eddi Din [679]

Answer:3.95 m/s

Explanation:

Given

mass of object $m=3.75 kg$

$\omega =1.55 rad/s$

radius of circle $=2.55 m$

initial Position $r=2.55 \hat{i}$

angular displacement $\theta _0=8.95 rad$

8.95 radian can be written as

$1.42 (2\pi )$

i.e. Particle is at first quadrant with $\theta =0.4242\pi \times \frac{180}{\pi }$

$\theta =76.36^{\circ}$

(c)velocity is $v=\omgea \times r$

$v=1.55\times 2.55=3.95 m/s$

8 0

2 years ago

A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-s

laiz [17]

Answer:

30298514.82 m/s

Explanation:

M = Mass of star = 2×10³ kg

r = Radius of star = 5×10³ m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$a=G\frac{M}{r^2}\\\Rightarrow a=6.67\times 10^{-11}\frac{2\times 10^{30}}{5\times 10^3}\\\Rightarrow a=2.7\times 10^{16}\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 2.7\times 10^{16}\times 0.017+0^2}\\\Rightarrow v=30298514.82\ m/s$

The object would be moving at a velocity of 30298514.82 m/s

5 0

2 years ago

Read 2 more answers

a motorist traveling at 18m/s approaches traffic lights when he is 30 m from the stop line they turn red it takes 0.7 s before h

Gelneren [198K]

Answer:

35.2m

Explanation:

S= V^2 - U^2 ÷ 2A

S= (0) - (18) ÷ 2(4.6)

S= 324 ÷ 92

S= 35.2m

7 0

2 years ago