Ask question

Ask question

All categories

1 year ago

12

A bullet is shot from a rifle with a speed of 720 m/s. What time is required for the bullet to strike a target 3240 meters away

?

2 answers:

Alik [6]1 year ago

7 0

4.5 seconds you devoured 3240/720=4.5

Nataliya [291]1 year ago

4 0

Answer:

Time taken, t = 4.5 seconds

Explanation:

Speed of the bullet, v = 720 m/s

Distance to be covered by the target, d = 3240 m

Let t is the time required for the bullet to strike a target. The time taken is calculated as :

$t=\dfrac{d}{v}$

$t=\dfrac{3240\ m}{720\ m/s}$

t = 4.5 seconds

So, the time required for the bullet to strike a target is 4.5 seconds. Hence, this is the required solution.

You might be interested in

A construction worker pushes a crate horizontally on a frictionless floor with a net force of 10\, \text 10N, start text, N, end

Strike441 [17]

Answer:

$k_f$ = 40J

Explanation:

The work and energy theorem says that:

$W_f =k_f-k_i$

where $W_f$ is the work of the force, $k_f$ the final kinetic energy and $k_i$ the initial kinetic energy.

Addittionally, the work of the force is calculate as force multiply by distance and if the crate inittialy is at rest, the initial kinetic energy is zero, so:

$Fd = k_f$

where F is the force and d the distance. Then, replacing values, we get:

$(10N)(4m) = k_f$

$40J = k_f$

it means that the system gain 40J of kinetic energy.

7 0

2 years ago

Two equal length of wire made of the same material but of different diameters have an effective resistance of 0.8 ohm when they

nata0808 [166]

Answer:

hfdfg

Explanation:

7 0

1 year ago

Two trucks with equal mass are attracted to each other with a gravitational force of 5.3 x 10 -4 N. The trucks are separated by

HACTEHA [7]

Answer:

7327 kg or 7.3 tons

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408*10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M = M_1 = M_2$ is the masses of the 2 objects. and R = 2.6m is the distance between them.

$F_G = 5.3*10^{-4}N$ is the attraction force.

$5.3*10^{-4} = 6.67408*10^{-11}\frac{M^2}{2.6^2}$

$7941265 = \frac{M^2}{2.6^2}$

$M^2 = 53682948.76$

$M = \sqrt{53682948.76} \approx 7327 kg$ or 7.3 tons

3 0

2 years ago

A mass is oscillating horizontally on a spring. At the locations A, B, C, D, and E, photogates are used to measure the speed of

svetoff [14.1K]

Complete Question

The complete question is is shown on the first uploaded

Answer:

The elastic potential energy at point B is $PE_{elastic} = 50J$

The kinetic energy at point D is $KE = 75J$

Explanation:

Looking at the given point we can observe that mechanically energy(i.e potential and kinetic energy ) is conserved and it value is $E_ {m} = 100J$

So at point B

$E_{m} = PE_{elastic} +KE$

$100 = PE_{elastic} + KE$

KE at point B is 50J

So $PE_{elastic} = 100 - 50 = 50J$

Now at point D

$E_{m} = PE_{elastic} +KE$

$PE_{elastic}$ at point D is 25J

So $KE = 100 - 25 = 75J$

4 0

1 year ago

A counter attendant in a diner shoves a ketchup bottle with a mass 0.30 kg along a smooth, level lunch counter. The bottle leave

balu736 [363]

Answer:

1.176 N

Explanation:

$m$ = mass of the bottle = 0.30 kg

$v_{o}$ = initial speed of the bottle = 2.8 m/s

$v$ = final speed of the bottle = 0 m/s

$d$ = stopping distance traveled = 1.0 m

$f$ = magnitude of frictional force acting on bottle

Using work-change in kinetic energy theorem

$- f d = (0.5) m (v^{2} - v_{o}^{2} )\\- f (1) = (0.5) (0.30) (0^{2} - 2.8^{2} )\\-f = - 1.176 \\f = 1.176$ N

direction :

frictional force acts in opposite direction of motion.

8 0

2 years ago