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Jet001 [13]
3 years ago
9

PLEASE HELP - In running a ballistics test at the police department, Officer Rios fires a 6.1g bullet at 350 m/s into a containe

r that stops it in 0.32 min. What average force stops the bullet?​
Physics
2 answers:
Arlecino [84]3 years ago
8 0

Answer:

–0.11 N.

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 6.1 g

Initial velocity (u) = 350 m/s

Time (t) = 0.32 mins

Final velocity (v) = 0 m/s

Force (F) =?

Next, we shall convert 0.32 mins to seconds. This can be obtained as follow:

1 min = 60 s

Therefore,

0.32 mins = 0.32 × 60 = 19.2 s

Next, we shall determine the acceleration of the the bullet. This can be obtained as follow:

Initial velocity (u) = 350 m/s

Time (t) = 19.2 s

Final velocity (v) = 0 m/s

v = u + at

0 = 350 + (a × 19.2)

0 = 350 + 19.2a

Collect like terms

0 – 350 = 19.2a

–350 = 19.2a

Divide both side by 19.2

a = –350 / 19.2

a = –18.23 m/s²

Next, we shall convert 6.1 g to Kg.

1000 g = 1 Kg

Therefore,

6.1 g = 6.1 g × 1 Kg / 1000 g

6.1 g = 0.0061 Kg

Finally, we shall determine the force. This can be obtained as follow:

Mass (m) = 0.0061 Kg

Acceleration (a) = –18.23 m/s²

Force (F) =?

F = ma

F = 0.0061 × –18.23

F = –0.11 N

NOTE: The negative sign indicates that the force is in opposite direction.

vfiekz [6]3 years ago
7 0

Answer:

–0.11 N.

Explanation:

Hope this helps

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5 0
3 years ago
- A cannon of 2000 kg fires a shell of 10 kg at
fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

p=0

After the shot, the momentum is:

p=MV+mv

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

0=MV+mv

And solving for V, we find the velocity of the cannon:

V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

v^2-u^2=2as

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

a=-0.25 m/s^2 is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m

7 0
4 years ago
The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32
lutik1710 [3]

Answer:

5N

Explanation:

Given parameters:

Original length = 22cm

Spring constant, K  = 50N/m

New length = 32cm

Unknown

Force applied  = ?

Solution:

The force applied on a spring can be derived using the expression below;

   Force  = KE

 k is the spring constant

 E is the extension

  extension = new length - original length

  extension  = 32cm  - 22cm  = 10cm

convert the extension from cm to m;  

   100cm  = 1m;

    10cm will give 0.1m

So;

  Force  = 50N/m x 0.1m  = 5N

8 0
3 years ago
If a honeybee is flying 7.0 m/s, what is the kinetic energy (Joules) if its mass is 0.1 g?
svetlana [45]

Answer:

KINETIC ENERGY=K.E=0.00245 Joules ≈ 0.003 Joules

Explanation:

mass=m=0.1g=0.0001kg

velocity=v=7.0m/s

kinetic energy=K.E.=?

as we know that

kinetic energy=\frac{1}{2}mv^2\\ putting values\\kinetic energy=\frac{1}{2}(0.0001kg)(7.0m/s)^2\\ kinetic energy=\frac{1}{2} 0.0049joules\\kinetic energy= 0.00245Newton meter or joules

hope it will help ^_^

5 0
3 years ago
A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?
allsm [11]

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side.  See more details further below.

Explanation:

Start with the formula for pressure

(pressure P) = (Force F) / (Area A)

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

F_g = m \cdot g

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3

The density of water is 997 kg/m^3, so the mass of the barrel is:

m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg

and so the weight is

F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N

and so the pressure is

P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.  

7 0
3 years ago
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