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miss Akunina [59]
3 years ago
11

This is for Criminal Law:

Physics
1 answer:
tigry1 [53]3 years ago
8 0
No it not because
Jill didn’t have actually have the gun
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The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t
slamgirl [31]

Answer:

The charge is 2.75\times10^{-13}\ C

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length L=5.0\times5.0\times10^{-4}\ m

We need to calculate the linear charge density

Using formula of linear charge density

E=\dfrac{2k\lambda}{r}

\lambda=\dfrac{Er}{2k}

Put the value into the formula

\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}

\lambda=1.1\times10^{-10}\ C/m

We need to calculate the charge

Using formula of charge

Q=\lambda\timesL

Put the value into the formula

Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})

Q=2.75\times10^{-13}\ C

Hence, The charge is 2.75\times10^{-13}\ C

5 0
3 years ago
. Một con lắc đơn có chiều dài dây treo 1m dao động điều hoà treo trong một xe chạy trên mặt phẳng nghiêng
lyudmila [28]

Answer:

C

Explanation:

hope for help ....im expert

6 0
3 years ago
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Radar use reflected microwaves to detect objects and measure their... what?
Sloan [31]
Height and depth..... for sure....

7 0
4 years ago
The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force
yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given  

N = W Cos30

N = (500) Cos30

N = 433 N

3 0
3 years ago
A 0.5 kg block of aluminum (caluminum=900j/kg⋅∘c) is heated to 200∘c. the block is then quickly placed in an insulated tub of co
Alex_Xolod [135]

To solve this problem, we should recall the law of conservation of energy. That is, the heat lost by the aluminium must be equal to the heat gained by the cold water. This is expressed in change in enthalpies therefore:

- ΔH aluminium = ΔH water

where ΔH = m Cp (T2 – T1)

The negative sign simply means heat is lost. Therefore we calculate for the mass of water (m):

- 0.5 (900) (20 – 200) = m (4186) (20 – 0)

m = 0.9675 kg

 

Using same mass of water and initial temperature, the final temperature T of a 1.0 kg aluminium block is:

- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)

- 900 T + 180,000 = 4050 T

4950 T = 180,000

T = 36.36°C

 

The final temperature of the water and block is 36.36°C

4 0
4 years ago
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