Compare the kinetic and potential energies of a 400 kg box being moved

A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni

dimaraw [331]

Answer:

The magnitude of the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})$

Put the value into the formula

$E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})$

$E=5.75\ N/C$

The direction is toward positive x- axis.

Hence, The magnitude of the electric field is 5.75 N/C towards positive x- axis.

7 0

2 years ago

A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck trave

Dennis_Churaev [7]

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

20 = 0 + 2 t

t = 20 /2 = 10 s .

Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

Distance travelled in first 10 s

S₁ = ut + 1/2 a t²

= 0 + .5 x 2 x 10²

= 100 m

Distance travelled in next 20 s

S₂= 20s x 20 m/s = 400 m

Distance travelled in last 5 s .

deceleration in last 5 s

v = u + at

0 = 20 m/s + a x 5

a = - 4 m/s²

v² = u² - 2 a s

0 = (20 m/s)² - 2 x 4 m/s² x s

s = 50 m

S₃ = 50 m

Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

3 0

3 years ago

A man at point A directs his rowboat due north toward point B, straight across a river of width 100 m. The river current is due

prohojiy [21]

Answer:

1.35208 m/s

Explanation:

Speed of the boat = 0.75 m/s

Distance between the shores = 100 m

Time = Distance / Speed

$Time=\frac{100}{0.75}=133.33\ s$

Time taken by the boat to get across is 133.33 seconds

Point C is 150 m from B

Speed = Distance / Time

$Speed=\frac{150}{\frac{100}{0.75}}=1.125\ m/s$

Velocity of the water is 1.125 m/s

From Pythagoras theorem

$c=\sqrt{0.75^2+1.125^2}\\\Rightarrow c=1.35208\ m/s$

So, the man's velocity relative to the shore is 1.35208 m/s

3 0

3 years ago

A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of

Leto [7]

Answer: $2.27\ m/s^2$

Explanation:

Given

Length of the race track $L=200\ m$

the radius of curvature of the track $r=29.5\ m$

time taken to run on track is $t=24.4\ s$

Speed of runner is

$\Rightarrow v=\dfrac{L}{t}=\dfrac{200}{24.4}\\\\\Rightarrow v=8.196\ m/s$

Centripetal acceleration is

$\Rightarrow a_c=\dfrac{v^2}{r}=\dfrac{8.196^2}{29.5}\\\\\Rightarrow a_c=2.27\ m/s^2$

5 0

3 years ago

| C₄H10<br> Number of hydrogen atoms?

Sunny_sXe [5.5K]

That's a molecule of Butane.

It's made out of 4 Carbon atoms and 10 Hydrogen atoms.

3 0

2 years ago