Can someone please help me it would mean alot

An electric field of 6.1x 10^5 N/C is directed along a charge 2.9 x 10^-19 C. What

Aloiza [94]

Answer:

Electric field is a function 1/r^2

Explanation:

4 0

3 years ago

A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a

Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

$mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}$

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}$

From Newton's second law

$F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}$

Squarring both sides

$112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m$

The height from which the student fell is 5.7141 m

5 0

3 years ago

Select the correct answer.

Serhud [2]

Answer:

b hope this helps

Explanation:

5 0

3 years ago

Read 2 more answers

A 6.89-nC charge is located 1.76 m from a 4.10-nC point charge. (a) Find the magnitude of the electrostatic force that one charg

iogann1982 [59]

Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive

Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:

$F=\frac{k*q1*q2}{d^{2}}$

Replacing the dat we obtain F=82 nN.

The force is repulsive because the points charged have the same sign.

5 0

3 years ago

In 1898, the world land speed record was set by Gaston Chasseloup-Laubat driving a car named Jeantaud. His speed was 39.24 mph (

FromTheMoon [43]

Answer:

the acceleration $a^{\to} = (0.0159 \ \ m/s^2 )i$

Explanation:

Given that:

the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.

So; the acceleration for the first 6 miles can be calculated by using the formula:

v₂² = v₁² + 2a (Δx)

Making acceleration a the subject of the formula in the above expression ; we have:

v₂² - v₁² = 2a (Δx)

$a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}$