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3 years ago

A bolt of lightning discharges 9.7 C in 8.9 x 10^-5 s. What is the average current during the discharge?

Physics

1 answer:

Anastaziya [24]3 years ago

3 0

Answer: $1.089\times 10^5\ A$

Explanation:

Given

Charge discharged $Q=9.7\C$

time taken $t=8.9\times 10^{-5}\ s$

Current is given as rate of change of discharge i.e.

$\Rightarrow I=\dfrac{Q}{t}\\\\\Rightarrow I=\dfrac{9.7}{8.9\times 10^{-5}}\\\\\Rightarrow I=1.089\times 10^5\ A$

Therefore, the average current is $1.089\times 10^5\ A$

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A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in ab

Darya [45]

Answer:

a. $F=126N$

b. $E_K=1323J$

Explanation:

Given:

$m=54kg$

$v=7 m/s$

$t= 3s$

The runner force average to find given the equations

$F=m*a$

$a=\frac{v}{t}$

$F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}$

$F=126N$

Work done by the system by this force so

$W=F*d$

$W=E_K$

$E_K=\frac{1}{2}*m*v^2$

$E_K=\frac{1}{2}*54kg*(7m/s)^2$

$E_K=1323J$

6 0

3 years ago

The lightest car in the world was built in London and had a mass of less than 10 kg. it's maximum speed was 25.0 km/h. Suppose t

Softa [21]

25km/h = 6.94 m/s
suvat
s=16
u=6.94
v=0
a=a
v^2=u^2+2as
(v^2-u^2)/2s = a =1.5ms^-2

6 0

3 years ago

Describe what would happen if you rubbed a mineral with a Mohs hardness value of 7 against a mineral with a value of 5?

chubhunter [2.5K]

The mineral with Mohs hardness would be scratched because the mineral with Mohs 7 hardness is stronger than the Mohs 5 mineral. Eventually, that mineral would turn into dust if you kept rubbing it.

8 0

3 years ago

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

5 0

3 years ago

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

3 years ago