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3 years ago

What would happen? Why?

Physics

1 answer:

koban [17]3 years ago

6 0

Lightning rods provide low resistance paths to the ground that’ll be used to conduct the enormous electrical currents when lightning strikes happen. So the system will attempt to carry the harmful electrical current away from it and safely to ground. The system will have the ability to handle enormous electrical currents associated with the lightning strikes, if they contact a material that isn’t a good conductor then the material will suffer massive heat damage. So the lightning rod system is considered the best conductor & this allows the current to flow to the ground without causing any massive heat damage.

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When falling through the air, which of the following objects will hit the ground first?

Pavel [41]

Answer:

a penny

Explanation:

5 0

3 years ago

Read 2 more answers

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

2 years ago

A wheelbarrow is a complex machine that combines which simply marchines?

koban [17]

It's most likely the combination of a bucket and the wheel.

4 0

3 years ago

Read 2 more answers

When a clock reads 5.0 seconds a cart's velocity in the negative x direction is 3.0 m/s. when the clock reads 6.0 seconds, the c

DIA [1.3K]

Impulse is the change of momentum before and after the collision:
I = mv₁ - mv₂ = m (v₁-v₂)
m mass
v velocity

Impuls also is defined as the average force during a short period of time:
I = F * t = m * a * t
F average force
t time difference
m mass
a average acceleration

Combined:
m(v₁- v₂) = m * a * t
a = (v₁- v₂) / t
= (4m/s - (-3m/s)) / 6s - 5s
= 7m/s / 1s
= 7 m/s²

7 0

3 years ago

An automobile of mass 2500 kg moving at 49.4 m/s is braked suddenly with a constant braking force of 8,868 N. How far does the c

BaLLatris [955]

Answer:

344.68 m

Explanation:

The computation of the far does the car travel before stopping is

Data provided in the question

Force = F = 8,868 N

mass = m = 2,500 kg

So,

accleration = a is

$= \frac{-F}{m}\\\\\= \frac{8868}{2500}$

a = -3.54 m/s^2

The initial speed = u = 49.4 m / s

final speed = v = 0

Based on the above information

Now applying the following formula

v^ 2- u^ 2= 2aS

Therefore

$S = \frac{v^ 2- u^ 2}{2a}\\\\\ = \frac{0- 49.4^ 2}{2\times -3.54}$

= 344.68 m

4 0

2 years ago