Which statement best describes what happens when more waves pass a certain point per second?

When a current of 0.015 A passes through human body at 240 volts p.d it<br><br>causes

geniusboy [140]

Explanation:

Given that,

Current, I = 0.015 A

Voltage, V = 240 volts

We need to find the resistance. Using Ohm's law we can find it as follows :

$V=IR\\\\R=\dfrac{V}{I}\\\\R=\dfrac{240}{0.015}\\\\R=16000\ \Omega$

So, When a current of 0.015 A passes through human body at 240 volts p.d it causes 16000 ohms of resistance.

3 0

3 years ago

Density is calculated by combining 2 units, therefore the unit we use to measure density is called a _____________________unit.

nalin [4]

Answer: Dependent Unit or System of Units

Explanation:

Density is calculated by dividing mass (Kg) by volume (L).

The unit of Density is Kg/L or one of their derivatives such as g/cm³.

8 0

2 years ago

QUICK giving brainlyest

Soloha48 [4]

The momentum of the second ball was 15 kg.m/s.

<h3>What is inelastic collision?</h3>

In which collision some amount of kinetic energy of the system is lost that called inelastic collision. In purely inelastic collision, two bodies stick together. But principle of conservation of linear momentum is obeyed.

In the given question,

Two balls collide and after collision, the final momentum of the system = 18 kg.m/s.

Initial velocity of 1st ball of mass 3 kg is 1 m/s.

So, Initial momentum of first ball = mass × velocity = (3 kg) × (1 m/s) = 3 kg.m/s.

According to Principle of conservation of linear momentum for this inelastic collision,

Initial momentum of first ball + initial momentum of second ball = final momentum of the system

⇒ initial momentum of second ball = final momentum of the system - Initial momentum of first ball

= 18 kg.m/s - 3 kg.m/s.

= 15 kg.m/s.

Hence, initial momentum of second ball = 15 kg.m/s.

Learn more about momentum here:

brainly.com/question/24030570

#SPJ2

5 0

1 year ago

Read 2 more answers

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

2 years ago

1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.

Drupady [299]

1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

$W=Fd$

where

F = 15 N is the magnitude of the force

d = 85 cm = 0.85 m is the displacement of the cart

Substituting in the formula, we get

$W=(15 N)(0.85 m)=12.75 J$

2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

$W=Fd=12.75 J$

However, in this case the displacement is

d = 120 cm = 1.20 m

Therefore, the magnitude of the force in this case is

$F=\frac{W}{d}=\frac{12.75 J}{1.20 m}=10.6 N$

3 0

2 years ago