A stone falls from ledge and takes 16 seconds to hit ground. The stone has an original velocity 0 m/s, How tall is the ledge?

How is ultraviolet light and infrared light alike? How are they different?

swat32

Alike - both part of the light spectrum, cannot be seen with the naked eye

Different - IR is less energetic than UV, UV has a shorter wavelength than IR

IR - infrared
UV - ultraviolet

5 0

4 years ago

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given b

Mnenie [13.5K]

Explanation:

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(t^3-2t^2+4t+4)}{dt}\\\\I=3t^2-4t+4$

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.

7 0

4 years ago

What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface

NARA [144]

Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

$g=\frac{GM}{(R+h)^2}$

where:

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

$R=6.37\cdot 10^6 m$ is the Earth's radius

Here,

$h=6.38\cdot 10^6 m$

So the acceleration due to gravity is

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2$

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

$F=mg=(70)(2.45)=171.5 N$

5 0

3 years ago

A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential differ

ra1l [238]

Answer:

The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

$\Delta U = q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

For a proton,

$q=+e=1.6\cdot 10^{-19}C$

And since $\Delta V=100 V$

The energy gained by the proton is

$\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J$

For an alpha particle,

$q=+2e=3.2\cdot 10^{-19}C$

Therefore, the energy gained is

$\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J$

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

$q=+e=1.6\cdot 10^{-19}C$

So the energy gained is the same as the proton:

$\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J$

6 0

3 years ago

How to use kinetic energy to calculate velocity

BartSMP [9]

As we know the formula of kinetic energy is

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

here given that

KE = 150,000 J

mass = 120 kg

we can use this to find speed

$150,000 = \frac{1}{2} * 120 * v^2$

$v^2 = 2500$

$v = 50 m/s$

So speed of above object is 50 m/s

7 0

3 years ago