PHYSICS COLLEGE

Instantaneous speed is...a) A speed of 1000 km/h
b) The speed attained at a particular instant in time.
c) The speed that can be reached in a particular amount of time.

PLEASE HURRY

Answers

Answer 1

Answer:

Answer:

The speed attained at a particular instant in time.

Explanation:

Instantaneous speed is the speed attained at a particular instant in time.

It is given by :

$v=(dx)/(dt)$

It is equal to the rate of change of speed.

It can be also defined as when the speed of an object is constantly changing, the instantaneous speed is the speed of an object at a particular moment (instant) in time.

Hence, the correct option is (b).

Related Questions

) An electron moving along the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, what is the direction of the magnetic field in this region
A rabbit is moving in the negative x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 13.9 m/s along the negative y-axis, all in 1.20 s. Determine the x-component and the y-component of the rabbit's acceleration. (Enter your answers in m/s2. Indicate the direction with the signs of your answers.) HINT
Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. What is thefinal angular speed of the rod?
A 2.4 kg toy oscillates on a spring completes a cycle every 0.56 s. What is the frequency of this oscillation?

A research Van de Graaff generator has a 2.00-m diameter metal sphere with a charge of 5.00 mC on it. (a) What is the potential near its surface?
(b) At what distance from its center is the potential 1.00 MV?
(c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV when the atom is at the distance found in part b?

Answers

Answer:

a) $V=49.5MV$

b) $r=49.5m$

c) $\Delta U=3*(49.5-1)=145.5 MeV$

Explanation:

a) The potential equation is given by:

$V=k(Q)/(r)$

k is the electrostatic constant ( $k=9.9*10^(9)Nm^(2)/C^(2)$ )

Q is the charge Q = 5mC

r is the radius of the sphere r = 1 m

$V=9.9*10^(9)(5*10^(-3))/(1)=49.5MV$

b) We solve it using the same equation.

Here we need to find r:

$r=k(Q)/(V)$

$r=9.9*10^(9)(5*10^(-3))/(1*10^(6))$

$r=49.5m$

c) The relation between difference potential and electrical energy is:

$\Delta U=\Delta Vq$

here q is 3e becuase oxygen atom has three missing electrons

Therefore:

$\Delta U=3*(49.5-1)=145.5 MeV$

I hope it heps you!

Which exerts more force, the Earth pulling on the moon or the moon pulling on the Earth? Explain.

Answers

Answer: the earth

Explanation: Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth. Over a very long time, the moon's rotations created fiction with the Earth's tugging back, until the moon's orbit and rotational locked with Earth.

and that's why the earth pulls the moon

Final answer:

The Earth pulling on the moon and the moon pulling on the Earth exert the same amount of force on each other due to Newton's third law of motion.

Explanation:

In terms of force, the Earth pulling on the Moon and the Moon pulling on the Earth exert the same amount of force on each other. This is because of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. So, while the Earth's gravitational force pulls the Moon towards it, the Moon's gravitational force also pulls the Earth towards it with an equal amount of force.

Newton's third law of motion states that for every action, there is an equal and opposite reaction. In the context of the gravitational interaction between the Earth and the Moon, the forces they exert on each other are equal in magnitude and opposite in direction.

The Earth pulls on the Moon with a gravitational force, and, according to Newton's third law, the Moon simultaneously pulls on the Earth with an equal gravitational force. These forces are sometimes referred to as "action and reaction pairs." The force that the Earth exerts on the Moon is often called the gravitational attraction of the Earth on the Moon, and vice versa.

Learn more about gravitational force here:

brainly.com/question/32609171

#SPJ2

What is the wavelength of the electromagnetic radiation needed to eject electrons from a metal?

Answers

Answer:

λ = hc/(eV + h $f_(0)$ )

Explanation:

Let the work function of the metal = ∅

the kinetic energy with which the electrons are ejected = E

the energy of the incident electromagnetic wave = hf

Then, we know that the kinetic energy of the emitted electron will be

E = hf - ∅

because the energy of the incident electromagnetic radiation must exceed the work function for electrons to be ejected.

This means that the energy of the incident e-m wave can be written as

hf = E + ∅

also, we know that the kinetic energy of the emitted electron E = eV

and the work function ∅ = h $f_(0)$

we can they combine all equations to give

hf = eV + h $f_(0)$

we know that f = c/λ

substituting, we have

hc/λ = eV + h $f_(0)$

λ = hc/(eV + h $f_(0)$ ) This is the wavelength of the e-m radiation needed to eject electrons from a metal.

where

λ is the wavelength of the e-m radiation

h is the Planck's constant = 6.63 x 10^-34 m^2 kg/s

c is the speed of e-m radiations in a vacuum = 3 x 10^8 m/s

e is the charge on an electron

V is the voltage potential on the electron

$f_(0)$ is the threshold frequency of the metal

A centrifuge is a common laboratory instrument that separates components of differing densities in solution. This is accomplished by spinning a sample around in a circle with a large angular speed. Suppose that after a centrifuge in a medical laboratory is turned off, it continues to rotate with a constant angular deceleration for 10.0s before coming to rest.Part A

If its initial angular speed was 3890rpm , what is the magnitude of its angular deceleration? (|?| in revs/s^2 )

Part B

How many revolutions did the centrifuge complete after being turned off?

Answers

Answer:

$a_r=389\ rev.s^(-2)$

$n=58350 rev$

Explanation:

Given:

time of constant deceleration, $t=10\ s$

initial angular speed, $N_i=3890\ rpm\$

Using equation of motion:

$N_f=N_i+a_r.t$

$0=3890+a_r* 10$

$a_r=389\ rev.s^(-2)$

Using eq. of motion for no. of revolutions, we have:

$n=N_i.t+(1)/(2) a_r.t^2$

$n=3890* 10+0.5* 389* 100$

$n=58350\ rev$

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 2.00 s later. A)How high was the bridge?B)How fast were the swimmers moving when they hit the water?

C)What would the swimmer's drop time be if the bridge were twice as high?

Answers

Answer: part a: 19.62m

part b: 19.62 m/s

part a: 2.83 secs

Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence

u=0

a=9.81 m/s2

t=2 secs

$s=ut+0.5at^2$

s=h

$h=0*2+0.5*9.81*2^2\nh=19.62 meters$

Part b

$v=u+at\nv=0+9.81*2\nv=19.62m/s$

Part c

now we have h=2*19.62=39.24

$39.24=0+0.5*9.81*t^2\nt^2=8\nt=2.83 secs$

The starter motor of a car engine draws a current of 170 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.930 s until the car engine starts How much charge passes through the starter motor?

Answers

Answer:

The charge that passes through the starter motor is $\Delta Q=158.1 C$ .

Explanation:

Known Data

Avogadro's Number $N_(A)=6.02x10^(23)$
Current, $I=170A=170(C)/(s)$
Charge in an electron, $q=1.60x10^(-19)C$
Time, $\Delta t=0.930s$
Diameter, $d=4.60mm=0.0046m$
Transversal Area, $A=((d)/(2))^(2) \pi=((0.0046m)/(2))^(2) \pi=1.66x10^(-5) m^(2)$
Volume, $V=Length*A=(1.2m)(1.66x10^(-5) m^(2))=1.99x10^(-5) m^(3)$

First Step: Find the number of the electrons per unit of volume in the wire

We use the formula $n=(N_(A))/(V)= (6.02x10^(23) electrons)/(1.99x10^(-5) m^(3)) =3.02x10^(28)el/ m^(3)$ .

Second Step: Find the drag velocity

We can use the following formula $v_(d)=(I)/(nqA)=(170C/s)/((3.02x10^(28)m^(-3))(1.60x10^(-19)C)(1.66x10^(-5) m^(2))) =2.11x10^(-3) m/s$

Finally, we use the formula $\Delta Q=(nAv_(d)\Delta t)q=(3.02x10^(28) m^(-3))(1.66x10^(-5) m^(2))(2.11x10^(-3) m/s)(0.930s)(1.60x10^(-19)C)=158.1 C$ .