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3 years ago

I'm stuck on number 4... help please?

Physics

2 answers:

lora16 [44]3 years ago

8 0

$V= \frac{\Delta S}{\Delta t} = \frac{5}{15} = \frac{1}{3} m/s$

Paha777 [63]3 years ago

3 0

Average speed = (distance covered) / (time to cover the distance) =

 (5m) / (15 sec) =

 (5/15) (m/s) = 1/3 m/s .

Average velocity =

 (displacement) / (time spent traveling) in the direction of the displacement

Average velocity = (5m) / (15 sec) left =

 (5/15) / (m/sec) left =

 1/3 m/s left.

A number without a direction is a speed, not a velocity.

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Express force in terms of base units

denis-greek [22]

Answer:

F = [MLT⁻²]

Explanation:

Force = ma

m (mass) = [M]

a (acceleration) = [LT⁻²]

F(force) = m x a = [MLT⁻²]

3 0

2 years ago

An experimentalist fires a beam of electrons, creating a visible path in the air that can be measured. The beam is fired along a

gtnhenbr [62]

Answer:

Velocity of the electron in the beam.

Radius of the circulating electrons due to the magnetic field.

Explanation:

We have a Mathematical expression for the force on a moving charge in a magnetic field as:

$F=q.v.B.sin \theta$ ...........................(1)

where:

q= charge on the particle in coulomb

v= velocity of the charge projected into the magnetic field

B= intensity of the magnetic field in tesla

$\theta$ = angle between the velocity and direction of magnetic field

For the forces on rotating mass we have the formula:

$F=m.\frac{v^2}{r}$ ..........................................(2)

where:

m= mass of the charged particle

v= velocity of projection of charge into the magnetic field

r= radius of the path traced by the charge in the magnetic field

From eq. (1) and (2) we can calculate the magnetic field .

Now,

Using Ampere's Law we have:

$B = \frac{\mu_0 .I}{2 \pi r}$

where:

I= current in the wire

$\mu_0=$ The permeability of free space.

r= radial distance from the current carrying wire( in this case it is same as the radius of the circular path)

4 0

3 years ago

A 2 kg rock is at the edge of a cliff 20 meters above a lake The rock becomes loose and falls toward the water below. Calculate

natima [27]

Answer:

The potential energy (P.E) at the top is 392 J

The kinetic energy (K.E) at the top is 0 J

The potential energy (P.E) at the halfway point is 196 J.

The kinetic energy (K.E) at the halfway point is 196 J.

Explanation:

Given;

mass of the rock, m = 2 kg

height of the cliff, h = 20 m

speed of the rock at the halfway point, v = 14 m/s

The potential energy (P.E) and kinetic energy (K.E) when its at the top;

P.E = mgh

P.E = (2)(9.8)(20)

P.E= 392 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the top of the cliff = 0

K.E = ¹/₂(2)(0)²

K.E = 0

The potential energy (P.E) and kinetic energy (K.E) at the halfway point;

P.E = mg(¹/₂h)

P.E = (2)(9.8)(¹/₂ x 20)

P.E = 196 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the halfway point = 14 m/s

K.E = ¹/₂(2)(14)²

K.E = 196 J.

4 0

2 years ago

Please someone knows the answer? it is physics

Assoli18 [71]

Answer:C

Explanation: I took the quiz if you don't believe me then your gonna miss the problem.

3 0

2 years ago

A particular circular track on level ground has a circumference of 400 m. a walker, traveling counterclockwise and starting at t

Sidana [21]

Answer:

Average speed: 0.5 m/s. Average velocity: 0

Explanation:

Average speed is given by:

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

where

d is the total distance covered (the length, of one lap of the track, so d = 400 m)

t is the time taken to cover that distance (so, t = 800 s)

Substituting,

$v=\frac{400}{800}=0.5 m/s$

Instead, average velocity is defined as

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

where this time,

d is the displacement, which is the vector connecting the starting point to the final point of the motion

t is still the time taken (800 s)

However, in this case the walker starts and finishes his trip at the same point: therefore, the displacement is zero (d=0), and this means that the average velocity is zero as well.

3 0

3 years ago