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1 year ago

A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *

Physics

1 answer:

Oksi-84 [34.3K]1 year ago

7 0

Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

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where again p is the phonon momentum, E is the photon energy and c is the speed of light. When you divide the photon energy foun

creativ13 [48]

Answer:

Yes

Explanation:

p = momentum of photon

E = energy of photon

c = velocity of light

Units of p = kg m /s

Units of E = kg m^2 / s^2

Units of E / p = {kg m^2 / s^2} / {kg m /s} = m/s

It is the unit of speed, so by the division of energy to the momentum, we get the speed. yes it is correct.

8 0

2 years ago

A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th

son4ous [18]

Answer:

Part a)

$E = 8436.7 N/C$

Part b)

$E = 8436.7 N/C$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = \frac{Q}{A}$

$Q = -3.8 nC$

$A = \pi(0.09)^2$

$A = 0.025 m^2$

$\sigma = \frac{-3.8\times 10^{-9}}{0.025}$

$\sigma = -1.5 \times 10^{-7} C/m^2$

now the electric field is given as

$E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}$

$E = 8436.7 N/C$

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

$E = 8436.7 N/C$

5 0

2 years ago

An alpha particle is identical to a(n) _____.

Alexxx [7]

Helium atom, in other words, it consistis of a particle having four protons and two neutrons.

3 0

2 years ago

Read 2 more answers

3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charg

Anni [7]

Answer:

$\mathbf{1.51\times10^{-15}N}$

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

$q_{paper}=-4.1\times10^{-15}C$

Now the distance from the charge is

$r=1cm=0.01m$

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

$mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}$

$\Rightarrow W=mg$

$=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}$

$=\mathbf{1.51\times10^{-15}N}$

6 0

1 year ago

A small child gives a plastic frog a big push at the bottom of a slippery 2.0 meter long, 1.0 meter high ramp, starting it with

valentinak56 [21]

Refer to the diagram shown below.

Because the ramp is slippery, ignore dynamic friction.
Let m = the mass of the frog.
g = 9.8 m/s²

The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J

Let v = the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J

Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s

Answer: 2.324 m/s

7 0

2 years ago