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3 years ago

A wave is traveling at the speed of light. If it has a frequency of 4 x 10^15 Hz, what is its wavelength?

Physics

1 answer:

maria [59]3 years ago

8 0

Answer:

7.5x10-^8m

Explanation:

the formulae is v=fh ,where f is frequency

h is wavelength

hence wavelength=v/f

since the speed of light is 3x10^8

then the wavelength =3 x 10^8/4 x 10^15

=7.5x10^-8m

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A car is moving with a speed of 15 ms. How long does it take to cover a distance of 1.2 km?

fgiga [73]

Answer: 80 s

Explanation:

Speed is expressed in v= d/t, derive the equation so we can have time.

First conert km into meters to cancel out both units and only seconds will remain.

1.2 km x 1000m/ 1km = 1200 m

t = 1200 m /15 m/s = 80 s

8 0

3 years ago

Two ions with masses of 5.29×10−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is

riadik2000 [5.3K]

Answer:

0.132 m

Explanation:

$m$ = mass of the ion = 5.29 x 10⁻²⁷ kg

$q$ = magnitude of charge on singly charged ion = 1.6 x 10⁻¹⁹ C

$r$ = radius of circular path followed by singly charged ion

$v$ = speed of the ion = 1.13 x 10⁶ m/s

$B$ = magnitude of the magnetic field = 0.283 T

Radius of the circular path is given as

$r = \frac{mv}{qB}$

$r = \frac{(5.29\times 10^{-27})(1.13\times 10^{6})}{(1.6\times 10^{-19})(0.283)}$

$r$ = 0.132 m

6 0

3 years ago

A racquet ball with mass m = 0.256 kg is moving toward the wall at v = 11.8 m/s and at an angle of θ = 29° with respect to the h

icang [17]

Answer:

Part a)

$P = 5.72 kg m/s$

Part b)

$\Delta P = 2.93 kg m/s$

Part c)

$F = 44.4 N$

Part d)

$\Delta P = 5.02 kg m/s$

Part e)

$\Delta t = 0.113 s$

Part f)

$\Delta K = 0$

Explanation:

As we know that initial velocity of the ball is given as

$v = 11.8 cos29 \hat i + 11.8 sin29 \hat j$

$v_i = 10.3 \hat i + 5.72 \hat j$

Now final velocity of the system is given as

$v_f = 10.3\hat i - 5.72\hat j$

Part a)

now magnitude of initial momentum is given as

$P = mv$

$P = 0.256(11.8)$

$P = 5.72 kg m/s$

Part b)

Change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.256(5.72 + 5.72)$

$\Delta P = 2.93 kg m/s$

Part c)

As we know that average force is defined as the rate of change in momentum

so here we have

$F = \frac{\Delta P}{\Delta t}$

$F = \frac{2.93}{0.066}$

$F = 44.4 N$

Part d)

Magnitude of change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.256(7.8 + 11.8)$

$\Delta P = 5.02 kg m/s$

Part e)

As we know that in 2nd case the force is same as the initial force

so we will have

$\frac{\Delta P}{\Delta t} = F$

$\frac{5.02}{\Delta t} = 44.4$

$\Delta t = 0.113 s$

Part f)

Since this is elastic collision so change in kinetic energy must be ZERO

$\Delta K = 0$

8 0

3 years ago

A 15kg object strikes the ground with 2000J of kinetic energy after freely falling from rest. How far above the ground was the o

arsen [322]

Answer:

Explanation:

Find the final velocity at which it struck the ground by using the kinetic energy formula.

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

$2000=\frac{1}{2} (15)v^2\\v=16.33$

Now use kinematics to solve for the vertical displacement. We were given the initial velocity and acceleration can be assumed to be 9.8

$V^2=V^2_i+2a(y_f-y_i)\\16.33^2=0^2+2(9.8)(y_f-y_i)$

Δy = 13.6 meters

8 0

3 years ago

Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mas

ASHA 777 [7]

M*U + 0 = m*v'1 + 2m*v'2
the zero means deuteron has no velocity
where v'1 and v'2 are the post-collision velocities.
The equatio becomes
U = v'1 + 2v'2
U = v'2- v'1 
v'2 = U + v'1 

U = v'1 + 2(U + v'1) = 2U + 3v'1 
U = -3V 
V = -U / 3 
The speed ratio is 1/3 

B) Since KE is proportional to the square of the speed, if the speed is 1/3, then KE is 1/9 

C) (1/3)ⁿ = 1/729 
3ⁿ = 729 
n = 6

5 0

4 years ago

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