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Talja [164]
4 years ago
6

Anyone wanna help ?

Physics
2 answers:
julia-pushkina [17]4 years ago
8 0
D

Distance/time
50m/10s
5m/s
So the answer is 5 meters per second
xz_007 [3.2K]4 years ago
5 0
10 seconds, 50 meters

10 seconds 1 second
----------------- = ---------------
50 meters 5 meters

1/5 is .2. Second/meters is second/meters. Combined it is answer A, .2 seconds/meter. The wording in the problem is unusual to be confusing
You might be interested in
1. A baseball is thrown vertically at 16.7 m/s. What is the maimum height of the baseball?
Anvisha [2.4K]

Answer:

14.2 m

Explanation:

Using conservation of energy:

PE at top = KE at bottom

mgh = ½ mv²

h = v² / (2g)

h = (16.7 m/s)² / (2 × 9.8 m/s²)

h = 14.2 m

Using kinematics:

Given:

v₀ = 16.7 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 14.2 m

7 0
4 years ago
When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
A very narrow beam of white light is incident at 40.80° onto the top surface of a rectangular block of flint glass 11.6 cm thick
DerKrebs [107]
Dispersion angle = 0.3875 degrees. 
Width at bottom of block = 0.09297 cm 
Thickness of rainbow = 0.07038 cm 
 Snell's law provides the formula that describes the refraction of light. It is:
 n1*sin(θ1) = n2*sin(θ2)
 where
 n1, n2 = indexes of refraction for the different mediums
 Î¸1, θ2 = angle of incident rays as measured from the normal to the surface. 
 Solving for θ2, we get
 n1*sin(θ1) = n2*sin(θ2)
 n1*sin(θ1)/n2 = sin(θ2)
 asin(n1*sin(θ1)/n2) = θ2 
 The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
 Red:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.641) = θ2
 asin(1.00029*0.653420604/1.641) = θ2
 asin(0.398299876) = θ2
 23.47193844 = θ2 
 Violet:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.667) = θ2
 asin(1.00029*0.653420604/1.667) = θ2
 asin(0.39208764) = θ2
 23.08446098 = θ2 
 So the dispersion angle is:
 23.47193844 - 23.08446098 = 0.38747746 degrees. 
 Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
 tan(θ) = X/11.6
 11.6*tan(θ) = X
 So for Red:
 11.6*tan(θ) = X
 11.6*tan(23.47193844) = X
 11.6*0.434230136 = X
 5.037069579 = X 
 And violet:
 11.6*tan(θ) = X
 11.6*tan(23.08446098) = X
 11.6*0.426215635 = X
 4.944101361 = X 
 So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm 
 The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
 cos(θ) = X/0.092968218
 0.092968218*cos(θ) = X 
 0.092968218*cos(40.80) = X
 0.092968218*0.756995056 = X
 0.070376481 = X 
 So the distance between the red and violet rays is 0.07038 cm.
7 0
3 years ago
Mr. Fuller assigned his science class a lab comparing the masses of objects to the nearest gram using a balance. The mass of a p
Tomtit [17]
The scale would need 10 aluminum cubes on one side. Figure out how many paper clips would be needed on the other side to balance this. You have to use more than one aluminum cube because you need to have enough cubes so that you get a whole number mass. 10 cubes gives you a total mass of 27 g for the aluminum.
5 0
3 years ago
If two coils placed next to one another have a mutual inductance of 3.00 mH, what voltage (in V) is induced in one when the 2.50
hodyreva [135]

Answer:

-0.1875 V.

Explanation:

Using

E₂ = MdI₁/dt........................ Equation 1

Where E₂ = Voltage induced in the second coil, M = mutual inductance of both coil, dI₁ = change in current in the first coil, dt = change in time.

Given: M = 3.00 mH = 0.003 H, dI₁ = (0-2.50) = -2.5 A, dt = 40 ms = 0.04 s.

Substitute into equation 1

E₂ = 0.003(-2.5)/0.04

E₂ = -0.1875 V.

Hence the induced emf = -0.1875 V.

3 0
3 years ago
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