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Luba_88 [7]
3 years ago
14

A block slides at constant speed down a ramp while acted on by three forces: its weight, the normal force, and kinetic friction.

Respond to each statement, true or false.
Physics
1 answer:
Afina-wow [57]3 years ago
4 0

Answer:

a) True

b)False

c)False

• Had to complete the question first.

A block slides at constant speed down a ramp while acted on by three forces: its weight, the normal force, and kinetic friction. Respond to each statement, true or false.

(a) The combined net work done by all three forces on the block equals zero.

(b) Each force does zero work on the block as it slides.

(c) Each force does negative work on the block as it slides.

Explanation:

Net work is the change in kinetic energy, which leads to final kinetic energy - our initial kinetic energy this is the formula for net work. This is the working energy theorem, a theorem that states that the net work on an object induces a change in the object's kinetic energy.

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1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com
Vsevolod [243]

Answer and explanation:

A.

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

B.

In muon rest frame it travels Zero meters

C.

Distance, d = Velocity, v * Time, s

where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s

d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m

D.

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

E.

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}

d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}

d' = 594\sqrt{1 - 0.81}

d' = 594 \times 0.4359

d' = 258.92m

F.

in the rest frame of someone standing on the mountain,

muon moves straight down

3 0
4 years ago
What action could a student take to show a transverse wave
ICE Princess25 [194]

Answer:

You can make a horizontale transverse by moving a slinky vertically up and down!

Explanation:

8 0
4 years ago
Read 2 more answers
A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging
bekas [8.4K]

Answer:

The answers to the question are

(a) 2.1 m/s

(b) 0.83 N

(c) 1.9 N

Explanation:

To solve the question, we list out the varibles

Length, l of string = 0.8 m

mass of rock, m = 0.12 kg

Angle with the verrticakl, θ = 45 °

a) To find the speed of the rock  when the string passes through the vertical position we have

From the first law of thermodynamics

Potential energy = kinetic energy

m×g×l×(1-cosθ) = 1/2×m×v²

That is v² = 2×g×l×(1-cosθ)

= 2×9.81×0.8×(1-cos45) = 4.597

or v = √4.597 = 2.1 m/s

(b) The tension in the string when it makes an angle of  45∘ with the vertical is given by

For balance between Tension and mass of rock is gigen by

∑Forces = 0, T - m×g×cosθ = 0

or T =  m×g×cosθ = 0.12×9.81×cos45 = 0.83 N

c) The tension in the string as it passes through the vertical

when passing through the vertical we have T - m×g = (m×v²)/r

or T = m×g + (m×v²)/r = mg(1+2(1-cosθ)) =0.981*0.12 (1+ 2(1-cos45)) =1.867 N

= 1.9 N

3 0
3 years ago
A surfer is able to stay on the surfboard as she rides the waves. Which explains the force(s) that enable her to do this? A. The
lyudmila [28]
ZC. The forward force of the surfboard's acceleration is balanced by the backward force of the surfer's mass.
3 0
4 years ago
Light of wavelength 633 nm falls on a single slit 0.5 mm wide and forms a diffraction pattern on a screen 1.0 m away from the sl
andrey2020 [161]

Answer:

The position of the first dark spot on the positive side of the central maximum is 1.26 mm.

Explanation:

Given that,

Wavelength of light is 633 nm.

Slit width, d = 0.5 mm

The diffraction pattern forms on a screen 1 m away from the slit. We need to find the position of the first dark spot on the positive side of the central maximum.

For destructive interference :

\dfrac{dY}{D}=n\lambda

Y is the distance of the minima from central maximum

Here, n = 1

Y=\dfrac{n\lambda D}{d}\\\\Y=\dfrac{1\times 633\times 10^{-9}\times 1}{0.5\times 10^{-3}}\\\\Y=0.00126\ m\\\\Y=1.26\times 10^{-3}\ m\\\\Y=1.26\ mm

So, the position of the first dark spot on the positive side of the central maximum is 1.26 mm.

4 0
3 years ago
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