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2 years ago

Why doesn't the skateboarder roll as high each time she goes up the ramp

Physics

2 answers:

Westkost [7]2 years ago

6 0

Answer:

The skater has mechanical/gravitational potential energy at the two meter mark. The skater gets to two meters high on the other end of the ramp. In terms of the conservation of energy, the skater will never go higher than two meter on the other end of the the ramp because energy can be neither created nor destroyed.

Explanation:

I hoping it is right!!!∪∧∪ ∪ω∪

Andrej [43]2 years ago

3 0

Answer:

the reason it wont go higher is because some of the energy converts into thermal energy

Explanation:

I'm the energy master

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Internal Forces are :

Ugo [173]

Hi Pupil Here's your answer ::

➡➡➡➡➡➡➡➡➡➡➡➡➡

They are always balanced forces since if they are applied on an object the resultant of tgese forces on the object is Zero

So, the option A is the correct. They are always balanced forces.

8 0

2 years ago

What science did the study of the night sky eventually become?

Andrews [41]

This study of science is called Astronomy; the study of the night sky, planets, galaxies, and other celestial objects

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3 years ago

Planet Tatoone is about 1.7 AU from its Sun. Approximately how long will it take for light to travel from the Sun to Tatoone in

Radda [10]

Answer:

The value is $t = 14.129 \ minutes$

Explanation:

From the question we are told that

The distance of planet Tatoone is $d = 1.7 \ AU = 1.7 *1.496* 10^{11}=2.543*10^{11} \ m$

The speed of light is $c = 3.0*10^{8} \ m/ s$

Generally the time taken is mathematically represented as

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

=> $t = \frac{2.543*10^{11}}{3.0*10^{8} }$

=> $t = 847.7 \ s$

Now converting to minutes

$t = \frac{847.7}{60}$

=> $t = 14.129 \ minutes$

8 0

2 years ago

For a given Prandtl-Meyer expansion, the upstream Mach number is 3 and the pressure ratio across the wave is P2/P1 = 0.4. Calcul

loris [4]

Answer:

The angle for the forward Mach line is 19.47°

The angle for the rearward Mach line is 5.21°

Explanation:

From table A-1 (Modern Compressible Flow: with historical perspective):

$\frac{P_{o} }{P_{1} } =36.73$ (M₁ = 3)

If Po₁ = Po₂

$\frac{P_{o2} }{P_{2} } =\frac{P_{o1} }{P_{1} } *\frac{P_{1} }{P_{2} } =36.73*\frac{1}{4} =91.825$

Table A-1:

$\frac{P_{o2} }{P_{2} } =91.825,M_{2} =3.63$

Table A-5:

v₁ = 49.76°

μ₁ = 19.47°

v₂ = 60.55°

μ₂ = 16°

θ = 60.55 - 49.76 = 10.79°

The angle for the forward Mach line is:

μ₁ = 19.47°

The angle for the rearward Mach line is:

θr = μ₂ - θ = 16 - 10.79 = 5.21°

3 0

2 years ago

Help with 1 2 and 3 please

geniusboy [140]

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. $I = \frac{q}{t}$ Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3. So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

$\frac{p1L1}{A1} = \frac{p2L2}{A2}\\$

We are looking for L2 so we can isolate using algebra to get:

$\frac{A2(\frac{P1L1}{A1}) }{P2} = L2$

If you fill in those values you get 0.0205

or 2.05 cm

6 0

2 years ago