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3 years ago

What distance is required for a train to stop if its initial velocity is 23 m/s and its

Physics

1 answer:

Varvara68 [4.7K]3 years ago

7 0

-- The train starts at 23 m/s and slows down by 0.25 m/s every second.

So it'll take (23/0.25) = 92 seconds to stop.

-- Its average speed during that time will be (1/2)(23+0) = 11.5 m/s

-- Moving at an average speed of 11.5 m/s for 92 sec, the train will cover

(11.5 m/s) x (92 sec) = <em>1,058 meters</em> .

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Consider the following statement:

vovangra [49]

Answer:

The statement "The magnetic field of a magnet comes out of the north pole and goes into the south pole" is imprecise

Explanation:

This is because the zero divergence equation (∇ · B = 0 ) is valid for any magnetic field, even if it is time dependent rather than static. Physically, it means that there are no magnetic charges otherwise we would have ∇ · B ∝ ρmag instead of ∇ · B = 0. Consequently, the magnetic field lines never begin or end anywhere in space; instead they form closed loops or run from infinity to infinity.

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3 years ago

The equation for momentum is p=mv. Which letter represents the momentum? Question 2 options: p m v

vovikov84 [41]

P stands for momentum

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3 years ago

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ???? 0 ν0 . Find the mi

Allisa [31]

Answer:

<h2>E = 2.8028*10⁻¹⁹ Joules</h2>

Explanation:

The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo

h = planck's constant

fo = threshold frequency

Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹

h = 6.626× 10⁻³⁴ m² kg / s

Substituting this value into the formula to get the energy E

E = 4.23×10¹⁴ * 6.626 × 10⁻³⁴

E = 28.028*10¹⁴⁻³⁴

E = 28.028*10⁻²⁰

E = 2.8028*10⁻¹⁹ Joules

7 0

3 years ago

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

noname [10]

Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=> $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$

=> $\rho_d = 50.6*10^{-12} * 1.5*10^{3}$

=> $\rho_d = 7.59*10^{-8} \ C/m$

Generally the surface charge density is mathematically represented as

$\rho_s = \frac{\rho_l }{2 \pi * r_2 }$ here $2 \pi r = (circumference \ of \ outer \ conductor )$

=> $\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }$

=> $\rho_s = 4.026 *10^{-6} \ C/m^2$

3 0

2 years ago

If the vehiche has a speed of 24.0 m/s at point a what is the force of the track on the vehicle at this point?

ASHA 777 [7]

Question:

A roller-coaster vehicle has a mass of 500 kg when fully loaded with passengers. If the vehicle has a speed of 20.0 m/s at point A, what is the force exerted by the track on the vehicle at this point?

<em>See attachment</em>

Answer:

33700 Newton

Explanation:

Given

$m = 500kg\\v = 24.0m/s\\r=10m$