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Sergeeva-Olga [200]
3 years ago
13

A jet travels 1000 kilometers in 5 hours. What is its average speed?

Physics
1 answer:
nlexa [21]3 years ago
7 0

Given:

Total distance = 1000 kilometer

Total time = 5 hours

To find:

Average speed = ?

Formula used:

v_{avg} = \frac{s}{t}

Where v_{avg} = average speed

s = total distance

t = total time

Solution:

Average speed of the jet is given by,

v_{avg} = \frac{s}{t}

Where v_{avg} = average speed

s = total distance

t = total time

v_{avg} = \frac{1000}{5}

v_{avg} = 200 km/ h

Thus, average speed of the jet is 200 km/h.

Hence, Option (A) is correct.


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The magnitudes of the charge densities on the inner and outer shells are now changed (keeping λinner = -λouter) so that the resu
elena-14-01-66 [18.8K]

Answer: a) Cnew=Cinitial ; b) λouter new=  2*λ outer initial

Explanation: In order to explain this question we have to take into account the expression of teh cylinder capacitor given by:

C/L= (2*π*εo)/ln (b/a)=  where b and a are the outer and inner radius, respectively. L is the length of the capacitor.

As you can se this formule depents  of geometrical characateristics  of the capacitor.

The capacitance is the same after change the densities of charge.

On the other hand,

The new charge in each cylinder ( inner and outer) is determined

The new potential is 2 times the initial one so

V new= 2* Vinitial

Also we know that

Vnew= Q/C= λnew*L/C;   C= constant

using this formule and considering  that V new is doubled then the charge per one meter length,  is also doubled .

This is as follow:

Vnew=  λnew*L/C=  

λnew =  (2*Vinitial)* C/L= 2  (λ initial)

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5 0
4 years ago
A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 8.1 m from this surface, the potenti
igomit [66]

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, V_s = 450 V

potential at radial distance, V_r = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}

450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r}  \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m

Therefore, the radius of the sphere is 4.05 m

4 0
3 years ago
How long will it take for the car to accelerate
solniwko [45]
You have to be 90 m away from the cat

Just divide 90 by 17.4 and you will get - 5.172 seconds

- It will take 5.17 seconds to accelerate uniformly to stop in exactly 2 m before the cat
8 0
3 years ago
A 0.20-kg block rests on a frictionless level surface and is attached to a horizontally aligned spring with a spring constant of
solmaris [256]

Answer:

0.57 ms^{-1}

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k = spring constant of the spring = 40 Nm⁻¹

A = amplitude of the simple harmonic motion = 4 cm = 0.04 m

m = mass of the block attached to spring = 0.20 kg

w = angular frequency of the simple harmonic motion

Angular frequency of the simple harmonic motion is given as w = \sqrt{\frac{k}{m} } \\w = \sqrt{\frac{40}{0.20} }\\w = 14.14 rads^{-1}

v = Speed of the block as it pass the equilibrium point

Speed of the block as it pass the equilibrium point is given as

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6 0
3 years ago
What is the best description of the relationship between emerging scientific ideas and open-mindedness?
Vilka [71]
The third one looks correct to me
3 0
3 years ago
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