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Burka [1]
3 years ago
13

Technician A says that a waste spark ignition system fires two spark plugs at the same time. Technician B says that the waste sp

ark ignition system uses ignition coils connected to companion cylinders. Which technician is correct?
a. Technician A only
b. Technician B only
c. Both technicians A and B
d. Neither technician A nor B
Physics
1 answer:
Alina [70]3 years ago
6 0

Answer:

d. neither technicians a and b

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The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what veloc
alukav5142 [94]

This question involves the concepts of orbital velocity and orbital radius.

The orbital velocity of ISS must be "7660.25 m/s".

The orbital velocity of the ISS can be given by the following formula:

v=\sqrt{\frac{GM}{R}}

where,

v = orbital velocity = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m

R = 67.86 x 10⁵ m

Therefore,

v=\sqrt{\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{67.86\ x\ 10^5\ m}}

<u>v = 7660.25 m/s</u>

Learn more about orbital velocity here:

brainly.com/question/541239

3 0
2 years ago
A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po
aleksandrvk [35]

Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5  / x² = 2 / ( .4 - x )²

Taking square root on both sides

2.236 / x = 1.414 / .4 - x

2.236 ( .4 - x ) = 1.414 x

.8944 - 2.236 x = 1.414 x

.8944 = 3.65 x

x = .245 m

24.5 cm

So the third charge should be at a distance of 24.5 cm from q₁ .

4 0
2 years ago
The input work done on a machine is 9.63 × 103 joules, and the output work is 3.0 × 103 joules. What is the percentage efficienc
notsponge [240]
Input work = 9.63×10³ J.
Output work = 3.0×10³ J

By definition,
Efficiency = (Output work)/(Input work)
                 = (3.0×10³)/(9.63×10³)
                 = 0.31 = 31%

Answer:  31%
5 0
3 years ago
Read 2 more answers
Help plzzz!!! I only have 10 minutes to turn in
Lostsunrise [7]
  • The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car.  When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

  • Kinetic energy = (1/2) (mass) (speed²)

And there we have it

  • The car's mass is 3,600 kg.
  • Its speed is 'v' m/s .
  • (1/2) (mass) (v²) =  5,406 Joules

(1/2) (3600 kg) (v²) = 5406 joules

1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

  • v² = (5406/1800) (Newton-meter/kg)
  • v² = (5406/1800) (kg-m²/s²  /  kg)
  • v² = (5406/1800)  (m²/s²)

= = = = =

v = √(5406/1800)  m/s

<em>v = 1.733 m/s</em>

4 0
3 years ago
(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroi
nydimaria [60]

Answer:

ks= 133.2 N/m

Explanation:

  • Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.
  • This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.
  • When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.
  • Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.
  • If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.
  • So, we can write the following equation:

       \Delta U + \Delta K = 0 (1)

  • where ΔU = -1/2*k*(Δx)²  (2)
  • and ΔK = 1/2*m*v² (3)
  • Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:

       k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)

5 0
2 years ago
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