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tatiyna
3 years ago
5

If object A has more mass than object B, what will object A need to accelerate at the same rate as object B?

Physics
1 answer:
Leni [432]3 years ago
3 0

Answer:

More force

Explanation:

Object A has more mass than object B

  For object A to accelerate at the same rate as object B, it will need more force.

According to Newton's second law of motion "the net force on a body is the product of its mass and acceleration".

  Net force  = mass x acceleration

Now, if a body has more mass and needs to accelerate at the same rate as another one with a lower mass, the force on it must be increased.

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Consider an embedded system which uses a battery with a 17.15-Amp-Hour capacity. What is the maximum average current draw (in mi
Marianna [84]

Answer:

<em>85.12 μAmp</em>

<em></em>

Explanation:

The battery power output = 17.15 Amp-hr

If the battery is to last 23 years, we have to calculate how many hours there are in 23 years

in one year there are 24 hours x 365 day = 8760 hrs

in 23 years there are 23 x 8760 = 201480 hours

maximum current to be drawn from the battery = (17.15 Amp-hr) ÷ (201480 hours) = 85.12 x 10^-6 Amp = <em>85.12 μA</em>

7 0
3 years ago
Common sports injury : ankle sprain , bone fracture , muscle strain , neck low back pain , shoulder pain. First aid needed
erastova [34]

Answer:

m

Explanation:

3 0
3 years ago
A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matt
anyanavicka [17]

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

6 0
3 years ago
If the child pulls with a force of 20 N for 12.0 m , and the handle of the wagon is inclined at an angle of 25 ∘ above the horiz
lilavasa [31]

Answer: 217.52 N

Explanation: The applied force is 20 N, the distance covered is 12.0 m and the angle is 25° above the horizontal.

Hence the formulae that defines work done is given by

W = Force × distance

But since the force has been inclined at an angle θ above the horizontal, the horizontal component of force is neccesary to produce the required motion to make the child do work on the wagon.

Hence

Work done = (horizontal component of force) × distance

Work done = F cos θ × distance

Work done = 20 cos 25 × 12 = 217.52 N

4 0
3 years ago
Given the equation p2 = a3, what is the orbital period, in days, for the planet venus? (venus is located 0.72 au from the sun?)
melisa1 [442]

The correct answer is 223 days.

The relationship between the duration of revolution and the separation between the sun is shown by Kepler's third law. Using the notions of circular motion and the gravitational and centripetal forces, we may obtain this equation.

According to Kepler's third rule, the semi-major axis of an orbit is linked to the orbital period of a planet around the sun as follows:

p² = a³

where an is the semi-major axis/distance to the star and p is the orbital period in years.

It is said that a = 0.72 AU for Venus.

P= √(0.72 AU)^3 = 0.61 years.

365 days in a year = 222.9 ≈ 223 days.

To learn more about Kepler's third rule refer the link:
brainly.com/question/1608361

#SPJ4

5 0
1 year ago
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