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Allushta [10]
3 years ago
14

Consider a stone in free fall on a planet with gravitational acceleration 3.4 m/s^2. Suppose you would like the stone to experie

nce the same magnitude of acceleration down an inclined plane on Earth. Neglecting friction and air resistance, which angle should the inclined plane have with respect to the horizontal ? (in deg)
Physics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

Angle of incline is 20.2978°

Explanation:

Given that;

Gravitational acceleration on a planet a = 3.4 m/s²

Gravitational acceleration on Earth g = 9.8 m/s²

Angle of incline = ∅

Mass of the stone = m

Force on the stone along the incline will be;

F = mgSin∅

F = ma

The stone has the same acceleration as that of the gravitational acceleration on the planet.

so

ma = mgSin∅

a = gSin∅

Sin∅ = a / g

we substitute

Sin∅ = (3.4 m/s²) / (9.8 m/s²)

Sin∅ = 0.3469

∅ = Sin⁻¹( 0.3469 )

∅ = 20.2978°

Therefore, Angle of incline is 20.2978°

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F=ma
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A light year is defined as the distance that light can travel in 1 year. What is the value of 1 light year in meter? Show your c
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d=9.462×10^15 meters

Explanation:

<u>Relation between distance, temps and velocity:</u>

d=v*t

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So:

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3 years ago
Correct Number Formats Numbers may be entered in several formats - including scientific notation and numerical expressions.
Illusion [34]

Answer:

a. Valid

b. Invalid

c. Invalid

d. Invalid

e. Valid

f.  Valid

g. Invalid

Explanation:

a.

Since, it follows all the four rules,

Therefore, this number is <u>Valid</u>

b.

It uses upper case E, instead of lower case e.

Therefore, it is <u>Invalid</u>

c.

It violates third rule for including unit.

Therefore, it is <u>Invalid</u>

d.

It violates the first rule. As, there is a space between 6 and e. While in the example given in question, there is no gap.

Therefore, it is <u>Invalid</u>

e.

Since, it follows all the four rules,

Therefore, this number is <u>Valid</u>

f.

Since, it follows all the four rules,

Therefore, this number is <u>Valid</u>

g.

It violates third rule for including unit.

Therefore, it is <u>Invalid</u>

6 0
4 years ago
If the torque required to loosen a nut that holds a wheel on a car has a magnitude of 55 n·m, what force must be exerted at the
erastova [34]

Either 175 N or 157 N depending upon how the value of 48° was measured from.    
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.    
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.    
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".     
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:  
U = F*cos(48)  
where  
U = Useful force  
F = Force applied    
So solving for F and calculating gives:  
U = F*cos(48)  
U/cos(48) = F  
117 N/0.669130606 = F  
174.8537563 N = F    
So 175 Newtons of force is required in this situation.    
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get  
U/cos(42) = F  
117/0.743144825 = F  
157.4390294 = F    
Or 157 Newtons is required for this case.
6 0
3 years ago
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