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Ronch [10]
3 years ago
14

If you're going 80 mph how long does it take to go 80 miles

Physics
1 answer:
Elza [17]3 years ago
6 0
1 mile. Is this a joke lol
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Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin
erastova [34]

Answer

(C).

When there is an angle between the two directions, the cosine of the angle must be considered.


Step by step Solution

The work done by a force is defined as the product of the force and the distance traveled in the direction of motion.

The first answer "Only the component of the force perpendicular to the motion is used to calculate the work" is wrong because, the force perpendicular to motion does no work.

The second choice "If the force acts in the same direction as the motion, then no work is done" is wrong because the work in the direction of the force is W=F\times d.

Fourth answer "A force at a right angle to the motion requires the use of the sine of the angle" is wrong because the sin(90)=0 meaning that there is no work done in the direction perpendicular to the motion.

The third answer" When there is an angle between the two directions, the cosine of the angle must be considered." is correct because the work is calculated using the force in the direction of the motion. The magnitude of this force is F\times d\times \cos(\theta).




4 0
3 years ago
Read 2 more answers
A transformer is to be used to provide power for a computer disk drive that needs 6.0 V (rms) instead of the 120 V (rms) from th
Ipatiy [6.2K]

Answer:

N_{2}=20 turns

Explanation:

The given case is a step down transformer as we need to reduce 120 V to 6 V.

number of turns on primary coil N_{P}= 400

current delivered by  secondary coil  I_{S}= 500 mA

output voltage = 6 V (rms)

we know that

I_{p}=\frac{V_{out}}{V_{in}\times I_{s}}

putting values we get

I_{p}=\frac{6}{120\times 0.5}

I_{p}= 0.1 A

to calculate number of turns in secondary

\frac{N_{2}}{400} =\frac{6}{120}

therefore, N_{2}=20 turns

5 0
3 years ago
A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​
Pani-rosa [81]

Explanation:

If the stones are unloaded from the boat, the weight of the boat will decrease. Therefore, the volume of the water displaced by the boat will also decrease. Due to this, the volume of the boat immersed in the water decreases. Hence, the level of the water around the boat will decrease.

5 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
3 years ago
The mass luminosity relation L  M 3.5 describes the mathematical relationship between luminosity and mass for main sequence sta
ivanzaharov [21]

Answer:

(a) <u>11.3 L</u>

(b) <u>10 M</u>

Explanation:

The mass-luminosity relationship states that:

Luminosity ∝ Mass^3.5

Luminosity = (Constant)(Mass)^3.5

So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:

Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)

where,

L = Luminosity of Sun

M = mass of Sun

(a)

It is given that:

Mass of Star = 2M

Therefore, eqn (1) implies that:

Luminosity of star/L = (2M/M)^3.5

Luminosity of Star = (2)^3.5 L

<u>Luminosity of Star =  11.3 L</u>

(b)

It is given that:

Luminosity of star = 3160 L

Therefore, eqn (1) implies that:

3160L/L = (Mass of Star/M)^3.5

taking ln on both sides:

ln (3160) = 3.5 ln(Mass of Star/M)

8.0583/3.5 = ln(Mass of Star/M)

Mass of Star/M = e^2.302

<u>Mass of Star = 10 M</u>

3 0
3 years ago
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