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Trava [24]
2 years ago
10

Mechanical advantage is always less than validity ratio​

Physics
1 answer:
nata0808 [166]2 years ago
7 0

That statement is true.

The question wasn't that specific, so I couldn't give you an answer...

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The average mass of a car in the US is 1.440 x 10^6 g. Express this mass in kg.
wolverine [178]

Answer:

Average mass of acar in the US (in kg) = 1440 kg

Explanation:

Average mass of a car in the US (in g) = 1.440 × 10⁶ g

Mass in kg:

\rm 1 \: g =  {10}^{ - 3}  \: kg \\  \\  \rm 1.440 \times 10^6 \ g = 1.440 \times 10^6  \times  {10}^{ - 3} \ kg \\  \\  \rm = 1.440 \times 10^{6 - 3} \ kg \\  \\  \rm = 1.440 \times 10^3 \ kg \\  \\ \rm = 1.440 \times 1000 \ kg \\  \\ \rm = 1440 \ kg

5 0
2 years ago
The speed of sound through diamond is about 12,000 m/s. The speed of sound through wood is about 3,300 m/s. Which statement expl
Damm [24]

Answer: They have different rigidities.

Explanation:

7 0
2 years ago
A stunt woman attempts to swing from the roof of a m 24 high building to the bottom of an identical building using a m 24 rope,
MakcuM [25]

Answer:

Height from ground is 8 m where string will break

Explanation:

Let the string makes some angle with the vertical after some instant of time

So here we have

T - mg cos\theta = \frac{mv^2}{R}

T = mg cos\theta + \frac{mv^2}{R}

now by energy conservation we have

\frac{1}{2}mv^2 = mg(R cos\theta)

mv^2 = 2mgR cos\theta

T = mg cos\theta + 2mgcos\theta

T = 3mg cos\theta

For string break down we have

T = 2mg = 3mgcos\theta

cos\theta = \frac{2}{3}

Now height from the ground is given as

h = R - Rcos\theta

h = 24 - 24(\frac{2}{3})

h = 8 cm

7 0
3 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
2 years ago
Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2
sleet_krkn [62]

-- Accelerating at the rate of 8 m/s², Andy's speed
   after 30 seconds is

                                   (8 m/s²) x (30.0 s)  =  240 m/s .

-- His average speed during that time is

                                 (1/2) (0 + 240 m/s)  =  120 m/s .

-- In 30 sec at an average speed of 120 m/s,
   Andy will travel a distance of
                                                 (120 m/s) x (30 sec) = 3,600 m

                                                                                = 3.6 km .

"But how ? ! ?", you ask.

How in the world can Andy leave a stop light and then
cover 3.6 km = 2.24 miles in the next 30 seconds ?

The answer is:  His acceleration of  8 m/s², or about  0.82 G
is what does it for him.

At that rate of acceleration ...

-- Andy achieves "Zero to 60 mph" in 3.35 seconds,
   and then he keeps accelerating.

-- He hits 100 mph in 5.59 seconds after jumping the light ...
   and then he keeps accelerating.

-- He hits 200 mph in 11.2 seconds after jumping the light ...
   and then he keeps accelerating.

-- After accelerating at 8 m/s² for 30 seconds, Andy and his
   car are moving at  537 miles per hour !
   We really don't know whether he keeps accelerating,
    but we kind of doubt it. 

A couple of observations in conclusion:

-- We can't actually calculate his displacement with the information given.
   Displacement is the distance and direction between the starting- and
   ending-points, and we're not told whether Andy maintains a straight line
   during this tense period, or is all over the road, adding great distance
   but not a lot of displacement.

-- It's also likely that sometime during this performance, he is pulled
   over to the side by an alert cop in a traffic-control helicopter, and
   never actually succeeds in accomplishing the given description. 
5 0
3 years ago
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