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3 years ago

Mass is measured using

1 answer:

7 0

By measuring the force that the object exerts on a scale (this is usually done with a dynamometer). This gives you the force W=m g. A some division, W/g, gives you the mass of the object.

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A ball is tossed up in the air. at its peak, it stops before beginning to fall. the ball at its peak has

Arte-miy333 [17]

The projectile (ball) reaches an instantaneous vertical speed (Vy) of zero at maximum height.

so, V(max height) = ¬Г(Vx)^2+(Vy)^2
in this case V(max height) = Vx, where Vy=0

The maximum height, Yf, can be solved using Vfy^2=Viy^2 + 2gy. At maximum height Vfy=0.

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3 years ago

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In order to protect underground pipelines and storage tanks made of iron, the iron is made the cathode of a voltaic cell through

emmasim [6.3K]

The answer would be that the process to protect the underground pipelines and storage tanks is cathodic protection.

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3 years ago

I was driving along at 20 m/s , trying to change a CD and not watching where I was going. When I looked up, I found myself 46 m

vampirchik [111]

Answer:

6.46393559312 m/s²

Explanation:

Time taken to cover 56 m

$t=\dfrac{56}{29}\ s$

Distance covered in 0.42 seconds

$0.42\times 20=8.4\ m$

From equation of linear motion

$s=ut+\frac{1}{2}at^2$

$8.4+20(\dfrac{56}{29}-0.42)+\dfrac{1}{2}a(\dfrac{56}{29}-0.42)^2=46\\\Rightarrow a=(46-8.4-20(\dfrac{56}{29}-0.42))\times\dfrac{2}{(\dfrac{56}{29}-0.42)^2}\\\Rightarrow a=6.46393559312\ m/s^2$

The minimum acceleration is 6.46393559312 m/s²

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3 years ago

Which is part of an AC generator?

Katen [24]

There is no such thing as an AC Generator, if its a generator, its DC. I think you are talking about an AC Alternator, in which case the answer would be -wrong answer, deleted-

6 0

3 years ago

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A subway train is traveling at 22.2 m/s when it approaches a slower train 50m ahead traveling in the same direction at 6.94 m/s.

Amiraneli [1.4K]

Answer:

Time that they collide = 4.99s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Explanation:

We will use the equations of motion to obtain the solution required

At time t = 0

speed of first train = 22.2 m/s

Initial space between the two trains = 50 m

Speed of second train = 6.94 m/s

For the first car, distance covered by the first train = y

y = distance covered between the beginning of the deceleration and the point where the the two trains hit one another.

u = initial velocity = 22.2 m/s

t = time taken for all this to happen

a = deceleration = - 2.1 m/s²

y = ut + (1/2)at²

y = 22.2t - 1.05t² (eqn 1)

For the second train,

At t = 0, y = 50 m

Let the new distance moved by the second train before collision = (y - 50)

u = initial velocity = 6.94 m/s

t = time taken = t

a = acceleration of the second train = 0 m/s² (constant velocity)

(y - 50) = ut + (1/2)at²

y - 50 = 6.94t

y = 6.94t + 50 (eqn 2)

substituting for y in eqn 2 using the expression obtained in eqn 1

y = 22.2t - 1.05t²

y = 6.94t + 50

22.2t - 1.05t² = 6.94t + 50

1.05t² - 15.26t + 50 = 0

Solving this quadratic equation

t = 4.99 s or 9.54 s

The position of the two trains are the same at those two times, but the first time is when they hit each other.

t = 4.99 s

At 4.99 s, the the velocity of the first train

v = u + at

v = 22.2 + (-2.1×4.99) = 11.721 m/s in the same direction as the second train.

Relative velocity at this point will be

= 11.721 - 6.94 = 4.781 m/s

Relative speed of the trains when they collide: The relative speed of The first train relative to the second, slower train at collision = 4.781 m/s

Hope this Helps!!!

4 0

3 years ago