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2 years ago

Potential energy calculation - MxGxH. Gravity equals 10. How much

Physics

1 answer:

Elena L [17]2 years ago

8 0

Answer:

Ep = 0.6095 [J]

Explanation:

As defined in the problem statement, potential energy is defined as the product of mass by gravity by height. But first we must convert all the values given to measures of the international system (SI)

g = gravity = 10 [m/s^2]

h = elevation = 40 [ft] = 12.19 [m]

m = mass = 5 [g] = 0.005 [kg]

Ep = potential energy [J]

Ep = 0.005*10*12.19 = 0.6095 [J]

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A plane comes in for a landing at a velocity of 80 meters per second west. As it touches down, it decelerates at a constant rate

azamat

Answer:

Answer D : about 1067 meters

Explanation:

There are two steps to this problem:

1) First find the time it takes the plane to stop using the equation for the acceleration:

$a=\frac{Vf-Vi}{t}$

Where Vf is the final velocity of the plane (in our case: zero )

Vi is the initial velocity of the plane (in our case: 80 m/s)

$a$ is the acceleration (in our case -3 m/s^2 - notice negative value because the velocity is decreasing)

$a=\frac{Vf-Vi}{t}\\-3=\frac{0-80}{t}\\t=\frac{-80}{-3} = \frac{80}{3}$

with units corresponding to seconds given the quantities involved in the calculation.

2) Second knowing the time it took the plane to stop, now use that time in the equation for the distance traveled under accelerated motion:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi= 80 (\frac{80}{3}) +\frac{1}{2} (-3) (\frac{80}{3}) ^{2}=1066.666666...$

Where the answer results in units of meters given the quantities used in the calculation.

We round this to 1067 meters

7 0

3 years ago

a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th

Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

Momentum = (mass) x (speed) = (362kg x 0.5m/s) = 181 kg-m/s

relative to the dock. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock. A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock. The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so

(300 x speed) = 36 kg-m/s .

Divide each side by 300: speed = 36/300 = 0.12 m/s , away from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum. The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction toward the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
(300 x speed) = 186

Divide each side by 300: speed = 186/300 = 0.62 m/s away from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from that at 0.62 m/s, and the person standing
on the dock sees you start to move away from him at 0.12 m/s, and that's the
same answer that I got earlier, in the frame of reference tied to the dock.

yay !

By the way ... thanks for the 6 points. The warm cloudy water
and crusty green bread are delicious.

4 0

3 years ago

When a rocket is traveling toward a mountain at 100 m/s, the sound waves from this rocket's engine approach the mountain at spee

Charra [1.4K]

Answer:

The correct option is C

Explanation:

From the question we are told that

The initial speed of the rocket is $v_i = 100 m/s$

The speed of the rocket engine sound is $V$

The final speed of the rocket is $v_f = 200 \ m/s$

The speed of the sound at $v_f$ would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).

A clear example when lightning strikes you will first see (that is because it travels at the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the lightning

Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates

3 0

3 years ago

How might you tell if a food contains an acid

MrRissso [65]

Answer:

it will taste sour

Explanation:

7 0

3 years ago

Is mercury a terrestrial or gaseous planet

liraira [26]

Mercury a terrestrial. It isn't made of gas.

8 0

3 years ago