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2 years ago

How does the mass affect the speed as it rolls down a slope?

1 answer:

6 0

As we use the Kinetic energy and the equation is 1/2mv^2, changing its mass will change its speed and its energy. So more mass, more speed more energy. also the gravitational potential energy; mass x gravity x height; more mass and more height more speed as it go down to the slope! Hope it helps!

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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel

Zanzabum

Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m

8 0

2 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for M, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

5 0

2 years ago

A stone of mass 5g is projected with a rubber catapult. if the catapult is stretched through a distance of 7cm by an average for

weqwewe [10]

Explanation:

solution: mass m = 5g = 0.005kg; extension e = 7cm = 0.07m; force f = 70 N; velocity = ?; using: work done in elastic material w = 1/2 fe = 1/2 ke2 = 1/2 mv2 - the kinetic energy of the moving stone. 1/2 fe =...

4 0

1 year ago

Robert hook used an early microscope to observed a cork sample. How did this help contribute to cell theory?

Lady bird [3.3K]

Answer:

Robert Hooke used an early microscope to observe a cork sample. How did this help contribute to cell theory? It helped to show that cells contain water. ... It helped to show that some cells are visible to the naked eye.

Explanation:

If your teacher checks if it was copied just put it in your on words

3 0

2 years ago

Storage of unlimited quantities of energy in batteries is possible. TrueFalse

dangina [55]

False

It is impossible to get infinite energy let alone to put it inside one battery

3 0

3 years ago

Read 2 more answers