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2 years ago

Can someone help me with universal gravation

1 answer:

6 0

States that particles are attracts with every other particle. wich force is directily proportional product of two masses and inversely proportional to the distance between the centers.

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A 500N person stands 2.5m from a wall against which a horizontal beam is attached. The beam is 6m long and weighs 200N. A cable

Vinil7 [7]

PART A)

Here by force balance along Y direction

$Tsin45 + F_y = 500 + 200$

Force balance along X direction

$Tcos45 = F_x$

now by torque balance

$Tsin45 (6) = 500 (2.5) + 200 (3)$

$T(4.24) = 1250 + 600$

$T = 436.3 N$

PART B)

now from above equations

$Tsin45 + F_y = 700$

$F_y = 391.5 N$

$F_x = Tcos45 = 308.5 N$

now net reaction force of wall is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = 498.4 N$

5 0

3 years ago

18. _______ is a subtype of the continental climate.

irina1246 [14]

C humid subtropical

3 0

3 years ago

The mass of a ship before launch is 55,000 metric tons. The ship is launched down a ramp and drops a total of 10 vertical meters

skelet666 [1.2K]

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C (3)

ΔT = 5.39x10⁹ / 4200 * 75000

Hope this helps

5 0

2 years ago

Hello, how to do qn 5?:)l

Hitman42 [59]

Answer:

Explanation:

Please see the attached picture for the full solution.

Since we are only concerned about the decrease in gravitational potential energy of the car, we look at the decrease in height of the car as it moves from point X to point Y, instead of the distance travelled by the car.

7 0

2 years ago

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di

Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $\frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} } = \frac{106x}{x+53}$

$26.5 = \frac{106x}{x+53} \\ \\ 79.5 x = 1404.5\\ \\ x = 17.67 miles/hour$

5 0

3 years ago

Can someone help me with universal gravation ​

Can someone help me with universal gravation