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SCORPION-xisa [38]
2 years ago
10

A 7800 W lift took 10 seconds to move a 5000 N weight. How far did the lift move this weight?

Physics
1 answer:
Eduardwww [97]2 years ago
7 0

Answer:

Both are moving at 30 km/h, so their speed is the same. ... enough fuel for the trip/how long it will take. 4 Weight is a force, and so is a vector. ... c At 10 seconds David's displacement is.

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The children at the local park think the slide is too slow. There is too much fiction. What could you do to decrease the amount
igor_vitrenko [27]
In order to decrease the friction on the slide,
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7 0
3 years ago
Water that is moving across Earth’s surface is called
Vanyuwa [196]

Answer:

I'm pretty sure the answer is runoff

5 0
2 years ago
Read 2 more answers
The energy stored per unit volume of the inductor is called the energy density of the magnetic field. Why?
Marrrta [24]

Answer:

yes because physics want that

7 0
2 years ago
This is for physical science, need help :/
marusya05 [52]

1).  From the frame of reference of a passenger on the airplane looking out of his window, the tree appears to be moving, at roughly 300 miles per hour toward the left of the picture.

2).  The SI unit best suited to measuring the height of a building is the meter.

3).  'Displacement' is the straight-line distance and direction from the start-point to the end-point, regardless of the path that was followed to get there.  

The ball started out in the child's hand, and it ended up 2 meters away from her in the direction of the wall.  So the displacement of the ball from the beginning to the end of the story is:  2 meters toward the wall.


6 0
2 years ago
On earth, two parts of a space probe weigh 14500 N and 4800 N. These parts are separated by a center-to-center distance of 18 m
Nastasia [14]

Answer:

F = 1.489*10^{-7}  N

Explanation: Weight of space probes on earth is given by:W= m*g

W= weight of the object( in N)

m= mass of the object (in kg)

g=acceleration due to gravity(9.81 \frac{m}{s^{2} })

Therefore,

m_{1} = \frac{14500}{9.81}

m_{1} = 1478.08  kg

Similarly,

m_{2} = \frac{4800}{9.81}

m_{2} = 489.29  kg

Now, considering these two parts as uniform spherical objects

Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.

Force between these two objects is given by:

F =  \frac{Gm_{1} m_{2}}{R^{2} }

G= gravitational constant (6.67 * 10^{-11} m^{3} kg^{-1} s^{-2})

m_{1} , m_{2}= masses of the object

R= distance between their centres (in m)(18 m)

Substituiting all these values into the above formula

F = 1.489*10^{-7}  N

This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.

7 0
3 years ago
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