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aivan3 [116]
2 years ago
5

1) A marble, rolling with speed 20 cm/s, rolls off the edge of a table that is 80 cm high.

Physics
1 answer:
Schach [20]2 years ago
4 0
1) A marble, rolling with speed 20 cm/s, rolls off the edge of a table that is 80 cm high.
(a) How long does it take to drop to the floor?
(b) How far, horizontally, from the table edge does the marble strike the floor?
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Can any kind soul help me please​
Alex

Answer:

I) 420000J

ii)

Explanation:

(I) so you can use the formula for quantity of heat then substitute the values given

formula-Q=mc∆9

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3 years ago
Which layer is made of flowing solid rock
ASHA 777 [7]
The mantle is correct
3 0
3 years ago
A block of wood has a mass of 120 g and a volume of 200cm what is the density of the word
Tanzania [10]
Density =mass/volume 120/200 =0.6 g/cm
8 0
3 years ago
An object 1 of mass m1 is separated by some distance d from an object 2 of mass 2m1 . An object 3 of mass m3 is to be placed bet
Harrizon [31]

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

If the potential energy of the three-object system is to be a maximum (closest to zero), should object 3 be placed closer to object 1, closer to object 2, or halfway between them?

 

Object 3 should be placed closer to object 1.

 

Object 3 should be placed on a halfway between object 2 and object 1.

 

Object 3 should be placed closer to object 2.

 

Solution

I think that Object 3 should be placed closer to object 2.

6 0
3 years ago
A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at
wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 =  \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s -  2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

4 0
3 years ago
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