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NISA [10]
3 years ago
7

Two objects have masses m and 5m, respectively. They both are placed side by side on a frictionless inclined plane and allowed t

o slide down from rest. Two objects have masses m and 5m, respectively. They both are placed side by side on a frictionless inclined plane and allowed to slide down from rest. A.The two objects reach the bottom of the incline at the same time.B.It takes the lighter object 5 times longer to reach the bottom of the incline than the heavier object. C.It takes the heavier object 10 times longer to reach the bottom of the incline than the lighter object. D.It takes the heavier object 5 times longer to reach the bottom of the incline than the lighter object. E.It takes the lighter object 10 times longer to reach the bottom of the incline than the heavier object.
Physics
1 answer:
Mama L [17]3 years ago
6 0

Answer:

A.The two objects reach the bottom of the incline at the same time.

Explanation:

-Since there is the incline is frictionless, there's no external force acting on the two objects except gravitational force(9.8N/kg).

-Gravitational acceleration ,9.8m/s^2, is equal for any two masses left in free fall despite the difference in massess.

-Therefore, the two bodies will reach the bottom of the incline at the exact same time.

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A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric
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The electric field produced by a single-point charge is given by

E(r)=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge


To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.


1) The first charge is q=-3.00 nC=-3.00 \cdot 10^{-9} C, and it is located at x=0, so its distance from the point x=0.200 m is

r=0.200 m-0=0.2 m

Therefore, the electric field is

E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.


2) The second charge is q=-5.50 nC=-5.5 \cdot 10^{-9}C and it is located at x=0.800 m, so its distance from the point is

r=0.800 m-0.200 m=0.6 m

Therefore, the electric field is

E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.


3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C

and the sign tells us that the field is directed toward negative x-direction.

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