The marble's volume is equal to 1.2 
Density= mass/volume
Mass= 3.05 Grams
Volume= 1.2 cm^3
3.05/1.2= 2.92 <span>g/mL</span>
Density= mass/volume
Mass= 3.05 Grams
Volume= 1.2 cm^3
3.05/1.2= 2.92 <span>g/mL</span>
The current in one wire is 5.00 a. what is the current in the other wire.
1 answer:
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Answer:
The impact occured at a distance of 2478.585 meters from the person.
Explanation:
(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:
<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>
Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete (), in seconds:
(1)
Where:
- Time spent by the sound in concrete, in seconds.
- Time spent by the sound in the air, in seconds.
By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:
(2)
Where:
- Speed of the sound in concrete, in meters per second.
- Speed of the sound in the air, in meters per second.
- Distance traveled by the sound, in meters.
If we know that ,
and
, then the distance travelled by the sound is:
The impact occured at a distance of 2478.585 meters from the person.
Answer:
(a) Electric field at 0.250 m is zero.
(b) Electric field at 2.90 m is zero.
(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.
(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.
Explanation:
Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.
Electric field inside and outside the metal sphere is :
E = 0 for r ≤ R ( inside )
= for r > R ( outside )
Here K is electric constant and r is the distance from the center of the metal sphere.
(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.
(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.
(c) Electric field at 3.10 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.
E = -
E = - 5.15 x 10³ V/m
(d) Electric field at 8.00 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.
E = -
E = - 0.77 x 10³ V/m