1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum
Answer:
Explanation:
1760 yd/mi / 120 yd/field = 14⅔ fields/mi
Answer:
2.72 km
Explanation:
(12.33 km)/ 1 hr * (1 hr)/ 60 min
0.2055 km/ min
distance=rate * time (assuming v is constant,
a=0)
(0.2055 km/ min)*(13.22 min)
2.72 km OR 2716.71 m
Answer:
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.
Explanation:
From coulomb's law, F = Eq
Thus,
F = E₁q₁
F = E₂q₂
Then
E₂q₂ = E₁q₁

where;
E₂ is the external electric field due to second test charge = ?
E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C
q₁ is the first test charge = 13 mC
q₂ is the second test charge = 23 mC
Substitute in these values in the equation above and calculate E₂.

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.
However, the direction of the external field is still to the right.
Answer:
A
Explanation:
So a pulse is a part of a mechanical wave, and mechanical waves are energy transfer trough some medium, in this case a stretched spring. So the correct answer is (A) energy only. The pulse cant be transferred into mass.