Answer:
A. Na2S + 2KCI - 2NaCl + KZS
Explanation:
Answer:
The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum. ... The collision would change the halfback's speed and thus his momentum.
Explanation:
Answer:

Explanation:
The equation relating the the enthalphy, pressure and temperature is expressed as

Where P is the pressure, H is the enthalphy, and T is the temperature.
since the given values are

if we insert values, we arrive at

Refer to the diagram shown below.
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
Answer:
Electric field E = kQ/r^2
Distance between charges = 6.30 - (-4.40) = 10.70m
Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.
Field from q1 at P = k(-9.50x^10^-6) / d^2
Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2
These fields are in opposite directions and are equal magnitudes if the resultant field = 0
k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2
9.50 / d^2 =8.40 / (10.70-d)^2
d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131
d/(10.70-d) = sqrt(1.1331) = 1.063
d = 1.063 ((10.70-d)
= 10.63 - 1.063d
2.063d = 10.63
d = 5.15m
The y coordinate where field is zero is 6.30 - 5.15 = 1.15m
Explanation: