Answer: The degree of the first term.
Explanation:
The function:
The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.
Let the left hand side be donated by -x.
Then,
Hence, the correct option is the degree of the first term indicates the left and right end points of the function.
Momentum = mass x velocity
= 1500 x 8
= 12000 kg m/s
Answer:
The average velocity for the time period beginning when t=1 and lasting 0.1 seconds = 16.40 ft/s.
Explanation:
Given that the height of the ball at time t is
The average velocity of an object is defined as the total displacement covered by the particle divided by the total time taken in covering that displacement.
If 
are the heights of the ball at time 
and 
, then the total displacement covered by the ball from time to is 
.
Thus, the average velocity of the ball for the time interval 
is given by
For the time interval, beginning when t = 1 second and lasting 0.1 seconds,
Answer:
3.9 × 10^7 J
Explanation:
Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J
Solution
Since the tank is half full, the height = 2.5m
Pressure = density × gravity × height
Pressure = 900 × 9.8 × 2.5
Pressure = 22050 Pascal
The cross sectional area of the pump will be area of a circle.
A = πr^2
A = π × 15^2
A = 706.858 m^2
Using the formula
Density = mass/volume
Mass = density × volume
Mass = 900 × 706.86 × 2.5
Mass = 1590.435
Energy = mgh
Energy = 1590.435 × 9.8 × 2.5
Energy = 38965657.8 J
Since the work done = energy
Therefore, the work done = 3.9 × 10^7 J
Answer:
a) -180.7 kN/C
b) -474.3 kN/C
c) 180.7 kN/C
Explanation:
For infinite planes the electric field is constant on each side, and has a value of:
E = σ / (2 * e0) (on each side of the plate the field points in a different direction, the fields point towards positive charges and away from negative charges)
The plate at -5 m produces a field of:
E1 = 2.6*10^-6 / (2 * 8.85*10^-12) = 146.8 kN/C into the plate
The plate at 3 m:
E2 = 5.8*10^-6 / (2 * 8.85*10^-12) = 327.5 kN/C away from the plate
At x < -5 m the point is at the left of both fields
The field would be E = 146.8 - 327.5 = -180.7 kN/C
At -5 m < x < 3 m, the point is between the plates
E = -146.8 - 327.5 = -474.3 kN/C
At x > 3 m, the point is at the right of both plates
E = -146.8 + 327.5 = 180.7 kN/C
