Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is
and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:
Answer:
13.9
Explanation:
Apparent weight is the normal force. Sum of the forces on the alloy when it is submerged:
∑F = ma
N + B − W = 0
N + ρVg − mg = 0
6.6 + (0.78 × 1000) V (9.8) − (0.750) (9.8) = 0
V = 9.81×10⁻⁵
If x is the volume of the first material, and y is the volume of the second material, then:
x + y = 9.81×10⁻⁵
(7.87×1000) x + (4.50×1000) y = 0.750
Two equations, two variables. Solve with substitution:
7870 (9.81×10⁻⁵ − y) + 4500 y = 0.750
0.772 − 7870 y + 4500 y = 0.750
0.0222 = 3370 y
y = 6.58×10⁻⁶
x = 9.15×10⁻⁵
The ratio of the volumes is:
x/y = 13.9
We are missing an important piece of information needed to answer this question: the number of kcal Charles losses per day. However, we can come up with a general equation in which kcal/day is the only independent variable.
We know that it takes 3500 kcal to lose one pound. To lose 5 pounds, Charles needs to lose 5 x 3500 kcal = 17,500 kcal.
To find how many days it takes Charles to lose 17,500 kcal (5 pounds), we must divide that amount by the number of kcal Charles loses per day.
Here is the equation to calculate that number
Number of days= 17500 / (kcal per day)
If given calories, remember that 1000 calories = 1 kcal, and .001 kcal = 1 cal
Explanation:
Upstroke is a mechanism which helps to raise the plunger and downstroke helps to help lower the plunger. On the up-stroke of the plunger, the lower valve opens and the upper valve is closed. ... Whereas, on the downstroke, the lower valve closes and the upper one opens.
Answer:
Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.
Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.
Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).
Explanation:
Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.
Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.
Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).
This is the answer
