Answer:
the ball's was lowered because of the gravitational force and because of the friction force's
How can the power of an object required to lift a mass up a height be increased ?
A .reduce the height
B .reduce the mass
C .increase the time
D .reduce the time
The amswer should be B. reduce the mass
The frequency of a wave is 2 Hz and the wavelength is 4 m. What is the wave speed?
Answer:
4/2=2m/s
Explanation:
Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.50-cm slide behind a lens of 20.0 cm focal length in the slide projector. A) How far from the lens should the slide be placed in order to shine on a screen 6.00 m away? B) How wide must the screen be to accommodate the projected image?
Answer:
A) d_o = 20.7 cm
B) h_i = 1.014 m
Explanation:
A) To solve this, we will use the lens equation formula;
1/f = 1/d_o + 1/d_i
Where;
f is focal Length = 20 cm = 0.2
d_o is object distance
d_i is image distance = 6m
1/0.2 = 1/d_o + 1/6
1/d_o = 1/0.2 - 1/6
1/d_o = 4.8333
d_o = 1/4.8333
d_o = 0.207 m
d_o = 20.7 cm
B) to solve this, we will use the magnification equation;
M = h_i/h_o = d_i/d_o
Where;
h_o = 3.5 cm = 0.035 m
d_i = 6 m
d_o = 20.7 cm = 0.207 m
Thus;
h_i = (6/0.207) × 0.035
h_i = 1.014 m
What will happen to the force felt between two charged objects if the distance between them is 1/3rd of the original distance
Answer:
New force = 9(initial force)
Explanation:
The force between two charges is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
Where
d is the original distance
Let d' is the new distance such that, r' = r/3
New force,
[tex]F'=\dfrac{kq_1q_2}{r'^2}\\\\F'=\dfrac{kq_1q_2}{(\dfrac{r}{3})^2}\\\\F'=9\times \dfrac{kq_1q_2}{r^2}\\\\F'=9F[/tex]
So, the new force becomes 9 times the initial force.
Which of the following is evidence that a chemical reaction has occurred?
