Answer:
the ball's was lowered because of the gravitational force and because of the friction force's
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?
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
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: