In a way it’s true because you can get a ticket for getting caught littering
Answer:
0.4 g/cm^3
Explanation:
The density of an object can be found using the following formula.
d= m/v
where m is the mass and v is the volume.
The mass of the metal is 6 grams and the volume is 15 centimeters^3
m=6 g
v= 15 cm^3
Substitute these into the formula.
d= 6 g/ 15 cm^3
Divide 6 g by 15 cm^3 (6/15=0.4)
d= 0.4 g/ cm^3
The density of the metal is 0.4 grams per cubic centimeter.
Answer:
Four times higher
Explanation:
F- G (m1 x m2)/ r^2
if r 1 = 2 and r 2 = 1 therefore F = G (m1 x m2)/ 1^2 is 4 times higher than
2^2 since G and m1 and m2 remained the same
Answer:
The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.
Explanation:
The formula for electric field is given as:
E = F/q
where,
E = Electric field
F = Electric Force
q = Charge Experiencing Force
Now, we apply this formula to all the cases given in question.
A) <u>A 2C charge acted on by a 4 N electric force</u>
F = 4 N
q = 2 C
Therefore,
E = 4 N/2 C = 2 N/C
B) <u>A 3 C charge acted on by a 5 N electric force</u>
F = 5 N
q = 3 C
Therefore,
E = 5 N/3 C = 1.67 N/C
C) <u>A 4 C charge acted on by a 6 N electric force</u>
F = 6 N
q = 4 C
Therefore,
E = 6 N/4 C = 1.5 N/C
D) <u>A 2 C charge acted on by a 6 N electric force</u>
F = 6 N
q = 2 C
Therefore,
E = 6 N/2 C = 3 N/C
E) <u>A 3 C charge acted on by a 3 N electric force</u>
F = 3 N
q = 3 C
Therefore,
E = 3 N/3 C = 1 N/C
F) <u>A 4 C charge acted on by a 2 N electric force</u>
F = 2 N
q = 4 C
Therefore,
E = 2 N/4 C = 0.5 N/C
The highest field is 3 N, which is found in part D.
<u>A 2 C charge acted on by a 6 N electric force</u>
