Answer:
The answer is 24cm
Explanation:
This problem bothers on the curved mirrors, a concave type
Given data
Object height h= 5cm
Object distance = 12cm
Focal length f=24cm
Let the image distance be v=?
Applying the formula we have
1/v +1/u= 1/f
Substituting our given data
1/v+1/12=1/24
1/v=1/24-1/12
1/v=1-2/24
1/v=-1/24
v= - 24cm
This implies that the image is on the same side as the object and it is real
Answer: a. F doubled
b. F reduced by one-quarter i.e
1/4*(F)
c. 1/9*(F)
d. F increased by a factor of 4 i.e 4*F
e. F reduces 3/4*(F)
Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is
F = k*(qA*qB)/d²
a. If qA is doubled therefore the force is doubled since they are directly proportional.
b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4
Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.
c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9
d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4
e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,
3/4*F
The answer to question one is A.
The answer to question two is A.
The answer to question three is D.
Answer:
Depends on how long the string is, how heavy the weight, and how high you let go of it.
But it will most likely hit you :)
Answer:
f = 614.28 Hz
Explanation:
Given that, the length of the air column in the test tube is 14.0 cm. It can be assumed that the speed of sound in air is 344 m/s. The test tube is a kind of tube which has a closed end. The frequency in of standing wave in a closed end tube is given by :
f = 614.28 Hz
So, the frequency of the this standing wave is 614.28 Hz. Hence, this is the required solution.
