We have that the see-saw will be balance if a weight of 
is added to Nural's side of the see-saw
From the Question we are told that
Wan’s mass is 
Nurul’s mass is 
Generally
The Will be balance when the weight on both sides of the see-saw are equal
In conclusion
The see-saw will be balance if a weight of is added to Nural's side of the see-saw
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Answer: 34.9 cm
Explanation:
You are given the following parameters;
Object distance U = 32 cm
Magnification M = - 12.0
According to formula for magnification;
M = V/U
Where V = image distance.
Substitute V and M into the formula
-12 = V/32
Cross multiply
V = -12 × 32
V = - 384
You can use the formula
1/f = 1/V + 1/U
Where f = focal length
Substitute V and U into the formula
1/f = - 1/384 + 1/32
Find the lowest common factor of the denominator at Left hand side
1/f = ( -1 + 12 ) / 384
1/f = 11/384
Reciprocate both sides
F = 384/11
F = 34.9 cm
He should therefore use the focal length of 34.9 cm
Answer:
2.5A , 22500 JOULE
Explanation:
Resistance (R)= 6 ohm
Voltage (V) = 15 volt
a)
Current (I) = ?
From ohm's law,
V=IR
15 = 6*I
I = 2.5 A
b)
time(t) = 10min = 600 second
heat energy (Q) = ?
from Joule's law of heating,
Q = I²*R*t
Q = 2.5² *6*600
Q = 22500 Joule
hence, heat energy produced is 22500 joule.
Given:
The initial velocity of the object, v=30 m/s
a_t=0
a_c≠0
The time period is Δt.
To find:
The right conclusion among the given choices.
Explanation:
a_t represents the tangential accleration on the object and a_c represents the centripetal acceleration on the object.
The centripetal acceleration is the acceleration that keeps the object in its circular path. The centripetal force only changes the direction of the velocity and not the magnitude.
Thus the magnitude of the velocity of the object remains the same after a time interval of Δt. But the direction of the velocity of the object will be changed and will be unknown after Δt seconds.
Final answer:
Thus the object will be traveling at 30 m/s in some unknown direction.
Therefore, the correct answer is option a.
