The Image distance and Magnification of The Image will be 30 cm and 3.
<h3>What is focal length?</h3>
The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.
Given data;
Focal length,f=?
Image distance,v=?
Object distance,u= 10 cm
Magnification,m= 2.85
The focal length is half of the radius;
f=R/2
f=30 Cm/2
f= 15 Cm
The mirror equation is found as;
The magnification of the lens is found as;
Hence, the image distance and magnification of The image will be 30 cm and 3.
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Answer:
0.36 A.
Explanation:
We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:
Resistor 1 (R₁) = 35 Ω
Resistor 2 (R₂) = 20 Ω
Equivalent Resistance (Rₑq) =?
Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:
Rₑq = (R₁ × R₂) / (R₁ + R₂)
Rₑq = (35 × 20) / (35 + 20)
Rₑq = 700 / 55
Rₑq = 12.73 Ω
Next, we shall determine the total resistance in the circuit. This can be obtained as follow:
Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω
Resistor 3 (R₃) = 15 Ω
Total resistance (R) in the circuit =?
R = Rₑq + R₃ (they are in series connection)
R = 12.73 + 15
R = 27.73 Ω
Finally, we shall determine the current. This can be obtained as follow:
Total resistance (R) = 27.73 Ω
Voltage (V) = 10 V
Current (I) =?
V = IR
10 = I × 27.73
Divide both side by 27.73
I = 10 / 27.73
I = 0.36 A
Therefore, the current is 0.36 A.
On Earth, the acceleration of gravity is 9.8 m/s² downward.
So any object with only gravity acting on it gains 9.8 m/s of
downward speed every second.
If the rock starts out moving upward at 10 m/s, then it will
continue upward for only (10/9.8) = 1.02 second, before
it stops rising and starts falling.
Its average speed during that time is (1/2) (10 + 0) = 5 m/s .
At an average speed of 5 m/s for 1.02 sec,
the rock rises
(5 m/s) x (1.02 sec) = 5.102 meters .
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass 
kg, charge +e = 
C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 
m/s. The proton comes momentarily to rest at a distance 
m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 
m apart?
Explanation:
The given data is as follows.
Mass of proton = 
kg
Charge of proton = 
Speed of proton = 
Distance traveled = 
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
= 
where, 
U = 
Putting the given values into the above formula as follows.
U =
= 
= 
Therefore, we can conclude that the electric potential energy of the proton and nucleus is .
1.Life science involves fields of discipline catering to living organisms such as we humans while physical science caters to non-living organisms.
2.Life science has more fields of discipline than physical science.
3.Physical science relies on laws and theories to explain concepts while life science relies on biological explanations and can also rely on theories.
