Answer:
609547.12 Pa ≈ 6.10×10^5 Pa
Explanation:
Step 1:
Data obtained from the question. This include the following:
Force (F) = 49.8 N
Radius (r) = 0.00510 m
Pressure (P) =..?
Step 2:
Determination of the area of the head of the nail.
The head of a nail is circular in nature. Therefore, the area is given by:
Area (A) = πr²
With the above formula we can obtain the area as follow:
Radius (r) = 0.00510 m
Area (A) =?
A = πr²
A = π x (0.00510)²
A = 8.17×10^-5 m²
Therefore the area of the head of the nail is 8.17×10^-5 m²
Step 3:
Determination of the pressure exerted by the hammer.
This is illustrated below:
Force (F) = 49.8 N
Area (A) = 8.17×10^-5 m²
Pressure (P) =..?
Pressure (P) = Force (F) /Area (A)
P = F/A
P = 49.8/8.17×10^-5
P = 609547.12 N/m²
Now, we shall convert 609547.12 N/m² to Pa.
1 N/m² = 1 Pa
Therefore, 609547.12 N/m² = 609547.12 Pa.
Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa
Answer:
The wires are connected to both terminals of the battery, so they form a closed loop. Most circuits have devices such as light bulbs that convert electrical energy to other forms of energy. ... When the switch is turned on, the circuit is closed and current can flow through it.
Explanation:
Answer:
Maximum Tension=224N
Minimum tension= 64N
Explanation:
Given
mass =8 kg
constant speed = 6m/s .
g=10m/s^2
Maximum Tension= [(mv^2/ r) + (mg)]
Minimum tension= [(mv^2/ r) - (mg)]
Then substitute the values,
Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N
Minimum tension= [8 × 6^2)/2 -(8×9.8)]
=64N
Hence, Minimum tension and maximum Tension are =64N and 2224N respectively
Answer:
When magnesium reacts with oxygen, it produces light bright enough to blind you temporarily. Magnesium burns so bright because the reaction releases a lot of heat. As a result of this exothermic reaction, magnesium gives two electrons to oxygen, forming powdery magnesium oxide (MgO).
The charge of the object must be 
Answer: Option C
<u>Explanation:</u>
Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.
Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

Here, given E = 4500 N/C and F = 0.05 N.
We need to find charge of the object (q)
By substituting the given values, we get
