Answers
Answer:
Force, F = 20240 N
Explanation:
It is given that,
Pressure exerted by the four tires of an automobile,
Area of each tire,
Area of 4 tires,
We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :
F = 20240 N
So, the weight of the automobile is 20240 N. Hence, this is the required solution.
Final answer:
The vehicle's weight can be calculated by rearranging the definition of pressure (Pressure = Force/Area) to solve for Force (Force = Pressure * Area), then multiplying by four to account for all four tires. Remember that the result will be in newtons, so to convert it to kilograms, divide by gravitational acceleration.
Explanation:
Your question revolves around the concept of pressure. Tire pressure is a type of air pressure which is a part of physics. To determine the weight of the automobile in which each of the four tires has an area of 0.023 m2 and is inflated to a gauge pressure of 2.2 x 105 Pa, we need to utilize the fundamental equation of pressure:
P = F/A
Where P is the pressure, F is the force (which in this case will be the weight of the car), and A is the area of each of the tires where they are in contact with the ground. Solving for the weight (F) results in:
F = P * A
In your case, because there are four tires we multiply the result by four, therefore:
F = 4 * (2.2 x 105 Pa) * (0.023 m2)
We have to multiply this by 4 to account for all four tires. Finally, your weight will be in newtons, to convert it to kg you will divide by gravitational acceleration (approx 9.8 m/s2).
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Answers
Answer:
Explanation:
Zeeman Effect -
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Answers
Answer: 4,438.96m
Explanation:
(kindly find attachment below)
From the attachment below, it can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A+B as the hypotenus, 3.2km as the opposite and the displacement B as the adjacent.
By using phythagoras theorem
H² = O² + A²
(5.38)² = (3.20)² + B²
28.944 = 10.24 + B²
B² = 28.944 - 10.24
B² = 18.7044
B = √18.7044
B = 4.439km to meter is 4.439 * 1000 = 4,438. 96m
Answer:
B = 4325 m
Explanation:
Resolving the displacement into x and y components.
Let north = positive y component
East = positive x component
So,
Rx = B
Ry = -3.20 km
Magnitude of the resultant displacement is
R = √(B^2 + (-3.20)^2)
R is given as R = 5.38 km
Making B the subject of formula;
B = √(R^2 - (-3.20)^2)
B = √(5.38^2 - (-3.20)^2)
B = 4.325 km
B = 4325 m
Answers
Answer:
slit width, b = 0.2671 mm
Given:
distance of screen from the slit, x = 60.0 cm
wavelength of light,
distance between 1st and 3rd minima, t = 3.10 mm =
Solution:
Calculation of the distance between 1st and 3rd minima:
b = 0.2671 mm
slit width, b = 0.2671 mm
a water molecule,
A. the electronegative atom becomes strongly positive
B. the hydrogen atom becomes partially positive
O C. the oxygen atom becomes partially negative
If answer is right WILL GIVE BRAINLIEST
D. the hydrogen atom becomes partially negative
Answers
Answer:
I'm leaning twards A
Explanation:
Answers
Answer:
a) W₁ = - 127 J, b) W₂ = 148.18 J, c) = 3.43 m/s and d)
= 3.43 m / s
Explanation:
The work is given the equation
W = F. d
Where the bold indicates vectors, we can also write this expression take the module of each element and the angle between them
W = F d cos θ
They give us displacement, let's use Newton's second law to find strength, like the block has an equal acceleration (a = g / 7). We take a positive sign down as indicated
W-T = m a
T = W -m a
T = mg -mg/7
T = mg 6/7
T = 3.6 9.8 6/7
T = 30.24 N
Now we can apply the work equation to our problem
a) the force of the cord is directed upwards, the displacement is downwards, so there is a 180º angle between the two
W₁ = F d cos θ
W₁ = 30.24 4.2 cos 180
W₁ = - 127 J
b) the force of gravity is directed downwards and the displacement is directed downwards, the angle between the two is zero (T = 0º)
W₂ = (mg) d cos 0º
W₂ = 3.6 9.8 4.2
W₂ = 148.18 J
c) kinetic energy
K = ½ m v²
Let's calculate speed with kinematics
² = vo² + 2 a y
v₀ = 0
a = g / 7
² = 2g / 7 y
= √ (2 9.8 4.2 / 7)
= 3.43 m/s
We calculate
K = ½ 3.6 3.43²
K = 21.18 J
d) the speed of the block and we calculate it in the previous part
= 3.43 m / s
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?
Answers
Answer:
A. 2.083 MV/m from anode to cathode.
B. 93648278.15 m/s
C. 2.5x10^-5 C and there are about 1.56x10^14 electrons
D. 4x10^-15 Joules
Explanation:
Voltage V across plate is 25 kV = 25x10^3 V
Distance apart x = 1.2 cm = 1.2x10^-2 m
A. Electric field strength is the potential difference per unit distance
E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m
= 2.083 MV/m
B. Energy of electron is electron charge times the voltage across
i.e eV
Charge on electron = 1.6x10^-19 C
Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules
Mass of electron m is 9.12x10^-31 kg
Kinetic energy of electron = 0.5mv^2
Where v is the speed
4x10^-15 = 0.5 x 9.12x10^-31 x v^2
v^2 = 8.77x10^15
v = 93648278.15 m/s
C. From Q = CV
Q = charge
C = capacitance = 1 nF 1x10^-9 F
V = voltage = 25x10^3 V
Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C
Total number of electrons = Q/e
= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons
D. To push electron from cathode to anode, I'll have to do a work of about
4x10^-15 Joules