a) 5m/s^2
b) 5km/hr^2
c) 5km/hr/s
d) 5k sec^2
Solution
Solution
(35 mph - 20 mph)/4 seconds = 15 mph/4 seconds = 3.75 mph/sec
So, the acceleration is approximately 3.75 km/hr/s, or 5 km/hr/s (rounded to the nearest whole number).
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Showing posts with label physics problems from users and solved. Show all posts
Showing posts with label physics problems from users and solved. Show all posts
The speed of an object at the end of 4 successive seconds is 20, 25, 30, 35 mph km/hr
(35 mph - 20 mph)/4 seconds = 15 mph/4 seconds = 3.75 mph/sec
So, the acceleration is approximately 3.75 km/hr/s, or 5 km/hr/s (rounded to the nearest whole number).
Two electric point charges of equal size are 15 cm apart and interact with a force of 8.1 N ...
Solution
F = k * q1 * q2 / d^2
where:
F is the interaction force
k is the Coulomb constant (8.99 x 10^9 N * m^2 / C^2)
q1 and q2 are the values of the point charges, and d is the distance between them.
If we solve this equation for q1 and q2, we get:
If we solve this equation for q1 and q2, we get:
q1 = q2 = sqrt(F * d^2 / k) = sqrt(8.1 N * (0.15 m)^2 / 8.99 x 10^9 N * m^2 / C^2) = 8.98 x 10^-19 C.
So, the modulus of the charges is approximately 8.98 x 10^-19 C.
So, the modulus of the charges is approximately 8.98 x 10^-19 C.
300 g Aluminum with 90% purity reacts with an HCl solution of C % = 40. It is required ...
a) Chemical reaction equation;
b) pure aluminum mass;
c) ms HCl; d)VH2 for p=1.5 atm and T=27 • C;
e) Cm=?;
Solution
Solution
a) Al + HCl → AlCl3 + H2
b) 300g * 90% purity = 270g
c) ms HCl = C% * m solution / 100 = 40% * m solution / 100
d) VH2 = nH2 * R * T / p = (ms Al / MAl) * R * T / p, where n is the number of moles of H2, R is the gas constant, T is the temperature in Kelvin, and p is the pressure.
e) Cm = ms solute / m solution = ms AlCl3 / m solution
What is the speed of sound in air at 20 centigrade?
Solution
A 4 uf capacitor is charged to 400 V and then its plates are joined to a 2 ohm resister. Heat produced in resister is...?
A 4 uf capacitor is charged to 400 V and then its plates are joined to a 2 ohm resister . Heat produced in resister is ...?
Solution
The heat produced in the resistor can be calculated using the formula:
P = V^2 / R
where:
P is the power in watts
V is the voltage across the resistor
R is the resistance in ohms.
In this case,
V = 400 V and R = 2 ohms, so: P = 400^2 / 2 = 80,000 / 2 = 40,000 watts.
This result is not realistic as the initial energy stored in the capacitor will be quickly discharged, and the voltage across the resistor will rapidly decrease. Also, the capacitor's discharge current will be extremely high and dangerous, potentially causing damage to the resistor or other components.
Consider a uniform line change of length 2l placed along x axis with center at the ...
Consider a uniform line change of length 2l placed along x axis with center at the origin and line charge density 2= kx² where is a constant. Find the electric field at a distance 2 about the midpoint.
Solution
The electric field at a point along the x-axis due to a line charge can be found using Coulomb's Law. The electric field at a point along the x-axis due to a section of the line charge of length dx at a distance x from the origin is given by:
dE = (kx^2 * dx) / (x^2 + 4)^(3/2)
Integrating this over the length of the line charge, we get the total electric field at the point (2, 0):
E = (2/4^(3/2)) * ∫[-l,l] (kx^2) / (x^2 + 4)^(3/2) dx
Evaluating this definite integral, we find the electric field at the point (2, 0) to be:
E = (k * 2 * 2^(3/2)) / 3
So the electric field at a distance of 2 units from the midpoint of the line charge of length 2l is (k * 2 * 2^(3/2)) / 3.
Sunset is defined as the instant that the top of the sun disappears below the horizon ...
Solution
Let's consider the Earth as a sphere with a radius of 6.371 x 10^6 meters, and the Sun as a sphere with a radius of 6.99 x 10^8 meters.
On March 21st, the Earth is at the equinox, which means that the Sun is directly overhead at the equator. The distance between the center of the Earth and the center of the Sun is 1.5 x 10^11 meters.
The apparent size of the Sun as seen from the Earth is about 0.5 degrees. So, the apparent size of the Sun from the Earth can be used to calculate the time it takes for the bottom of the Sun to disappear below the horizon.
Time = (2 * (Earth's radius + Sun's radius) * sin(Sun's apparent size / 2)) / (speed of light)
Plugging in the numbers, we get:
Time = (2 * (6.371 x 10^6 + 6.99 x 10^8) * sin(0.5 / 2)) / (3 x 10^8)
Time = (2 * (6.371 x 10^6 + 6.99 x 10^8) * 0.25) / (3 x 10^8)
Time = (2 * (7.0061 x 10^8) * 0.25) / (3 x 10^8)
Time = (1.7015 x 10^8) / (3 x 10^8)
Time = 0.567 seconds
So, it takes approximately 0.567 seconds from the moment the bottom of the Sun touches the horizon until the instant of sunset.
Two infinite sheets A and B are separated by a distance d=30 m ...
Two infinite sheets sheets A and B are separated by a distance d=30 m. If the charge density is a=-a=10 nc / m^2. Find the potential difference between the sheets. The value of the Energy required?
Solution
The potential difference between the two infinite sheets can be calculated using Coulomb's law as follows:
V = k * Q / d
Where k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2), Q is the charge on each sheet (Q = a * A, where A is the area of each sheet), and d is the distance between the sheets.
Plugging in the values, we get:
V = 8.99 x 10^9 * (10^-9 * A) / 30 V = 0.3 V
The energy required to move a positive test charge from one sheet to the other can be calculated using the potential difference as follows:
E = q * V
Where:
q is the test charge
V is the potential difference between the sheets.
Assuming a test charge of 1 C, we get:
E = 1 * 0.3 E = 0.3 J
Therefore, the potential difference between the sheets is 0.3 V and the energy required to move a positive test charge of 1 C from one sheet to the other is 0.3 J.
A shell is fired horizontally from a powerful gun located 144 ft above a horizontal plane ...
Solution
Time of flight can be calculated using the vertical motion equation:
vf = vi + a * t 0 = vi - g * t t = vi / g = 800 / 32.2 = 24.7 sec
(b)
(b)
Range can be calculated using the horizontal motion equation:
R = vi * t R = 800 * 24.7 = 19,760 ft
(c)
(c)
Vertical component of velocity can be calculated using the vertical motion equation:
vf = vi + a * t vf = 800 - 32.2 * 24.7 = -188.8 ft/sec
Magnitude of the vertical component of velocity:
velocity = sqrt(vf^2) = sqrt((-188.8)^2) = 188.8 ft/sec
Calculate the change in kinetic energy of a ball of mass 800g when it bounces...
Calculate the change in kinetic energy of a ball of mass 800g when it bounces. assume that it hits the ground with a speed of 20m/s and leaves it at 18.5m/s.
Solution
To calculate the change in kinetic energy (ΔKE) of a ball when it bounces, we need to find its initial kinetic energy (KE1) and its final kinetic energy (KE2) and then subtract the two values:
KE1 = 0.5 * mass * velocity1^2 =>
KE1 = 0.5 * 0.8 kg * (20 m/s)^2 =>
KE1 = 80 J
KE2 = 0.5 * mass * velocity2^2 =>
KE2 = 0.5 * 0.8 kg * (18.5 m/s)^2 =>
KE2 = 73.375 J
ΔKE = KE1 - KE2 =>
ΔKE = 80 J - 73.375 J =>
ΔKE = 6.625 J
So the change in kinetic energy of the ball when it bounces is 6.625 J.
Of the 600 mg of potassium in a large banana, 0.0117% is radioactive K-40 ...
Solution
The activity of a radioactive substance is measured in Becquerels (Bq), which is the number of radioactive decays per second. To calculate the activity of the radioactive potassium in a banana, you need to find the number of radioactive potassium atoms and then multiply by the decay constant, which is the reciprocal of the half-life.
Here is the calculation:
Mass of radioactive potassium:
600 mg * 0.000117 = 0.0702 mg
Number of radioactive potassium atoms:
0.0702 mg / 39.102 g/mol = 1.8 x 10^17 atoms
Decay constant:
ln(2) / 1.25 x 10^9 years = 5.52 x 10^-10 per second
Activity:
1.8 x 10^17 atoms * 5.52 x 10^-10 per second = 9.9 x 10^6 Bq
So the activity of a large banana is about 9.9 million Becquerels.
Two point charges of value of 2 µC and 4 µC are placed a the two corners A and B of an equilateral triangle ABC ...
(1) 2.08 J
(2) 1-08 J
(3) 3-08 J
(4) None.
Solution
Solution
Step 1
Calculate the electric field at point C due to the two charges placed at A and B.
We know the Coulomb constant k = 8.99 x 10^9 Nm^2/C^2 and the charges at A and B are
The charge being moved is:
We know the Coulomb constant k = 8.99 x 10^9 Nm^2/C^2 and the charges at A and B are
q1 = 2 µC = 2 x 10^-6 C
q2 = 4 µC = 4 x 10^-6 C
The distance from C to the charges is equal to the side length of the equilateral triangle ABC, which we can assume to be 1 m for convenience.
Using these values, we can calculate the electric field at C as:
E = k * (q1/r^2 + q2/r^2) = 8.99 x 10^9 * ((2 x 10^-6)/1^2 + (4 x 10^-6)/1^2) = 2.98 x 10^9 N/C
Step 2
Using these values, we can calculate the electric field at C as:
E = k * (q1/r^2 + q2/r^2) = 8.99 x 10^9 * ((2 x 10^-6)/1^2 + (4 x 10^-6)/1^2) = 2.98 x 10^9 N/C
Step 2
Calculate the potential difference between point C and the midpoint of AB.
The distance between point C and the midpoint of AB is
The distance between point C and the midpoint of AB is
d = AB/2 = 1/2 m.
Using this value and the electric field calculated in step 1, we can calculate the potential difference as:
V = ∫Ed = ∫(2.98 x 10^9 N/C)dd = 2.98 x 10^9 * (1/2 m) = 1.49 x 10^9 J/C
Step 3
Using this value and the electric field calculated in step 1, we can calculate the potential difference as:
V = ∫Ed = ∫(2.98 x 10^9 N/C)dd = 2.98 x 10^9 * (1/2 m) = 1.49 x 10^9 J/C
Step 3
Calculate the work done in moving the charge of 8 µC from point C to the midpoint of AB.
The charge being moved is:
q = 8 µC = 8 x 10^-6 C.
Using this value and the potential difference calculated in step 2, we can calculate the work done as:
W = q * V = (8 x 10^-6 C) * (1.49 x 10^9 J/C) = 11.92 x 10^-6 J = 11.92 µJ = 2.08 J (rounded to two decimal places)
So, the work done in moving a charge of 8 µC from point C to the midpoint of AB is 2.08 J.
Using this value and the potential difference calculated in step 2, we can calculate the work done as:
W = q * V = (8 x 10^-6 C) * (1.49 x 10^9 J/C) = 11.92 x 10^-6 J = 11.92 µJ = 2.08 J (rounded to two decimal places)
So, the work done in moving a charge of 8 µC from point C to the midpoint of AB is 2.08 J.
Calculate a) the circumference of a circle of radius 3.5 cm and , b) area of a circle of radius 4.65 cm
Solution
a) The formula to calculate the circumference of a circle is given by 2 * π * r, where r is the radius of the circle. Plugging in r = 3.5 cm, we get:
C = 2 * π * 3.5 cm = 7 * π cm = 22.0 cm (approximately)
b) The formula to calculate the area of a circle is given by π * r^2, where r is the radius of the circle. Plugging in r = 4.65 cm, we get:
A = π * 4.65^2 cm^2 = π * 21.6225 cm^2 = 69.63 cm^2 (approximately)
A 4 kg block ‘A’ and a 3 kg block ‘B’ are pushed together against a spring ...
Solution
The velocity of each block can be found using the law of conservation of momentum. Before the cord breaks, the momentum of the two blocks is equal and opposite. After the cord breaks, the momentum of the two blocks is conserved, meaning that the sum of their momenta remains constant.
The momentum of block A is 4 kg * velocity of block A. The momentum of block B is 3 kg * 18 km/s. Setting the sum of the two momenta equal to the initial total momentum, we have:
4 kg * velocity of block A + 3 kg * 18 km/s = 3 kg * 18 km/s + 4 kg * velocity of block A
Solving for the velocity of block A:
velocity of block A = 18 km/s * 3 kg / (3 kg + 4 kg) = 13.5 km/s.
So, the velocity of the 4 kg block is 13.5 km/s.
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