Explanation:
To answer how far and how long the object travels, we need to know either the length of the table or the kinetic coefficient of friction between copper and glass. We also need to know the kinetic coefficient of friction between the copper object and the steel incline. This information wasn't provided, so for sake of illustration, I'll assume both coefficients are the same, and that μk = 0.35.
Let's divide the path into four sections to stay organized:
a) object moves up incline
b) object is in free fall
c) object slides across table
d) object falls onto books
a)
First, the object slides up the incline. There are four forces acting on the object. Weight pulling down, normal force perpendicular to the incline, tension up the incline, and friction down the incline.
Sum of the forces normal to the incline:
∑F = ma
N - W cos θ = 0
N = mg cos θ
Sum of the forces parallel to the incline:
∑F = ma
T - F = ma
T - Nμ = ma
T - mgμ cos θ = ma
Now sum of the forces in the y direction on the hanging mass:
∑F = ma
T - W = M(-a)
T = Mg - Ma
Substituting:
Mg - Ma - mgμ cos θ = ma
Mg - mgμ cos θ = (m + M) a
a = g (M - mμ cos θ) / (m + M)
Given m = 33.1 g, M = 50.0 g, θ = 30°, μ = 0.35, and g = 9.8 m/s²:
a = 4.71 m/s²
So the velocity it reaches at the top of the incline is:
v² = v₀² + 2a(x - x₀)
v² = 0² + 2(4.71)(1.5)
v = 3.76 m/s
The time to reach the top of the incline is:
x = x₀ + v₀ t + ½ at²
1.5 = 0 + (0) t + ½ (4.71) t²
t = 0.798 s
The horizontal distance traveled is:
x = 1.5 cos 30°
x = 1.299 m
And the vertical distance traveled is:
y = 1.5 sin 30°
y = 0.750 m
b)
In the second stage, the object is in free fall. v₀ = 3.76 m/s and θ = 30°. The object lands on the table at the point where its speed is a minimum. This is at the highest point of the trajectory, when the vertical velocity is 0.
v = at + v₀
0 = (-9.8) t + (3.76 sin 30°)
t = 0.192 s
The horizontal distance traveled is:
x = x₀ + v₀ t + ½ at²
x = 0 + (3.76 cos 30°) (0.192) + ½ (0) (0.192)²
x = 0.625 m
The height it reaches is:
v² = v₀² + 2a(y - y₀)
(0)² = (3.76 sin 30°)² + 2(-9.8)(y - 0)
y = 0.180 m
c)
The object is now on the glass table. As it slides, there are three forces acting on it. Normal force up, gravity down, and friction to the left.
In the y direction:
∑F = ma
N - W = 0
N = mg
In the x direction:
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
The time it takes to reach the end of the table is:
v = at + v₀
0 = (-9.8×0.35) t + (3.76 cos 30°)
t = 0.949 s
And the distance it travels is:
v² = v₀² + 2a(x - x₀)
(0)² = (3.76 cos 30°)² + 2(-9.8×0.35) (x - 0)
x = 1.546 m
d)
Finally, the object falls onto the books. There are 9 books, each 5 cm thick, so the height is 45 cm. As the book falls, there are two forces acting on it: drag up and gravity down.
∑F = ma
D - W = ma
0.05 - (0.0331)(9.8) = 0.0331 a
a = -8.29 m/s²
Adding the vertical distances from parts a and b, we know the height of the table is 0.930 m. So the time it takes the object to land is:
y = y₀ + v₀ t + ½ at²
0.45 = 0.93 + (0) t + ½ (-9.8) t²
t = 0.313 s
Adding up the times from all four parts, the total time is:
t = 0.798 + 0.192 + 0.949 + 0.313
t = 2.25 s
Adding up the horizontal distances, the total distance traveled is:
x = 1.299 + 0.625 + 1.546
x = 3.47 m
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. A 21.0-V potential difference is applied to these plates. (a) Calculate the electric field between the plates.
Answer:
12353 V m⁻¹ = 12.4 kV m⁻¹
Explanation:
Electric field between the plates of the parallel plate capacitor depends on the potential difference across the plates and their distance of separation.Potential difference across the plates V over the distance between the plates gives the electric field between the plates. Potential difference is the amount of work done per unit charge and is given here as 21 V. Electric field is the voltage over distance.
E = V ÷ d = 21 ÷ 0.0017 = 12353 V m⁻¹
The bending of light as it moves from one medium to another with differing indices of refraction is due to a change in what property of the light? A) amplitude B) period C) frequency D speed E) Color
Answer:
D] speed
Explanation:
A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and its initial displacement is s(0) = 9 cm. Find its position function, s(t).
Answer:
The position function is [tex]s_{t}=2t^3+5t^2-5t+9[/tex].
Explanation:
Given that,
Acceleration [tex]a =12t+10[/tex]
Initial velocity [tex]v_{0} = -5\ cm/s[/tex]
Initial displacement [tex]s_{0}=9\ cm[/tex]
We know that,
The acceleration is the rate of change of velocity of the particle.
[tex]a = \dfrac{dv}{dt}[/tex]
The velocity is the rate of change of position of the particle
[tex]v=\dfrac{dx}{dt}[/tex]
We need to calculate the the position
The acceleration is
[tex]a_{t} = 12t+10[/tex]
[tex]\dfrac{dv}{dt} = 12t+10[/tex]
[tex]a_{t}=dv=(12t+10)dt[/tex]
On integration both side
[tex]\int{dv}=\int{(12t+10)}dt[/tex]
[tex]v_{t}=6t^2+10t+C[/tex]
At t = 0
[tex]v_{0}=0+0+C[/tex]
[tex]C=-5[/tex]
Now, On integration again both side
[tex]v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt[/tex]
[tex]s_{t}=2t^{3}+5t^2-5t+C[/tex]
At t = 0
[tex]s_{0}=0+0+0+C[/tex]
[tex]C=9[/tex]
[tex]s_{t}=2t^3+5t^2-5t+9[/tex]
Hence, The position function is [tex]s_{t}=2t^3+5t^2-5t+9[/tex].
To find the position function of a particle given acceleration a(t) = 12t + 10, one integrates twice. The first integral gives the velocity function, the second gives the position or displacement function. These are determined to be v(t) = 6t^2 + 10t - 5 and s(t) = 2t^3 +5t^2 - 5t + 9, respectively.
Explanation:The subject of this problem involves calculating the position function, or displacement, of a particle given an acceleration function, initial velocity, and initial position. In this case, acceleration is given by a(t) = 12t + 10.
To find the velocity function, v(t), you integrate the acceleration function: ∫a(t) dt = ∫(12t + 10) dt = 6t^2 + 10t + C1, where C1 is the constant of integration. We know that v(0) = −5 cm/s, so C1 is -5: the full velocity function is v(t) = 6t^2 + 10t - 5.
We then integrate the velocity function to find the position function, s(t): ∫v(t) dt = ∫(6t^2 + 10t - 5) dt = 2t^3 +5t^2 - 5t + C2. Given s(0) = 9 cm, we find C2 = 9. The position function s(t) is therefore s(t) = 2t^3 +5t^2 - 5t + 9.
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Hydro-Quebec transmits power from hydroelectric dams in the far north of Quebec to the city of Montreal at 735kV. The lines are 935 km long and are 3.50 cm in diameter. Given the resistivity of copper is 1.68 x 10^-8 Ω.m. a) find the resistance of one of the lines, and b) the current carried by the wire.
Answer:
a)
16.33 Ω
b)
45009.18 A
Explanation:
a)
L = length of the line = 935 km = 935000 m
d = diameter of the line = 3.50 cm = 0.035 m
ρ = resistivity of the line = 1.68 x 10⁻⁸ Ω.m
Area of cross-section of the line is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.035)²
A = 0.000961625 m²
Resistance of the line is given as
[tex]R=\frac{\rho L}{A}[/tex]
inserting the values
R = (1.68 x 10⁻⁸) (935000)/(0.000961625)
R = 16.33 Ω
b)
V = potential difference across the line = 735 kv = 735000 Volts
i = current carried by the wire
Using ohm's law, current carried by the wire is given as
[tex]i=\frac{V}{R}[/tex]
i = 735000/16.33
i = 45009.18 A
A meteorite has a speed of 90.0 m/s when 850 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.25 m. (a) What is its speed just before striking the sand?
Final answer:
The speed of the meteorite just before striking the sand is 0 m/s.
Explanation:
To find the speed of the meteorite just before striking the sand, we can use the principle of conservation of energy. Let's consider the initial energy when the meteorite is 850 km above the Earth's surface and the final energy when it comes to rest in the sand. Initially, the meteorite has gravitational potential energy and no kinetic energy. Finally, when it comes to rest in the sand, it has no potential energy and only kinetic energy.
Using the equation for gravitational potential energy, PE = mgh, where m is the mass of the meteorite, g is the acceleration due to gravity (9.8 m/s²), and h is the height above the Earth's surface, we can calculate the initial potential energy. PE = (2.0 × 10¹³ kg)(9.8 m/s²)(850,000 m) = 1.61 × 10¹⁹ J.
The final kinetic energy of the meteorite just before striking the sand can be calculated using the equation KE = (1/2)mv², where m is the mass of the meteorite and v is its velocity. We know that the meteorite comes to rest in a distance of 3.25 m, so its final velocity is 0. Using this information, we can solve for the final kinetic energy. 0 = (1/2)(2.0 × 10¹³ kg)v². By rearranging the equation, we find v = 0 m/s.
Therefore, the speed of the meteorite just before striking the sand is 0 m/s.
A 140 g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.537 m. The ball makes 2.27 revolutions in a second. What is its centripetal acceleration?
Answer:
109.13 rad/s^2
Explanation:
m = 140 g = 1.4 kg, r = 0.537 m, f = 2.27 rps
The centripetal acceleration is given by
a = r ω^2
a = r x (2 π f)^2
a = 0.537 x ( 2 x 3.14 x 2.27)^2
a = 109.13 rad/s^2
A 7.00 g bullet moving horizontally at 200 m/s strikes and passes through a 150 g tin can sitting on a post. Just after the impact, the can has a horizontal speed of 180 cm/s. What was the bullet’s speed after leaving the can?
Answer:
160 m/s
Explanation:
Momentum is conserved:
mu = mv + MV
(7.00) (200) = (7.00)v + (150) (1.8)
v = 160 m/s
161.43m/s
Explanation:
Using the principle of conservation of linear momentum i.e
Total momentum before impact is equal to total momentum after impact.
=> Momentum of bullet before impact + Momentum of tin before impact
=
Momentum of bullet after impact + Momentum of tin after impact
i.e
[tex]m_{B}[/tex] [tex]u_{B}[/tex] + [tex]m_{T}[/tex] [tex]u_{T}[/tex] = [tex]m_{B}[/tex] [tex]v_{B}[/tex] + [tex]m_{T}[/tex] [tex]v_{T}[/tex]
Where;
[tex]m_{B}[/tex] = mass of bullet = 7.00g = 0.007kg
[tex]m_{T}[/tex] = mass of tin can = 150g = 0.15kg
[tex]u_{B}[/tex] = initial velocity of bullet before impact = 200m/s
[tex]u_{T}[/tex] = initial velocity of tin can before impact = 0m/s (since the can is stationary)
[tex]v_{B}[/tex] = final velocity of the bullet after impact
[tex]v_{T}[/tex] = final velocity of the tin can after impact = 180cm/s = 1.8m/s
Substitute these values into the equation above;
=> [tex]m_{B}[/tex] [tex]u_{B}[/tex] + [tex]m_{T}[/tex] [tex]u_{T}[/tex] = [tex]m_{B}[/tex] [tex]v_{B}[/tex] + [tex]m_{T}[/tex] [tex]v_{T}[/tex]
=> (0.007 x 200) + (0.15 x 0) = (0.007 x [tex]v_{B}[/tex]) + (0.15 x 1.8)
=> 1.4 + 0 = 0.007[tex]v_{B}[/tex] + 0.27
=> 1.4 = 0.007[tex]v_{B}[/tex] + 0.27
=> 0.007[tex]v_{B}[/tex] = 1.4 - 0.27
=> 0.007[tex]v_{B}[/tex] = 1.13
Solve for [tex]v_{B}[/tex]
=> [tex]v_{B}[/tex] = 1.13 / 0.007
=> [tex]v_{B}[/tex] = 161.43m/s
Therefore, the speed of the bullet after impact (leaving the can) is 161.43m/s
A car (mass = 1200 kg) is traveling at 31.1 m/s when it collides head-on with a sport utility vehicle (mass = 2830 kg) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?
Answer:
13.18 m/s
Explanation:
Let the velocity of sports utility car is
-u as it is moving in opposite direction.
mc = 1200 kg, uc = 31.1 m/s
ms = 2830 kg, us = - u = ?
Using conservation of momentum
mc × uc + ms × us = 0
1200 × 31.1 - 2830 × u = 0
u = 13.18 m/s
A 2 kg ball travellng to the right with a speed of 4 m/s collidees with a 5 kg ball traveling to the left with a speed of 3 m/s. Take right to be the positive direction. What is the total momentum of the two balls before they collide? What is the total momentum of the two balls after they collide?
Explanation:
Momentum before collision:
(2 kg) (4 m/s) + (5 kg) (-3 m/s) = -3 kg m/s
No external forces act on the balls, so momentum is conserved. Therefore, momentum after collision is also -3 kg m/s.
The total momentum of the two balls before the collision is -7 kg*m/s. The total momentum of the two balls after the collision is also -7 kg*m/s.
Explanation:To find the total momentum before the collision, we need to calculate the individual momentum of each ball and then add them together. To compute an object's momentum, multiply its mass by its velocity.
For the first ball, the momentum is 2 kg * 4 m/s (since it is travelling to the right) = 8 kg*m/s. For the second ball, the momentum is 5 kg * (-3 m/s) (since it is travelling to the left) = -15 kg*m/s. Adding these two momenta together, we get a total momentum of 8 kg*m/s + (-15 kg*m/s) = -7 kg*m/s.
After the collision, if the balls stick together and move as one, the speed of the combined mass can be calculated using the principle of conservation of momentum. Since momentum is conserved in a collision, the total momentum after the collision will be the same as before. In this case, since the total momentum before the collision was -7 kg*m/s, the total momentum after the collision will also be -7 kg*m/s.
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A bullet is shot at an angle of 32° above the horizontal on a level surface. It travels in the air for 6.4 seconds before it strikes the ground 92m from the shooter. What was the maximum height reached by the bullet?
Answer:
H = 4.12 m
Explanation:
As we know that horizontal range is the distance moved in horizontal direction
Since horizontal direction has no acceleration
so here we have
[tex]Range = v_x T[/tex]
here we know that
[tex]v_x = vcos32[/tex]
so from above formula
[tex]92 = (vcos32)(6.4)[/tex]
[tex]v = 16.95 m/s[/tex]
now we have maximum height is given as
[tex]H = \frac{(vsin32)^2}{2g}[/tex]
[tex]H = \frac{(16.95 sin32)^2}{2(9.8)}[/tex]
[tex]H = 4.12 m[/tex]
A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction force will be required?
Answer:
22000 N
Explanation:
Convert velocity to SI units:
98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s
Draw a free body diagram. There are three forces acting on the car. Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.
I'm going to assume the friction force is pointed down the bank. If I get a negative answer, that'll just mean it's actually pointed up the bank.
Sum of the forces in the radial direction (+x):
∑F = ma
N sin θ + F cos θ = m v² / r
Sum of the forces in the y direction:
∑F = ma
N cos θ - F sin θ - W = 0
To solve the system of equations for F, first solve for N and substitute.
N = (W + F sin θ) / cos θ
Substituting:
((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r
(W + F sin θ) tan θ + F cos θ = m v² / r
W tan θ + F sin θ tan θ + F cos θ = m v² / r
W tan θ + F (sin θ tan θ + cos θ) = m v² / r
W tan θ + F sec θ = m v² / r
F sec θ = m v² / r - W tan θ
F = m v² cos θ / r - W sin θ
F = m (v² cos θ / r - g sin θ)
Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:
F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)
F = 21577 N
Rounding to two sig-figs, you need at least 22000 N of friction force.
Global warming will produce rising sea levels partly due to melting ice caps but also due to the expansion of water as average ocean temperatures rise. To get some idea of the size of this effect, calculate the change in length of a column of water 1.00 km high for a temperature increase of 1.00ºC. Note that this calculation is only approximate because ocean warming is not uniform with depth. (answer in ×10^{-3} −3 m)
The change in length of a column of water 1.00 km high for a temperature increase of 1.00ºC is approximately 0.21 meters.
Explanation:To calculate the change in length of a column of water for a temperature increase of 1.00ºC, we need to use the coefficient of thermal expansion for water. The coefficient of thermal expansion for water is approximately 0.00021 per degree Celsius.
To find the change in length, we multiply the height of the column (1.00 km) by the coefficient of thermal expansion (0.00021) and the temperature increase (1.00ºC).
Change in length = 1.00 km × 0.00021 × 1.00ºC = 0.00021 km = 0.21 m
Therefore, the change in length of a column of water 1.00 km high for a temperature increase of 1.00ºC is approximately 0.21 meters.
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If you weigh 665 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 18.0 km ? Take the mass of the sun to be ms = 1.99×10^30 kg , the gravitational constant to be G = 6.67×10^−11 N⋅m2/kg^2 , and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2 .
Your weight on a neutron star can be calculated based on your given weight on Earth using the Universal Law of Gravitation equation. This allows us to determine the gravitational force or weight for any celestial body given we know its mass and radius.
Explanation:The topic at hand is fundamentally related to Physics since it pertains to the gravitational effects of a massive body. From the question, we are given your weight on Earth which is 665 N. The weight on Earth can be related to the weight of the neutron star by the equation of universal gravitation:
W = mg = GMm/r²
Where:
M = mass of the star or planet or rather the celestial bodym = mass of the object or the person (in this case)G = universal gravitational constantr = distance from the center of the celestial body (radius)Given that the diameter of the neutron star is 18km, its radius becomes 9km = 9000m. Substituting the given values into the formula yields a theoretical weight on a neutron star.
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We find this acceleration to be approximately 4.09×[tex]10^{12} m/s^2[/tex]. Thus, your weight would be around 2.77×[tex]10^{14}[/tex] N on the neutron star.
To calculate your weight on the surface of a neutron star, you need to find the acceleration due to gravity on that neutron star using Newton's Universal Law of Gravitation.
The formula for gravitational force is:
F = G x (m1 x m2) /[tex]R^2[/tex]
Where G is the gravitational constant, m1 and m2 are the masses, and R is the radius of the object.
Given:
Mass of the neutron star, M = 1.99×[tex]10^{30}[/tex] kg
Radius, R = 18.0 km = 18,000 meters
Gravitational constant, G = 6.67×[tex]10^{-11} N⋅m^2/kg^2[/tex]The acceleration due to gravity on the neutron star, g_s, can be calculated as:
g_s = G x (M / [tex]R^2[/tex])
Substituting in the given values:
g_s = (6.67×[tex]10^{-11}[/tex] N⋅[tex]m^2/kg^2[/tex] x 1.99×[tex]10^{30}[/tex] kg) / [tex](18,000 m)^2[/tex]
g_s ≈ 4.09×[tex]10^{12}[/tex] m/[tex]s^2[/tex]
Your weight on Earth is given as 665 N. Since weight (W) is mass (m) times the acceleration due to gravity (g):
m = W / g = 665 N / 9.810 m/[tex]s^2[/tex] ≈ 67.8 kg
So, your weight on the neutron star would be:
Weight on neutron star = mass x g_s = 67.8 kg x 4.09×[tex]10^{12} m/s^2[/tex] ≈ 2.77×[tex]10^{14}[/tex] N
The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If 1 gram of this isotope is present initially, how long will it take for 80% of the lead to decay? (Round your answer to two decimal places.)
Final answer:
The question inquires about the time needed for 80% of Pb-209 to decay, knowing its half-life is 3.3 hours. By applying the exponential decay formula, we calculate that approximately 7.39 hours are required for 80% of Pb-209 to decay.
Explanation:
The question involves the concept of radioactive decay and specifically asks how long it will take for 80% of Pb-209 to decay, given that its half-life is 3.3 hours. To find the time required for 80% of the lead to decay, we use the half-life formula and the property that radioactive decay is an exponential process. Since 80% decay means 20% remains, we set up the equation based on the exponential decay formula: N = N0(1/2)(t/T), where N is the remaining amount of substance, N0 is the initial amount, t is the time elapsed, and T is the half-life of the substance.
Substituting the given values and solving for t, we find:
N0 = 1 gram (100% initially)
N = 0.2 grams (20% remains)
T = 3.3 hours
Thus, the equation becomes 0.2 = 1(1/2)(t/3.3). Solving for t gives us the time required for 80% decay.
After calculations, the result is that it takes approximately 7.39 hours for 80% of the Pb-209 to decay. This showcases the practical application of exponential decay and half-life in determining the amount of a radioactive substance that remains after a given period.
A block of mass 0.240 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.096 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)
Answer:
10.19 m
Explanation:
Energy is conserved, so elastic energy stored in spring = gravitational energy of block.
1/2 kx² = mgh
h = kx² / (2mg)
h = (5200 N/m) (0.096 m)² / (2 × 0.240 kg × 9.8 m/s²)
h = 10.19 m
A tiny object carrying a charge of +35 μC and a second tiny charged object are initially very far apart. If it takes 32 J of work to bring them to a final configuration in which the +35 μC object i is at x = 1.00 mm, y = 1.00 mm, and the other charged object is at x = 1.00 mm, y = 3.00 mm (Cartesian coordinate system), find the magnitude of the charge on the second object. (k = 1/4πε 0 = 8.99 × 109 N · m2/C2)
The magnitude of the charge on the second object is 0.025 μC and its sign is negative because it is required work to bring the two charges together, suggesting these are opposite charges and repel each other.
Explanation:The problem can be solved using the formula for the work done on a charge moving in an electric field, which is determined by the formula W = k * q1 * q2 / r, where k is the Coulomb's constant (8.99 × 10⁹ N · m²/C²), q1 and q2 are the charges, and r is the distance between them. From the problem, we know W = 32J, q1 = +35 μC, and r = 2.00 mm (the difference in the y-coordinates). Solving for q2 gives q2 = W * r / (k * q1) = 32J * 2.00 x 10⁻³m / (8.99 × 10⁹ N · m²/C² * 35 x 10⁻⁶C) = approximately -0.025 μC. Therefore, the magnitude of the charge on the second object is 0.025 μC, and its sign is negative because it takes work to bring the two charges together.
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A 1200 kg car traveling north at 10 m/s is rear-ended by a 2000 kg truck traveling at 30 m/s. What is the total momentum before and after the collision?
Answer:
The total momentum before and after collision is 72000 kg-m/s.
Explanation:
Given that,
Mass of car = 1200 kg
Velocity of car = 10 m/s
Mass of truck = 2000 kg
Velocity of truck = 30 m/s
Using conservation of momentum
The total momentum before the collision is equal to the total momentum after collision.
[tex]m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})V[/tex]
Where, [tex]m_{1}[/tex]=mass of car
[tex]v_{1}[/tex] =velocity of car
[tex]m_{1}[/tex]=mass of truck
[tex]v_{1}[/tex] =velocity of truck
Put the value into the formula
[tex]1200\times10+2000\times30=(1200+2000)V[/tex]
[tex]V=\dfrac{1200\times10+2000\times30}{(1200+2000)}[/tex]
[tex]V = 22.5\ m/s[/tex]
Now, The total momentum before collision is
[tex]P=m_{1}v_{1}+m_{2}v_{2}[/tex]
[tex]P=1200\times10+2000\times30[/tex]
[tex]P=72000\ kg-m/s[/tex]
The total momentum after collision is
[tex]P=(m_{1}+m_{2})v_{2}[/tex]
[tex]P=(1200+2000)\times22.5[/tex]
[tex]P= 72000 kg-m/s[/tex]
Hence, The total momentum before and after collision is 72000 kg-m/s.
How does the magnitude of the electrical force between a pair of charged particles change when they are brought to half their original distance of separation? 1. Reduces to one quarter of original value 2. Doubles 3. Doesn’t change 4. Reduces to one half of original value 5. Quadruple
According to Coulomb's Law, the magnitude of the electric force between two charged particles is inversely proportional to the square of the distance between them. Therefore, if the distance between the two particles is halved, the electric force becomes four times as strong, i.e., it quadruples.
Explanation:The phenomenon you're asking about is governed by Coulomb's Law. This law states that the magnitude of the electric (or Coulomb) force between two charged particles is directly proportional to the product of the magnitudes of their charges, and inversely proportional to the square of the distance between them.
So, if the distance between these two charges is halved, the electric force between them would be affected by the square of that change in distance. Therefore, instead of being half as strong, the electric force becomes four times as strong, i.e., it quadruples. Hence, the correct answer to your question is: it quadruples (option 5). The inverse square law, which this is an example of, is a fundamental principle in physics that applies to other forces too, like gravity.
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A 81 kg man lying on a surface of negligible friction shoves a 79 g stone away from himself, giving it a speed of 4.5 m/s. What speed does the man acquire as a result?
Answer:
The speed of the man is 4.54 m/s.
Explanation:
Given that,
Mass of man=8100 g
Mass of stone = 79 g
Speed = 4.5 m/s
We need to calculate the speed of the man
Using momentum of conservation
[tex](m_{1}+m_{2})V=m_{1}v_{1}+m_{2}v_{2}[/tex]
Where,
[tex]m_{1}[/tex]=mass of man
[tex]m_{2}[/tex]=mass of stone
[tex]v_{1}[/tex]=velocity of man
[tex]v_{2}[/tex]=velocity of stone
Put the value in the equation
As the stone is away from the man
So, the speed of stone is zero
[tex]8179\times4.5=8100\times v+0[/tex]
[tex]v=\dfrac{8179\times4.5}{8100}[/tex]
[tex]v = 4.54\ m/s[/tex]
Hence, The speed of the man is 4.54 m/s.
A 6.0-μF air-filled capacitor is connected across a 100-V voltage source. After the source fully charges the capacitor, the capacitor is immersed in transformer oil (of dielectric constant 4.5). How much ADDITIONAL charge flows from the voltage source, which remained connected during the process?
Answer:
[tex]2.1\cdot 10^{-3} C[/tex]
Explanation:
The initial charge stored on the capacitor is given by
[tex]Q_0 =C_0 V[/tex]
where
[tex]C_0 = 6.0 \mu F = 6.0 \cdot 10^{-6}F[/tex] is the initial capacitance
V = 100 V is the potential difference across the capacitor
Solving the equation,
[tex]Q_0 = (6.0 \cdot 10^{-6}F)(100 V)=6.0 \cdot 10^{-4}C[/tex]
The charge stored in the capacitor when inserting the dielectric is
[tex]Q = k Q_0[/tex]
where
k = 4.5 is the dielectric constant
Substituting,
[tex]Q=(4.5)(6.0 \cdot 10^{-4}C)=2.7\cdot 10^{-3}C[/tex]
So the additional charge is
[tex]\Delta Q=Q-Q_0 = 2.7 \cdot 10^{-3}C - 6.0 \cdot 10^{-4}C=2.1\cdot 10^{-3} C[/tex]
The additional charge that flows into the capacitor once it is immersed in transformer oil (of dielectric constant 4.5) and kept connected to the source is 2100 μC.
Explanation:The question involves a capacitor that is connected across a 100-V voltage source and then submerged in transformer oil. The capacitor initially charges to capacity while in the air. When a capacitor is then immersed in a material with a dielectric constant, its ability to store charge improves. Here, the dielectric constant of the transformer oil is given as 4.5, suggesting that the capacitor's capacity to store charge will increase 4.5 times as compared to when it was in air.
The additional charge that flows into the capacitor can be calculated using the formula for the charge in a capacitor, Q = CV. The initial charge (Q1) on the capacitor when it was in air would be Q1 = CV1 = 6.0 μF * 100 V = 600 μC. After immersing in transformer oil, the capacitance would increase by a factor of 4.5, giving a new capacitance C2 = 4.5 * 6.0 μF = 27.0 μF. The new charge (Q2) would be Q2 = CV2 = 27.0 μF * 100 V = 2700 μC. Hence the additional charge that's flown from the source would be Q2 - Q1 = 2700 μC - 600 μC = 2100 μC.
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An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Express your answer to two significant figures and include the appropriate units. What average power is required to stop him? Express your answer to two significant figures and include the appropriate units.
Explanation:
It is given that,
Mass of the football player, m = 92 kg
Velocity of player, v = 5 m/s
Time taken, t = 10 s
(1) We need to find the original kinetic energy of the player. It is given by :
[tex]k=\dfrac{1}{2}mv^2[/tex]
[tex]k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2[/tex]
k = 1150 J
In two significant figure, [tex]k=1.2\times 10^3\ J[/tex]
(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0
i.e. [tex]P=\dfrac{W}{t}=\dfrac{\Delta K}{t}[/tex]
[tex]P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}[/tex]
P = 115 watts
In two significant figures, [tex]P=1.2\times 10^2\ Watts[/tex]
Hence, this is the required solution.
What current is produced if 1473 sodium ions flow across a cell membrane every 3.4ju8? 0 5.81 pA O 694 pA O 7.76 pA 5.99 pA
Answer:
69.4 pA
Explanation:
n = number of sodium ions = 1473
e = magnitude of charge on each sodium ion = 1.6 x 10⁻¹⁹ C
t = time taken to flow across the membrane = 3.4 x 10⁻⁶ sec
Total Charge on sodium ions is given as
q = n e eq-1
Current produced is given as
[tex]i = \frac{q}{t}[/tex]
Using eq-1
[tex]i = \frac{ne}{t}[/tex]
Inserting the values
[tex]i = \frac{(1473)(1.6\times 10^{-19})}{3.4\times 10^{-6}}[/tex]
i = 69.4 x 10⁻¹² A
i = 69.4 pA
Ramon and Sally are observing a toy car speed up as it goes around a circular track. Ramon says, “The car’s speeding up, so there must be a net force parallel to the track.” “I don’t think so,” replies Sally. “It’s moving in a circle, and that requires centripetal acceleration. The net force has to point to the center of the circle.” Do you agree with Ramon, Sally, or neither
Answer:
Neither
Explanation:
In this situation, the net force acting on the toy car moving in the circle has two components:
- There is a component which is tangential (parallel) to the circle - we can understand this by the fact that the car is speeding up: this means that its tangential speed is changing, so it has a tangential acceleration, therefore there must be a component of the force tangential to the circle (parallel to the circle)
- There is a component which is radial to the circle, pointing towards the centre - this is called centripetal force. This is due to the fact that the car is constantly changing direction of motion: so, there must be a force that causes this change in direction of the car, and this force points towards the centre of the circle, and it is called centripetal force.
A marble is dropped from a toy drone that is 25 m above the ground, and slowly rising with a vertical velocity of 0.8 m/s. How long does it take the marble to reach the ground?
Answer:
2.18 s
Explanation:
H = - 25 m ( downwards)
U = - 0.8 m/s
g = - 9.8 m/s^2
Let time taken is t.
Use second equation of motion
H = u t + 1/2 g t^2
- 25 = - 0.8 t - 1/2 × 9.8 × t^2
4.9 t^2 + 0.8 t - 25 = 0
By solving we get
t = 2.18 s
An electron travels undeflected in a path that is perpendicular to an electric feld of 8.3 x 10 v/m. It is also moving perpendicular to a magnetic field with a magnitude of 7.3 x 103 T. If the electric field is turned off, at what radius would the electron orbit? O 124 x 10*m 889 x 104 m O 9.85 x 104m O 1.06 x 10o m
Answer:
[tex]8.6\cdot 10^{-18} m[/tex]
Explanation:
Initially, the electron is travelling undeflected at constant speed- this means that the electric force and the magnetic force acting on the electron are balanced. So we can write
q E = q v B
where
q is the electron's charge
[tex]E=8.3\cdot 10 V/m[/tex] is the electric field magnitude
v is the electron's speed
[tex]B=7.3\cdot 10^3 T[/tex] is the magnitude of the magnetic field
Solving for v,
[tex]v=\frac{E}{B}=\frac{8.3 \cdot 10 V/m}{7.3\cdot 10^3 T}=0.011 m/s[/tex]
Then the electric field is turned off, so the electron (under the influence of the magnetic field only) will start moving in a circle of radius r. Therefore, the magnetic force will be equal to the centripetal force:
[tex]qvB= m \frac{v^2}{r}[/tex]
where
[tex]q=1.6\cdot 10^{-19} C[/tex] is the electron's charge
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the electron's mass
Solving for r, we find the radius of the electron's orbit:
[tex]r=\frac{mv}{qB}=\frac{(9.11\cdot 10^{-31} kg)(0.011 m/s)}{(1.6\cdot 10^{-19} C)(7.3\cdot 10^3 T)}=8.6\cdot 10^{-18} m[/tex]
An us bomber is flying horizontally at 300 mph at an altitude of 610 m. its target is an iraqi oil tanker crusing 25kph in the same direction and same vertical plane. what horizontal distance behind the tanker must the pilot observe before he releases the shell to score a direct hit
Answer:
=1419.19 meters.
Explanation:
The time it takes for the shell to drop to the tanker from the height, H =1/2gt²
610m=1/2×9.8×t²
t²=(610m×2)/9.8m/s²
t²=124.49s²
t=11.16 s
Therefore, it takes 11.16 seconds for a free fall from a height of 610m
Range= Initial velocity×time taken to hit the tanker.
R=v₁t
Lets change 300 mph to kph.
=300×1.60934 =482.802 kph
Relative velocity=482.802 kph-25 kph
=457.802 kph
Lets change 11.16 seconds to hours.
=11.16/(3600)
=0.0031 hours.
R=v₁t
=457.802 kph × 0.0031 hours.
=1.41918 km
=1.41919 km × 1000m/km
=1419.19 meters.
Distance is a numerical representation of the distance between two objects or locations. The horizontal distance behind the tanker will be 1419.19 m.
What is the distance?Distance is a numerical representation of the distance between two objects or locations. The distance can refer to a physical length or an estimate based on other factors in physics or common use.
The given data in the problem is ;
v₁ is the horizontal flying velocity = 300 m/h
H is the altitude = 610 m
v₂ is he cruise velocity= 25km/h
If t is the time taken the shell drops to the tanker from the height h is found by the formula;
[tex]\rm H =\frac{1}{2} gt^2 \\\\ \rm t=\sqrt{\frac{H}{2g} } \\\\ \rm t=\sqrt{2gh} \\\\ \rm t=\sqrt{2\times 610 \times 9.81} \\\\ \rm t=11.16 \sec[/tex]
The velocity of bomber obtained after unit conversion;
[tex]V_{12}=300\times 1.60 = 482.802[/tex]
Relative velocity is defined as the velocity of an object with respect to the other object.
Relative velocity=482.802 kph-25 kph=457.802 kph
In one hour there are 3600 seconds then the conversion is found by;
On changing 11.16 seconds to hours we found;
[tex](\frac{11.16}{3600} )=0.0031 \ hours.[/tex]
The range is the horizontal distance which is given by ;
Range= Initial velocity×time taken to hit the tanker.
[tex]\rm R= v \times t \\\\ \rm R= 457.8\times 0.0031\\\\ \rm R= 1.41918\ Km \\\\ \rm R=1419.19 m[/tex]
Hence the horizontal distance behind the tanker will be 1419.19 m.
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A 350-g air track cart is traveling at 1.25 m/s and a 280-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the center of mass of the two carts?
Answer:
The speed of the center of mass of the two carts is 0.103 m/s
Explanation:
It is given that,
Mass of the air track cart, m₁ = 350 g = 0.35 kg
Velocity of air track cart, v₁ = 1.25 m/s
Mass of cart, m₂ = 280 g = 0.28 kg
Velocity of cart, v₂ = -1.33 m/s (it is travelling in opposite direction)
We need to find the speed of the center of mass of the two carts. It is given by the following relation as :
[tex]v_{cm}=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]
[tex]v_{cm}=\dfrac{0.35\ kg\times 1.25\ m/s+0.28\ kg\times (-1.33\ m/s)}{0.35\ kg+0.28\ kg}[/tex]
[tex]v_{cm}=0.103\ m/s[/tex]
Hence, this is the required solution.
How far from a converging lens with a focal length of 16 cm should an object be placed to produce a real image which is the same size as the object? Express your answer using two significant figures.
Answer:
32 cm
Explanation:
f = focal length of the converging lens = 16 cm
Since the lens produce the image with same size as object, magnification is given as
m = magnification = - 1
p = distance of the object from the lens
q = distance of the image from the lens
magnification is given as
m = - q/p
- 1 = - q/p
q = p eq-1
Using the lens equation, we get
1/p + 1/q = 1/f
using eq-1
1/p + 1/p = 1/16
p = 32 cm
To create a real image with the same size as the object using a converging lens, the object should be placed at a distance equal to twice the focal length of the lens.
Given the focal length (f) of the lens is 16 cm, the object should be placed at a distance of 2 * 16 cm = 32 cm from the lens.
1. The lens formula is given by:
1/f = 1/v - 1/u
Where:
- f is the focal length of the lens.
- v is the image distance.
- u is the object distance.
2. In this case, we want a real image of the same size as the object, which means that the magnification (M) is equal to 1. Magnification is given by:
M = -v/u
Since M = 1, we have v = -u.
3. For a real image formed by a converging lens, v is negative. So, we have v = -u = -2f.
4. Substituting the focal length (f = 16 cm) into the equation, we find that u = -2 * 16 cm = -32 cm. The negative sign indicates that the object is placed on the same side as the object. Therefore, the object should be placed 32 cm from the lens to produce the desired image.
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A certain particle at rest has a lifetime of 2.52 mu s. what must be the speed of the particle for its lifetime observed to be 8.32 mu s? 0.985c 0.953c 0.913c 0.302c None of the above
Answer:
0.953c
Explanation:
T₀ = lifetime of the particle at rest = 2.52 μs
T = Observed lifetime of the particle while in motion = 8.32 μs
v = speed of particle
c = speed of light
Using the formula for time dilation
[tex]T = \frac{T_{o}}{\sqrt{1-(\frac{v}{c})^{2}}}[/tex]
inserting the values
[tex]8.32 = \frac{2.52}{\sqrt{1-(\frac{v}{c})^{2}}}[/tex]
v = 0.953c
Final answer:
To calculate the speed at which a particle's lifetime appears dilated to 8.32 μs, one uses the time dilation formula from special relativity. The calculated speed is approximately 0.995c, which does not match the provided choices. A careful review of the calculation and rounding effects might lead to one of the options being the nearest correct answer.
Explanation:
To find the required speed for the particle's lifetime to be observed as 8.32 μs, we need to apply the concept of time dilation from Einstein's theory of special relativity. Time dilation is given by the equation τ' = τ / √(1 - v²/c²), where τ' is the dilated lifetime, τ is the proper lifetime (lifetime at rest), v is the velocity of the particle, and c is the speed of light.
The problem gives us τ' = 8.32 μs and τ = 2.52 μs. We substitute these values into the equation and solve for v:
8.32 μs = 2.52 μs / √(1 - v²/(3.00 × 10⁸ m/s)²)
We first square both sides, rearrange the equation, and then take the square root to find v:
(8.32 μs / 2.52 μs)² = 1 / (1 - v²/c²)
(3.30)² = 1 / (1 - v²/c²)
10.89 = 1 + v²/c²
v²/c² = 10.89 - 1
v²/c² = 9.89
v = √(9.89) ∑ c
v ≈ 0.995c
Thus the required speed of the particle for its lifetime to be observed as 8.32 μs is approximately 0.995c. However, this speed is not listed among the multiple-choice options, suggesting either a calculation error or that none of the provided answers are correct. It's important to review the calculation carefully and consider rounding effects that might lead to one of the provided answers being the nearest correct choice.
A 19.4 cm pendulum has a period of 0.88 s. What is the free-fall acceleration at the pendulum's location?
Answer:
Acceleration due to gravity value
[tex]=9.89m/ {s}^{2} [/tex]
Explanation:
Time period of simple pendulum is given by the expression
[tex]T=2\pi\sqrt{\frac{l}{g}} \\ [/tex]
Here we have
T = 0.88 s
l = 19.4 cm = 0.194 m
Substituting
[tex]0.88=2\pi\sqrt{\frac{0.194}{g}}\\\\\frac{0.194}{g}=0.0196\\\\g=9.89m/s^2 \\ [/tex]
Acceleration due to gravity value
[tex]=9.89m/s^2 \\ [/tex]