Answer:
8.64 m/s
Explanation:
v1 = 8 m/s, v2 = ?, P2 - P1 = 5338 Pa, density of water, d = 1000 kg/m^3
By the use of Bernoulli's theorem
P 1 + 1/2 x d x v1^2 = P2 + 1/2 x d x v2^2
P2 - P1 = 1/2 x d x (v2^2 - v1^2)
5338 = 0.5 x 1000 x (v2^2 - 64)
10.676 = v2^2 - 64
v2^2 = 74.676
v2 = 8.64 m/s
A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) the ball’s initial velocity and (b) the height it reaches.
Answer:
Initial velocity = [tex]29.43m/s[/tex] b) Height it reaches = 44.145 m
Explanation:
Using the first equation of motion we have
[tex]v=u+at[/tex]
here
v is the final velocity
u is the initial velocity
a is the acceleration of the object
t is time
When the ball reaches it's highest point it's velocity will become 0 as it will travel no further
Also the acceleration in our case is acceleration due to gravity ([tex]-9.8m/s^{2}[/tex]) as the ball moves in it's influence alone with '-' indicating downward direction
Thus applying the values we get
[tex]0=u-(9.81)m/s^{2}\times 3\\\\u=19.43m/s[/tex]
b)
By 3rd equation of motion we have
[tex]v^{2}-u^{2}=2gs[/tex]
here s is the distance covered
Applying the value of u that we calculated we get
[tex]s=\frac{-u^{2}}{2g}\\\\s=\frac{-29.43^{2}}{-2\times 9.81}\\\\s=44.145m[/tex]
A cosmic ray proton moving toward the Earth at 3.5x 10^7 ms experiences a magnetic force of 1.65x 10^-16 N. What is the strength of the magnetic field if there is a 45° angle between it and the proton's velocity?
Answer:
Magnetic field, [tex]B=4.16\times 10^{-5}\ T[/tex]
Explanation:
It is given that,
Velocity of proton, [tex]v=3.5\times 10^7\ m/s[/tex]
Magnetic force, [tex]F=1.65\times 10^{-16}\ N[/tex]
Charge of proton, [tex]q=1.6\times 10^{-19}\ C[/tex]
We need to find the strength of the magnetic field if there is a 45° angle between it and the proton's velocity. The formula for magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
[tex]B=\dfrac{F}{qv\ sin\theta}[/tex]
[tex]B=\dfrac{1.65\times 10^{-16}}{1.6\times 10^{-19}\times 3.5\times 10^7\times sin(45)}[/tex]
B = 0.0000416 T
[tex]B=4.16\times 10^{-5}\ T[/tex]
Hence, this is the required solution.
The strength of the magnetic field, given the force on the proton, its charge, and velocity, and the angle between the velocity and the magnetic field, is approximately 4.21 x 10^-5 T.
Explanation:The strength of a magnetic field can be calculated using the formula for the force exerted on a moving charge in a magnetic field, which is F = q * v * B * sin(θ). Here, F is the magnetic force, q is the charge of the particle, v is the speed of the particle, B is the strength of the magnetic field, and θ is the angle between the velocity and the magnetic field direction.
In this problem, the magnetic force (F) is given as 1.65 x 10^-16 N, the charge of a proton (q) is +1.6 x 10^-19 C, the speed of the proton (v) is 3.5 x 10^7 m/s, and the angle between the velocity and the magnetic field direction (θ) is 45 degrees. Hence we can rearrange the formula to find the magnetic field (B), getting B = F / (q * v * sin(θ)).
Replacing the values into the equation gives: B = 1.65 x 10^-16 N / (+1.6 x 10^-19 C * 3.5 x 10^7 m/s * sin(45°)) which gives the strength of the magnetic field as approximately 4.21 x 10^-5 T.
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A 1200 kg car traveling north at 14 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?
Final answer:
The Physics question involves calculating the total momentum before and after a traffic collision. Total momentum before the collision is 76800 kg·m/s northward. Assuming a perfectly inelastic collision where the vehicles stick together, the total momentum remains unchanged after the collision.
Explanation:
To calculate the total momentum before the collision, we use the formula: momentum = mass × velocity. For the 1200 kg car traveling north at 14 m/s, its momentum is 1200 kg × 14 m/s = 16800 kg·m/s northward.
For the 2000 kg truck traveling at 30 m/s, its momentum is 2000 kg × 30 m/s = 60000 kg·m/s northward. The total momentum before the collision is the sum of these two momenta, so 76800 kg·m/s northward.
Assuming the collision is perfectly inelastic and the two vehicles stick together, the law of conservation of momentum states that the total momentum before the collision must be equal to the total momentum after the collision.
Therefore, the total momentum after the collision will also be 76800 kg·m/s northward.
Batman (mass = 88.8 kg) jumps straight down from a bridge into a boat (mass = 530 kg) in which a criminal is fleeing. The velocity of the boat is initially +14.5 m/s. What is the velocity of the boat after Batman lands in it?
Answer:
The velocity of the boat after Batman lands in it is 2.08 m/s.
Explanation:
Given that,
Mass of batman [tex]m_{1}= 88.8\ kg[/tex]
Mass of boat [tex]m_{2}=530\ kg[/tex]
Initial velocity = 14.5 m/s
We need to calculate the velocity of boat
Using conservation of momentum
[tex]m_{1}u=(m_{1}+m_{2})v[/tex]...(I)
Where, [tex]m_{1}[/tex]=mass of batman
[tex]m_{2}[/tex] =mass of boat
u=initial velocity
v = velocity of boat
Put the value in the equation
[tex]88.8\times14.5=530+88.8\times v[/tex]
[tex]v=\dfrac{88.8\times14.5}{530+88.8}[/tex]
[tex]v=2.08\ m/s[/tex]
Hence, The velocity of the boat after Batman lands in it is 2.08 m/s.
A charged capacitor is connected to an inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time t = 0 the capacitor is fully charged. At a given instant later the charge on the capacitor is measured to be Q = 3 μC and the current in the circuit is equal to 75 μA. What is the maximum charge of the capacitor?
Answer:
[tex]Q = 8.61 \times 10^{-4} C[/tex]
Explanation:
Since in LC oscillation there is no energy loss
so here we can say that
initial total energy of capacitor = energy stored in capacitor + energy stored in inductor at any instant of time
so we can say
[tex]\frac{Q^2}{2C} = \frac{q^2}{2C} + \frac{1}{2}Li^2[/tex]
now we have
[tex]q = 3\mu C[/tex]
[tex]i = 75 \mu A[/tex]
now we have
[tex]Q^2 = q^2 + (LC) i^2[/tex]
we also know that
[tex]2\pi f = \frac{1}{\sqrt{LC}}[/tex]
[tex]2\pi(1.6) = \frac{1}{\sqrt{LC}}[/tex]
[tex]LC = 9.89 \times 10^{-3}[/tex]
now from above equation
[tex]Q^2 = (3\mu C)^2 + (9.89 \times 10^{-3})(75 \mu A)[/tex]
[tex]Q = 8.61 \times 10^{-4} C[/tex]
To find the maximum charge of the capacitor in the LC circuit, set up a cosine equation using the initial charge and current value.
Explanation:To find the maximum charge of the capacitor in the LC circuit, we need to utilize the relationship between the charge on the capacitor and the current in the circuit. At time t = 0, the capacitor is fully charged, so the initial charge is Q = 3 µC. The current in the circuit is equal to 75 µA. Knowing these values, we can set up a cosine equation to find q(t). By solving the equation, we can find the maximum charge of the capacitor.
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A cosmic ray proton moving toward the Earth at 5.00×107 m/s experiences a magnetic force of 1.70×10−16 N . What is the strength of the magnetic field if there is a 45º angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface?
Explanation:
It is given that,
Speed of proton, [tex]v=5\times 10^7\ m/s[/tex]
Magnetic force, [tex]F=1.7\times 10^{-16}\ N[/tex]
(1) The strength of the magnetic field if there is a 45º angle between it and the proton’s velocity. The magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
[tex]B=\dfrac{F}{qv\ sin\theta}[/tex]
[tex]B=\dfrac{1.7\times 10^{-16}\ N}{1.6\times 10^{-19}\ C\times 5\times 10^7\ m/s\ sin(45)}[/tex]
B = 0.00003 T
or
B = 0.03 mT
The magnitude of Earth's magnitude of 25 to 65 Tesla. The value obtained in part (a) is not consistent with the known strength of the Earth’s magnetic field on its surface.
A cat named SchrÖdinger walks along a uniform plank that is 4.00 m long and has a mass of 7.00 kg. The plank is supported by two sawhorses, one that is 0.44 m from the left end of the plank, and the other that is 1.50 m from the right end. When the cat reaches the very right end of the plank, the plank starts to tip. What is Schrodinger’s mass? Note: Just when the plank begins to tip, the normal force on the plank from the sawhorse on the left will go to zero.
Answer:
2.3 kg
Explanation:
L = length of the plank = 4 m
M = mass of the plank = 7 kg
m = mass of cat = ?
[tex]F_{c}[/tex] = Weight of the cat = mg
[tex]F_{p}[/tex] = Weight of the plank = Mg = 7 x 9.8 = 68.6 N
ED = 2 m
CD = 1.5 m
EC = ED - CD = 2 - 1.5 = 0.5 m
Using equilibrium of torque about sawhorse at C
[tex]F_{p}[/tex] (EC) = [tex]F_{c}[/tex] (CD)
(68.6) (0.5) = (mg) (1.5)
(68.6) (0.5) = (m) (1.5) (9.8)
m = 2.3 kg
Determine the length of a copper wire that has a resistance of 0.172 ? and cross-sectional area of 7.85 × 10-5 m2. The resistivity of copper is 1.72 × 10-8 ?·m.
Final answer:
To determine the length of a copper wire given its resistance, cross-sectional area, and the resistivity of copper, use the formula L = (R·A)/ρ. The length of the copper wire, based on the provided resistance of 0.172 Ω, cross-sectional area of 7.85 × [tex]10^-^5[/tex] m², and resistivity of 1.72 × [tex]10^-^8[/tex]Ω·m, is calculated to be 7.85 meters.
Explanation:
The question involves determining the length of a copper wire given its resistance, cross-sectional area, and the resistivity of copper. The formula to calculate the length of the wire is derived from Ohm's law, which in terms of resistivity (ρ) is expressed as R = ρL/A, where R is the resistance, L is the length of the wire, A is the cross-sectional area, and ρ is the resistivity of the material.
We are provided with the resistance (R) as 0.172 Ω, the cross-sectional area (A) as 7.85 × [tex]10^-^5[/tex] m², and the resistivity (ρ) of copper as 1.72 × [tex]10^-^8[/tex] Ω·m. By rearranging the formula to solve for L, we get L = (R·A)/ρ. Substituting the given values into the formula, we find the length of the copper wire.
Let's calculate: L = (0.172 Ω × 7.85 × [tex]10^-^5[/tex] m^2) / (1.72 × [tex]10^-^8[/tex] Ω·m) = 7.85 m. Therefore, the length of the copper wire is 7.85 meters.
At a certain instant after jumping from the airplane A, a skydiver B is in the position shown and has reached a terminal (constant) speed vB = 52 m/s. The airplane has the same constant speed vA = 52 m/s, and after a period of level flight is just beginning to follow the circular path shown of radius ρA = 2330 m. (a) Determine the velocity and acceleration of the airplane relative to the skydiver. (b) Determine the time rate of change of the speed vr of the airplane and the radius of curvature ρr of its path, both as observed by the nonrotating skydiver.
(a) The velocity of the airplane relative to the skydiver is 52 m/s, and the acceleration is directed radially inward. (b) The time rate of change of the speed [tex]\(v_r\)[/tex] of the airplane, as observed by the skydiver, is 0 m/s², and the radius of curvature [tex]\(\rho_r\)[/tex] of its path is 2330 m.
Explanation:(a) The velocity of the airplane relative to the skydiver is the vector difference of their individual velocities. Since both have the same constant speed of 52 m/s, the relative velocity is 52 m/s. The acceleration of the airplane, as observed by the skydiver, is directed radially inward due to the circular motion.
(b) The time rate of change of speed [tex](\(v_r\))[/tex] is the radial component of acceleration, which is 0 m/s² since the airplane is moving at a constant speed. The radius of curvature [tex](\(\rho_r\))[/tex] of its path is given as 2330 m, representing the circular path's curvature.
These results are derived from principles of relative motion and circular motion, providing insights into the dynamics of the airplane-skydiver system.
A proton moving to the right at 6.2 × 105 ms-1 enters a region where there is an electric field of 62 kNC-1 directed to the left. Describe qualitatively the motion of the proton in this filed. What is the time taken by the proton to come back to the point where it entered the field? (Use the standard values of the mass and charge of a proton)
Answer:
2.06 x 10^-7 s
Explanation:
For going opposite to the applied electric field
u = 6.2 x 10^5 m/s, v = 0 , q = 1.6 x 19^-19 C, E = 62000 N/C,
m = 1.67 x 106-27 Kg, t1 = ?
Let a be the acceleration.
a = q E / m = (1.6 x 10^-19 x 62000) / (1.67 x 10^-27) = 6 x 10^12 m/s^2
Use first equation of motion
v = u + a t1
0 = 6.2 x 10^5 - 6 x 10^12 x t1
t1 = 1.03 x 10^-7 s
Let the distance travelled is s.
Use third equation of motion
V^2 = u^2 - 2 a s
0 = (6.2 x 10^5)^2 - 2 x 6 x 10^12 x s
s = 0.032 m
Now when the proton moves in the same direction of electric field.
Let time taken be t2
use second equation of motion
S = u t + 1/2 a t^2
0.032 = 0 + 1/2 x 6 x 10^12 x t2^2
t2 = 1.03 x 10^-7 s
total time taken = t = t1 + t2
t = 1.03 x 10^-7 + 1.03 x 10^-7 = 2.06 x 10^-7 s
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 48 ft/s2. What is the distance covered before the car comes to a stop
Answer:
Distance covered by the car is 56.01 feet.
Explanation:
It is given that,
Initial velocity, u = 50 mi/h = 73.33 ft/s
Constant deceleration of the car, [tex]a=-48\ ft/s^2[/tex]
Final velocity, v = 0
Let s is the distance covered before the car comes to rest. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{0^2-(73.33\ ft/s^2)^2}{2\times -48\ ft/s^2}[/tex]
s = 56.01 ft
So, the distance covered before the car comes to a stop is 56.01 feet. Hence, this is the required solution.
The distance covered by the car as it comes to rest is 17.1 m.
To calculate the distance covered before the car come to stop, we use the formula below.
Formula:
v² = u²+2as............... Equation 1Where:
s = distance covered by the carv = final velocityu = initial velocitya = acceleration of the carmake s the subject of the equation
s = (v²-u²)/2a........................Equation 2From the question,
Given:
v = 50 mi/h = (50×0.447) = 22.35 m/su = 0 m/sa = 48 ft/s² = (48×0.3048) = 14.63 m/s²Substitute these values into equation 2
s = (22.35²-0²)/(2×14.63)s = 17.1 mHence, the distance covered by the car as it comes to rest is 17.1 m.
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A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of 4.2 m/s, hoping to land on the roof of an adjacent bullding Air resistance is negligible. The horlzontal distance between the two buildings is D, and the roof of the adjacent building is 2.1 m below the jumping off polint Find the maximum value for D. (g 9.80ms
Answer:
Maximum separation between buildings, D = 1.806 m
Explanation:
Vertical motion of criminal
Initial speed = 0 m/s
Acceleration = 9.81 m/s²
Displacement = 2.1 m
We need to find time when he moves vertically 2.1 m.
We have
s = ut + 0.5at²
2.1 = 0 x t + 0.5 x 9.81 x t²
t = 0.43 s.
Horizontal motion of criminal
Initial speed = 4.2 m/s
Acceleration =0 m/s²
Time = 0.43 s
We need to find displacement.
We have
s = ut + 0.5at²
s = 4.2 x 0.43 + 0.5 x 0 x 0.43²
s = 1.806 m
So maximum separation between buildings, D = 1.806 m
A 100-lb load is suspended by two chains in a room. The angle between each and the horizontal ceiling is 45°. What is the magnitude of the force each chain must be support?
Magnitude of force on each chain suspended with 100 ib load in a room, must be support is 314.54 N.
What is static equilibrium state of hanging load?If the load, is hanging on chain or rope, and is in the equilibrium state, then the sum of all the horizontal component of it must is equal to zero.
Given information-
The load suspended by two chains in a room is 100-ib.
The angle between each chain and the horizontal ceiling is 45°.
The image attached below shows the given situation.
First convert the 100-ib into unit of N as,
[tex]F=100\rm ib=100\times4.4482\rm N\\F=444.82N[/tex]
As the load suspended by two chains in a room is 100-ib and the angle between each chain and the horizontal ceiling is 45°. Thus, the vertical component of the,
[tex]\sin (45)=\dfrac{F_y}{444.82}\\F_y=314.54\rm N[/tex]
Hence, the magnitude of the force each chain must be support is 314.54 N.
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Thus, the magnitude of the force each chain must support is approximately 70.7 lb.
To find the magnitude of the force each chain must support, you can use the principles of equilibrium.
Given:
A 100-lb load is suspended by two chains.The angle between each chain and the horizontal ceiling is 45°.Since the system is in equilibrium, the vertical forces must balance the weight of the load, and the horizontal components of the forces must cancel each other out.
1. Determine the vertical component of the force in each chain:
Each chain supports part of the total load. Let [tex]\( F \)[/tex] be the force in each chain.
Since the chains are symmetric, each chain supports half the load.
The vertical component of the force in each chain is given by:
[tex]\[ F_{\text{vertical}} = F \sin(45^\circ) \][/tex]
where [tex]\( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)[/tex].
2. Calculate the vertical component for each chain:
For equilibrium, the sum of the vertical components of the forces in the two chains must equal the total load:
[tex]\[ 2 \times F \sin(45^\circ) = 100 \text{ lb} \][/tex]
Substitute [tex]\( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)[/tex] :
[tex]\[ 2 \times F \times \frac{\sqrt{2}}{2} = 100 \][/tex]
Simplify:
[tex]\[ F \sqrt{2} = 100 \][/tex]
Solve for [tex]\( F \)[/tex]:
[tex]\[ F = \frac{100}{\sqrt{2}} \][/tex]
Rationalize the denominator:
[tex]\[ F = \frac{100 \sqrt{2}}{2} = 50 \sqrt{2} \][/tex]
3. Approximate the force:
Using [tex]\( \sqrt{2} \approx 1.414 \)[/tex]:
[tex]\[ F \approx 50 \times 1.414 = 70.7 \text{ lb} \][/tex]
A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?
The ball thrown by a major-league pitcher to a catcher 17.0m away, with a speed of 41.0m/s, would drop approximately 0.84 meters due to the gravitational acceleration.
Explanation:The subject of this question involves physics, specifically the concept of projectile motion. When a pitcher throws a ball with a horizontal speed, the ball also experiences a vertical motion due to gravity. Therefore, even though the ball is thrown horizontally, it will gradually drop as it travels towards the catcher. To figure out how far it drops, we can use the physics equation for motion under constant acceleration: distance = 0.5 * acceleration * time2.
Here, the acceleration is due to gravity, which is approximately 9.8m/s2. The time can be calculated by dividing the horizontal distance (17.0 m) by the horizontal speed (41.0 m/s), giving us approximately 0.4146 s. Substituting these into the equation: 0.5 * 9.8m/s2 * (0.4146s)2 = 0.84 meters, the drop of the ball is approximately 0.84 meters.
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A simple pendulum is made from a 0.75-m-long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest speed?
Answer:
It takes 0.43 seconds before the pendulum attains the maxium speed.
Explanation:
L= 0.75m
g= 9.8 m/s²
T= 2π * √(L/g)
T=1.73 sec
T(vmax) = T/4
T(vmax) = 0.43 sec
The time elapsed before a simple pendulum attains its greatest speed is given by t = T/4 = π√(L/g)/2, where T is the period and L is the length of the pendulum.
Explanation:When a simple pendulum is released from rest, it oscillates back and forth. The time it takes for the pendulum to reach its greatest speed depends on the length of the pendulum. The formula for the period of a simple pendulum is given by:
T = 2π√(L/g)
Where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. To find the time it takes for the pendulum to reach its greatest speed, we can divide the period by 4. This is because the pendulum reaches its greatest speed when it is at the bottom of its swing, which is halfway through one period.
So, the time elapsed before the pendulum attains its greatest speed is:
t = T/4 = (2π√(L/g))/4 = π√(L/g)/2
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A person fires a 38 gram bullet straight up into the air. It rises, then falls straight back down, striking the ground with a speed of 345 m/s. The bullet embeds itself into the ground a distance of 8.9 cm before coming to a stop. What force does the ground exert on the bullet? Express your answer in newtons.
Answer:
Force exerted = 25.41 kN
Explanation:
We have equation of motion
v² = u²+2as
u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s
0² = 345²+2 x a x 0.089
a = -668679.78 m/s²
Force exerted = Mass x Acceleration
Mass of bullet = 38 g = 0.038 kg
Acceleration = 668679.78 m/s²
Force exerted = 25409.83 N = 25.41 kN
In a particular case of Compton scattering, a photon collides with a free electron and scatters backwards. The wavelength after the collision is exactly double the wavelength before the collision. What is the wavelength of the incident photon?
Answer:
hence initial wavelength is [tex]\lambda =4.86\times10^{-12}m[/tex]
Explanation:
shift in wavelength due to compton effect is given by
[tex]\lambda ^{'}-\lambda =\frac{h}{m_{e}c}\times(1-cos\theta )[/tex]
λ' = the wavelength after scattering
λ= initial wave length
h= planks constant
m_{e}= electron rest mass
c= speed of light
θ= scattering angle = 180°
compton wavelength is
[tex]\frac{h}{m_{e}c}= 2.43\times10^{-12}m[/tex]
[tex]\lambda '-\lambda =2.43\times10^{-12}\times(1-cos\theta )[/tex]
[tex]\lambda '-\lambda =2.43\times10^{-12}\times(1+1 )[/tex] ( put cos 180°=-1)
also given λ'=2λ
putting values and solving we get
[tex]\lambda =4.86\times10^{-12}m[/tex]
hence initial wavelength is [tex]\lambda =4.86\times10^{-12}m[/tex]
Consider a grain of table salt which is made of positive
and negative ions (Na+and Cl−). Suppose each of these ions carries a charge of 1.60x10^−19 C and are 5.29x10^−11mapart. What is the magnitude of the electrostatic force
between them?
Answer:
Force, [tex]F=8.23\times 10^{-8}\ N[/tex]
Explanation:
It is given that,
Each ion in Na⁺ and Cl⁻ has a charge of, [tex]q=1.6\times 10^{-19}\ C[/tex]
Distance between two ions, [tex]d=5.29\times 10^{-11}\ m[/tex]
We need to find the electrostatic force. It is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{(1.6\times 10^{-19})^2}{(5.29\times 10^{-11})^2}[/tex]
[tex]F=8.23\times 10^{-8}\ N[/tex]
So, the magnitude of electrostatic force between them is [tex]F=8.23\times 10^{-8}\ N[/tex]. Hence, this is the required solution.
Express 79 m in units of (a) centimeters
(b) feet
(c) inches and
(d) miles.
Answer: a) 7,00 centimeters
(b) 259. 19 feet
(c) 3110.28 inches
(d) 0.049 miles
Explanation:
(a) We know that 1 meter = 100 centimeters
Therefore,
[tex]79\ m= 7,900\text{ centimeters}[/tex]
(b)Since 1 meter = 3.28084 feet
Then, [tex]79\ m= 79\times3.28084=259.18636\approx 259.19\text{ feet}[/tex]
(c) Since, 1 feet = 12 inches.
[tex]79\ m=259.19\text{ feet}=259.19\times12=3110.28\text{ inches}[/tex]
(d) [tex]\text{Since 1 feet= }\dfrac{1}{5280}\text{ mile}[/tex]
[tex]79\ m=259.19\text{ feet}=\dfrac{259.19}{5280}= 0.0490890151515\approx0.049\text{ miles}[/tex]
A proton and an alpha particle (q = +2e, m = 4 u) are fired directly toward each other from far away, each with an initial speed of 0.010c. What is their distance of closest approach, as measured between their centers?
Answer:
Distance of closest approach, [tex]r=1.91\times 10^{-14}\ m[/tex]
Explanation:
It is given that,
Charge on proton, [tex]q_p=e[/tex]
Charge on alpha particle, [tex]q_a=2e[/tex]
Mass of proton, [tex]m_p=1.67\times 10^{-27}\ kg[/tex]
Mass of alpha particle, [tex]m_a=4m_p=6.68\times 10^{-27}\ kg[/tex]
The distance of closest approach for two charged particle is given by :
[tex]r=\dfrac{k2e^2(m_p+m_a)}{2m_am_pv_p^2}[/tex]
[tex]r=\dfrac{9\times 10^9\times 2(1.6\times 10^{-19})^2(1.67\times 10^{-27}+6.68\times 10^{-27})}{2\times 6.68\times 10^{-27}\times 1.67\times 10^{-27}(0.01\times 3\times 10^8)^2}[/tex]
[tex]r=1.91\times 10^{-14}\ m[/tex]
So, their distance of closest approach, as measured between their centers [tex]1.91\times 10^{-14}\ m[/tex]. Hence, this is the required solution.
A vertical straight wire carrying an upward 24-A current exerts an attractive force per unit length of 88 X 104N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?
Answer:
The current flows in the second wire is [tex]1.3\times10^{10}\ A[/tex]
Explanation:
Given that,
Upward current = 24 A
Force per unit length[tex]\dfrac{F}{l} =88\times10^{4}\ N/m [/tex]
Distance = 7.0 cm
We need to calculate the current in second wire
Using formula of magnetic force
[tex]F=ILB[/tex]
[tex]\dfrac{F}{l}=\dfrac{\mu I_{1}I_{2}}{2\pi r}[/tex]
Where,
[tex]\dfrac{F}{l}[/tex]=force per unit length
I₁= current in first wire
I₂=current in second wire
r = distance between the wires
Put the value into the formula
[tex]88\times10^{4}=\dfrac{4\pi\times10^{-7}\times24\times I_{2}}{2\pi \times7\times10^{-2}}[/tex]
[tex]I_{2}=\dfrac{88\times10^{4}\times7\times10^{-2}}{2\times\times10^{-7}\times24}[/tex]
[tex]I_{2}=1.3\times10^{10}\ A[/tex]
Hence, The current flows in the second wire is [tex]1.3\times10^{10}\ A[/tex]
Consider two cubes, one of aluminum and one of copper. If each cube measures 8 cm along an edge, calculate the mass of each cube.
Answer:
≈1,39 and 4.56 kg.
Explanation:
for more details see the attachment. Note, the ρ(Cu) means 'Density of copper', ρ(Al) means 'Density of aluminium'.
Which of the following are true (choose all that apply)? Sound can travel through a vacuum. -Sound can travel through water. Light can travel through a vacuum Sound can travel through air. Light can travel through air Light can travel through water
Answer:
option (B) - true
option (C) - true
option (D) - true
option (E) - true
option (F) - true
Explanation:
A uniform non-conducting ring of radius 2.2 cm and total charge 6.08 µC rotates with a constant angular speed of 2.01 rad/s around an axis perpendicular to the plane of the ring that passes through its center. What is the magnitude of the magnetic moment of the rotating ring?
Answer:
The magnitude of the magnetic moment of the rotating ring is [tex]2.96\times10^{-9}\ Am^2[/tex]
Explanation:
Given that,
Radius = 2.2 cm
Charge [tex]q = 6.08\times10^{-6}\ C[/tex]
Angular speed = 2.01 rad/s
We need to calculate the time period
[tex]T = \dfrac{2\pi}{\omega}[/tex]
Now, The spinning produced the current
Using formula for current
[tex] I = \dfrac{q}{T}[/tex]
We need to calculate the magnetic moment
Using formula of magnetic moment
[tex]M = I A[/tex]
Put the value of I and A into the formula
[tex]M=\dfrac{q}{T}\times A[/tex]
[tex]M=\dfrac{q\times\omega}{2\pi}\times \pi\times r^2[/tex]
Put the value into the formula
[tex]M=\dfrac{6.08\times10^{-6}\times2.01\times(2.2\times10^{-2})^2}{2}[/tex]
[tex]M=2.96\times10^{-9}\ Am^2[/tex]
Hence, The magnitude of the magnetic moment of the rotating ring is [tex]2.96\times10^{-9}\ Am^2[/tex]
The magnetic moment of the rotating ring with a radius of 2.2 cm, a total charge of 6.08 µC, and an angular velocity of 2.01 rad/s is approximately 1.176 x 10⁻⁵ A⋅m².
Explanation:The magnitude of the magnetic moment (μ) of a rotating charged ring can be found using the formula given by the product of the charge (Q) and the angular velocity (ω) and the radius (r) squared, as μ = Qωr²/2. For a uniform non-conducting ring rotating about an axis perpendicular to its plane, this relationship can be used to determine the magnetic moment generated by its rotation.
To calculate the magnetic moment for the given scenario:
Carrying out the multiplication gives the magnetic moment:
μ = (6.08 x 10-6 C)(2.01 rad/s)(0.000484 m2)/2 = 0.01176 x 10-6 C⋅m2/s
This results in a magnetic moment of approximately μ ≈ 1.176 x 10-5 A⋅m2.
A flywheel in the form of a uniformly thick disk of radius 1.93 m has a mass of 92.1 kg and spins counterclockwise at 419 rpm. Calculate the constant torque required to stop it in 1.25 min.
Answer:
Torque = 99.48 N-m²
Explanation:
It is given that,
Radius of the flywheel, r = 1.93 m
Mass of the disk, m = 92.1 kg
Initial angular velocity, [tex]\omega_i=419\ rpm=43.87\ rad/s[/tex]
Final angular speed, [tex]\omega_f=0[/tex]
We need to find the constant torque required to stop it in 1.25 min, t = 1.25 minutes = 75 seconds
Torque is given by :
[tex]\tau=I\times \alpha[/tex]...........(1)
I is moment of inertia, for a solid disk, [tex]I=\dfrac{mr^2}{2}[/tex]
[tex]\alpha[/tex] is angular acceleration
[tex]I=\dfrac{92.1\ kg\times (1.93\ m)^2}{2}=171.53\ kgm^2[/tex]..............(2)
Now finding the value of angular acceleration as :
[tex]\omega_f=\omega_i+\alpha t[/tex]
[tex]0=43.87+\alpha \times 75[/tex]
[tex]\alpha =-0.58\ m/s^2[/tex]..........(3)
Using equation (2) and (3), solve equation (1) as :
[tex]\tau=171.53\ kgm^2\times -0.58\ m/s^2[/tex]
[tex]\tau=-99.48\ N-m^2[/tex]
So, the torque require to stop the flywheel is 99.48 N-m². Hence, this is the required solution.
The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ/mol. Given the following information: heat of sublimation for Cs is +76 kJ/mol, bond dissociation energy for 12Cl2 is +121 kJ/mol, Ei1 for Cs is +376 kJ/mol, and Eea for Cl(g) is −349 kJ/mol. What is the magnitude of the lattice energy for CsCl?
Answer : The magnitude of the lattice energy for CsCl is, 667 KJ/mole
Explanation :
The steps involved in the born-Haber cycle for the formation of [tex]CsCl[/tex] :
(1) Conversion of solid calcium into gaseous cesium atoms.
[tex]Cs(s)\overset{\Delta H_s}\rightarrow Cs(g)[/tex]
[tex]\Delta H_s[/tex] = sublimation energy of calcium
(2) Conversion of gaseous cesium atoms into gaseous cesium ions.
[tex]Ca(g)\overset{\Delta H_I}\rightarrow Ca^{+1}(g)[/tex]
[tex]\Delta H_I[/tex] = ionization energy of calcium
(3) Conversion of molecular gaseous chlorine into gaseous chlorine atoms.
[tex]Cl_2(g)\overset{\frac{1}{2}\Delta H_D}\rightarrow Cl(g)[/tex]
[tex]\Delta H_D[/tex] = dissociation energy of chlorine
(4) Conversion of gaseous chlorine atoms into gaseous chlorine ions.
[tex]Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)[/tex]
[tex]\Delta H_E[/tex] = electron affinity energy of chlorine
(5) Conversion of gaseous cations and gaseous anion into solid cesium chloride.
[tex]Cs^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow CsCl(s)[/tex]
[tex]\Delta H_L[/tex] = lattice energy of calcium chloride
To calculate the overall energy from the born-Haber cycle, the equation used will be:
[tex]\Delta H_f^o=\Delta H_s+\Delta H_I+\Delta H_D+\Delta H_E+\Delta H_L[/tex]
Now put all the given values in this equation, we get:
[tex]-443KJ/mole=76KJ/mole+376KJ/mole+121KJ/mole+(-349KJ/mole)+\Delta H_L[/tex]
[tex]\Delta H_L=-667KJ/mole[/tex]
The negative sign indicates that for exothermic reaction, the lattice energy will be negative.
Therefore, the magnitude of the lattice energy for CsCl is, 667 KJ/mole
The magnitude of Lattice energy of CsCl is -676 kJ/mol.
Given Here,
Enthalpy of sublimation of Cs [tex]\rm \bold{ \Delta H(sub ) }[/tex] = +76 kJ/mo
Ionization Energy for Potassium IE(Cs) = +376 kJ/mol
Electron affinity for Chlorine is EA(Cl) = −349 kJ/mol.
Bond dissociation energy of Chlorine, BE(Cl) = +121 kJ/mol
Enthalpy of formation for CsCl, [tex]\rm \bold{ \Delta H(f) }[/tex] = −436.5 kj/mol .
The lattice energy of KCl can be calculated from the formula,
[tex]\rm \bold {U(CsCl) = \Delta Hf(CsCl) - [ \Delta H(sub) + IE(Cs) +BE(Cl_2) + EA(Cl)]}[/tex]
U( CsCl) = -436 - [+76 +376 +121 kJ/mol -349]
U( CsCl) = -676 kJ/mol
Hence we can calculate that the magnitude of Lattice energy of CsCl is -676 kJ/mol.
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A spacecraft of mass 1500 kg orbits the earth at an altitude of approximately 450 km above the surface of the earth. Assuming a circular orbit, what is the attractive force that the earth exerts to keep the spacecraft+ in orbit? Answer: (a) 1.28x 10^7 (N) (b) 2.99 x 10^7 (N) (c) 3.56 x 10^7 (N). (d) 4.11 x 10^7 (N) (e) 5.06x 10^7 (N)
Answer:
1.28 x 10^4 N
Explanation:
m = 1500 kg, h = 450 km, radius of earth, R = 6400 km
Let the acceleration due to gravity at this height is g'
g' / g = {R / (R + h)}^2
g' / g = {6400 / (6850)}^2
g' = 8.55 m/s^2
The force between the spacecraft and teh earth is teh weight of teh spacecraft
W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N
The windpipe of a typical whooping crane is about 4.6 ft long. What is the lowest resonant frequency of this pipe, assuming that it is closed at one end? Assume a temperature of 33°C. ans in hz
Answer:
Frequency, f = 481.8 Hz
Explanation:
Given that,
Length of windpipe, l = 4.6 ft = 0.182 m
We need to find the lowest resonant frequency of this pipe at 33 degrees Celcius. Firstly, we will find the speed of sound at 33 degrees Celcius as :
[tex]v=331+0.6T[/tex]
[tex]v=331+0.6\times 33[/tex]
v = 350.8 m/s
At resonance, wavelength is equal to 4 times length of pipe i.e.
λ = 4 l
We need that, [tex]f=\dfrac{v}{\lambda}[/tex]
[tex]f=\dfrac{350.8\ m/s}{4\times 0.182\ m}[/tex]
f = 481.8 Hz
So, the resonant frequency of the windpipe is 481.8 Hz. Hence, this is the required solution.
The lowest resonant frequency of the whooping crane's windpipe, assuming it is closed at one end and a temperature of 33°C, is approximately 62.82 Hz.
Explanation:The lowest resonant frequency of the whooping crane's windpipe, which can be thought of as a pipe that is closed at one end, can be found using the formula for the fundamental frequency of such a pipe: f = v/λ, where f is the frequency, v is the speed of sound, and λ is the wavelength. In this case, the speed of sound is dependent on the temperature and can be approximated as v = 331.4 + 0.6*T m/s, where T is the temperature in degrees Celsius.
At a temperature of 33°C, the speed of sound, v, is approximately 351.8 m/s. Note that we need to convert the length of the windpipe from feet to meters, with 4.6 ft being approximately 1.4 m. The fundamental wavelength for a pipe closed at one end is four times the length of the pipe (λ=4L), so in this case, λ=4*1.4 m = 5.6 m.
Substituting these values into the frequency formula gives us f = v/λ = 351.8 m/s / 5.6 m = 62.82 Hz. So, the lowest resonant frequency of the whooping crane's windpipe is approximately 62.82 Hz.
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A ballet dancer has a maximum net torque of 45 Nm, and the moment of inertia of his body is 30 kg mA2. if he starts twirling from rest and keeps accelerating for 95 seconds, how fast will he be rotating? a) b) if he holds rocks in his hands, which variables in this problem would change, in which direction, and why?
Answer:
(a) 142.5 rad/s
Explanation:
Torque = 45 Nm
Moment of inertia = 30 kgm^2
w0 = 0, t = 95 second
(a) Let it is rotating with angular speed w.
Torque = moment of inertia x angular acceleration
45 = 30 x α
α = 1.5 rad/s^2
Use first equation of motion for rotational motion
w = w0 + α t
w = 0 + 1.5 x 95
w = 142.5 rad/s
(b) if he hold rocks, then the moment of inertia change and then angular acceleration change and then final angular velocity change.
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Answer:
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Explanation: