Answer:
Change in momentum = 2.197 kgm/s
Explanation:
Momentum = MV
Initial momentum = MU
Final momentum = MV
Computation:
⇒ Change in momentum = MV - MU
⇒ Change in momentum = M (V - U)
⇒ Change in momentum = 0.13(-10.3 - 6.6)
⇒ Change in momentum = 0.13(16.9)
⇒ Change in momentum = 2.197 kgm/s
A ball is thrown straight upward and falls back to Earth. Suppose a y-coordinate axis points upward, and the release point is the origin. Instantaneously at the top its flight, which of these quantities are zero
a. Displacment
b. Speed
c. Velocity
d. Accerlation
Explanation:
A ball is thrown straight upward and falls back to Earth. It means that it is coming to the initial position. Displacement is given by the difference of final position and initial position. The displacement of the ball will be 0. As a result velocity will be 0.
Acceleration is equal to the rate of change of velocity. So, its acceleration is also equal to 0.
Hence, displacement, velocity and acceleration are zero.
If you slide down a rope, it's possible to create enough thermal energy to burn your hands or your legs where they grip the rope. Suppose a 30 kg child slides down a rope at a playground, descending 2.5 m at a constant speed.
How much thermal energy is created as she slides down the rope?
Answer:
Q = 735 J
Explanation:
In this exercise we must assume that all the mechanical energy of the system transforms into cemite energy.
Initial energy
Em₀ = U = m g h
final energy
[tex]Em_{f}[/tex] = Q
Em₀ = Em_{f}
m g h = Q
let's calculate
Q = 30 9.8 2.5
Q = 735 J
A 73 kg swimmer dives horizontally off a 462 kg raft initially at rest. If the diver's speed immediately after leaving the raft is 5.54 m/s, what is the corresponding raft speed
Answer:
Corresponding raft speed = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)
Explanation:
Law of conservation of momentum gives that the momentum of the diver and the raft before the dive is equal to the momentum of the diver and the raft after the dive.
And since the raft and the diver are initially at rest, the momentum of the diver after the dive is equal and opposite to the momentum experienced by the raft after the dive.
(Final momentum of the diver) + (Final momentum of the raft) = 0
Final Momentum of the diver = (mass of the diver) × (diving velocity of the diver)
Mass of the diver = 73 kg
Diving velocity of the diver = 5.54 m/s
Momentum of the diver = 73 × 5.54 = 404.42 kgm/s
Momentum of the raft = (mass of the raft) × (velocity of the raft)
Mass of the raft = 462 kg
Velocity of the raft = v
Momentum of the raft = 462 × v = (462v) kgm/s
404.42 + 462v = 0
462v = -404.42
v = (-404.42/462) = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)
Hope this Helps!!!
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the floor. A man weighing 700 N could climb up to 7.0 m before slipping. What is the coefficient of static friction between the floor and the ladder
Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37
An airplane is flying on a bearing of N 400 W at 500 mph. A strong jet-stream speed wind of 100 mph is blowing at S 500 W.
Required:
a. Find the vector representation of the plane and of the wind.
b. Find the resultant vector that represents the actual course of the plane.
c. Give the resulting speed and bearing of the plane.
Answer:
A. a (-321.393, 383.022) b (-76.40, -64.278)
B. (-397.991, 318.744)
C. a. resulting speed 509.9mph b. bearing of the plane = 51.6°
Explanation:
A child of mass 46.2 kg sits on the edge of a merry-go-round with radius 1.9 m and moment of inertia 130.09 kg m2 . The merrygo-round rotates with an angular velocity of 2.4 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 0.779 m from the center. Now what is the angular velocity of the merry-go-round
Answer:
The angular velocity is [tex]w_f = 4.503 \ rad/s[/tex]
Explanation:
From the question we are told that
The mass of the child is [tex]m_c = 46.2 \ kg[/tex]
The radius of the merry go round is [tex]r = 1.9 \ m[/tex]
The moment of inertia of the merry go round is [tex]I_m = 130.09 \ kg \cdot m^2[/tex]
The angular velocity of the merry-go round is [tex]w = 2.4 \ rad/s[/tex]
The position of the child from the center of the merry-go-round is [tex]x = 0.779 \ m[/tex]
According to the law of angular momentum conservation
The initial angular momentum = final angular momentum
So
[tex]L_i = L_f[/tex]
=> [tex]I_i w_i = I_fw_f[/tex]
Now [tex]I_i[/tex] is the initial moment of inertia of the system which is mathematically represented as
[tex]I_i = I_m + I_{b_1}[/tex]
Where [tex]I_{b_i}[/tex] is the initial moment of inertia of the boy which is mathematically evaluated as
[tex]I_{b_i} = m_c * r[/tex]
substituting values
[tex]I_{b_i} = 46.2 * 1.9^2[/tex]
[tex]I_{b_i} = 166.8 \ kg \cdot m^2[/tex]
Thus
[tex]I_i =130.09 + 166.8[/tex]
[tex]I_i = 296.9 \ kg \cdot m^2[/tex]
Thus
[tex]I_i * w_i =L_i= 296.9 * 2.4[/tex]
[tex]L_i = 712.5 \ kg \cdot m^2/s[/tex]
Now
[tex]I_f = I_m + I_{b_f }[/tex]
Where [tex]I_{b_f}[/tex] is the final moment of inertia of the boy which is mathematically evaluated as
[tex]I_{b_f} = m_c * x[/tex]
substituting values
[tex]I_{b_f} = 46.2 * 0.779^2[/tex]
[tex]I_{b_f} = 28.03 kg \cdot m^2[/tex]
Thus
[tex]I_f = 130.09 + 28.03[/tex]
[tex]I_f = 158.12 \ kg \ m^2[/tex]
Thus
[tex]L_f = 158.12 * w_f[/tex]
Hence
[tex]712.5 = 158.12 * w_f[/tex]
[tex]w_f = 4.503 \ rad/s[/tex]
A rock falls from a vertical cliff that is 4.0 m tall and experiences no significant air resistance as it falls. At what speed will its gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy
Answer:
About 6.26m/s
Explanation:
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
Divide both sides by mass:
[tex]gh=\dfrac{1}{2}v^2[/tex]
Since the point of equality of kinetic and potential energy will be halfway down the cliff, height will be 4/2=2 meters.
[tex](9.8)(2)=\dfrac{1}{2}v^2 \\\\v^2=39.4 \\\\v\approx 6.26m/s[/tex]
Hope this helps!
The gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.
Given data:
The height of vertical cliff is, h = 4.0 m.
Since, we are asked for speed by giving the condition for gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy. Then we can apply the conservation of energy as,
Kinetic energy = Gravitational potential energy
[tex]\dfrac{1}{2}mv^{2}=mgh[/tex]
Here,
m is the mass of rock.
v is the speed of rock.
g is the gravitational acceleration.
Solving as,
[tex]v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 4.0}\\\\v =8.85 \;\rm m/s[/tex]
Thus, we can conclude that the gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.
Learn more about the conservation of energy here:
https://brainly.com/question/15707891
A student is investigating the relationship between sunlight and plant growth for her science expieriment. Determine which of the following tables is set up correctly
The question is incomplete as it does not have the options which have been provided in the attachment.
Answer:
Option-D
Explanation:
In the given question, the effect of the sunlight on the growth of the plant has been studied. The values provided in the Option-D can be considered correct as the values are measured in the decimal value up to two decimal value.
The values are measured after the first week, second week, and the initial readings. The difference in the values provided in Option-D does not show much difference as well as are up to two decimal places.
Thus, Option-D is the correct answer.
when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.2525 N. What were the initial charges on the spheres
Answer:
q1 = 7.6uC , -2.3 uC
q2 = 7.6uC , -2.3 uC
( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )
Explanation:
Solution:-
- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.
- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:
[tex]F = k\frac{|q_1|.|q_2|}{r^2}[/tex]
Where,
k: The coulomb's constant = 8.99*10^9
- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.
- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.
- Therefore, the force of attraction between the spheres would be:
[tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1
- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).
- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,
[tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]
- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:
[tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex] .. Eq2
- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:
[tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]
[tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]
A 3.10-mm-long, 430 kgkg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 69.0 kgkg construction worker stands at the far end of the beam.What is the magnitude of the gravitational torque about the point where the beam is bolted into place?
Answer:
Explanation:
Given that,
The length of the beam L = 3.10m
The mass of the steam beam [tex]m_1[/tex] = 430kg
The mass of worker [tex]m_2[/tex] = 69.0kg
The distance from the fixed point to centre of gravity of beam = [tex]\frac{L}{2}[/tex]
and our length of beam is 3.10m
so the distance from the fixed point to centre of gravity of beam is
[tex]\frac{3.10}{2}=1.55m[/tex]
Then the net torque is
[tex]=-W_sL'-W_wL\\\\=-(W_sL'+W_wL)[/tex]
[tex]W_s[/tex] is the weight of steel rod
[tex]=430\times9.8=4214N[/tex]
[tex]W_w[/tex] is the weight of the worker
[tex]=69\times9.8\\\\=676.2N[/tex]
Torque can now be calculated
[tex]-(4214\times1.55+676.2\times3.9)Nm\\\\-(6531.7+2637.18)Nm\\\\-(9168.88)Nm[/tex]
≅ 9169Nm
Therefore,the magnitude of the torque is 9169NmA car traveling with velocity v is decelerated by a constant acceleration of magnitude a. It takes a time t to come to rest. If its initial velocity were doubled, the time required to stop would
Answer:
If the initial speed is doubled the time is also doubled
Explanation:
You have that a car with velocity v is decelerated by a constant acceleration a in a time t.
You use the following equation to establish the previous situation:
[tex]v'=v-at[/tex] (1)
v': final speed of the car = 0m/s
v: initial speed of the car
From the equation (1) you solve for t and obtain:
[tex]t=\frac{v-v'}{a}=\frac{v}{a}[/tex] (2)
To find the new time that car takesto stop with the new initial velocity you use again the equation (1), as follow:
[tex]v'=v_1-at'[/tex] (3)
v' = 0m/s
v1: new initial speed = 2v
t': new time
You solve the equation (3) for t':
[tex]0=2v-at'\\\\t'=\frac{2v}{a}=2t[/tex]
If the initial speed is doubled the time is also doubled
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107 N/m2). By what volume has 1.0 m3 of water from the surface of the lake been compressed if it is forced down to this depth? The bulk modulus of water is 2.3 × 109 Pa.
Answer:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Explanation:
The bulk modulus is represented by the following differential equation:
[tex]K = - V\cdot \frac{dP}{dV}[/tex]
Where:
[tex]K[/tex] - Bulk module, measured in pascals.
[tex]V[/tex] - Sample volume, measured in cubic meters.
[tex]P[/tex] - Local pressure, measured in pascals.
Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:
[tex]-\frac{K \,dV}{V} = dP[/tex]
This resultant expression is solved by definite integration and algebraic handling:
[tex]-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP[/tex]
[tex]-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}[/tex]
[tex]\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}[/tex]
[tex]\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }[/tex]
The final volume is predicted by:
[tex]V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }[/tex]
If [tex]V_{o} = 1\,m^{3}[/tex], [tex]P_{o} - P_{f} = -10132500\,Pa[/tex] and [tex]K = 2.3\times 10^{9}\,Pa[/tex], then:
[tex]V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }[/tex]
[tex]V_{f} \approx 0.996\,m^{3}[/tex]
Change in volume due to increasure on pressure is:
[tex]\Delta V = V_{o} - V_{f}[/tex]
[tex]\Delta V = 1\,m^{3} - 0.996\,m^{3}[/tex]
[tex]\Delta V = 0.004\,m^{3}[/tex]
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
g You have a suction cup that creates a circular region of low pressure with a 30 mm diameter. It holds the pressure to 78 % of atmospheric pressure. What "holding force" does the suction cup generate in N
Answer:
F=49.48 N
Explanation:
Given that
Diameter , d= 30 mm
Holding pressure = 70 % P
P=Atmospherics pressure
We know that
P= 1 atm = 10⁵ N/m²
The force per unit area is known as pressure.
[tex]P=\dfrac{F}{A}[/tex]
[tex]F=P\times A[/tex]
[tex]F=0.7\times 10^5\times \dfrac{\pi}{4}\times 0.03^2\ N[/tex]
Therefore the force will be 49.48 N.
F=49.48 N
A 4.00 kg ball is moving at 4.00 m/s to the EAST and a 6.00 kg ball is moving at 3.00 m/s to the NORTH. The total momentum of the system is:___________.A. 14.2 kg m/s at an angle of 48.4 degrees SOUTH of EAST.B. 48.2 kg m/s at an angle of 24.2 degrees SOUTH of EAST.C. 48.2 kg m/s at an angle of 48.4 degrees NORTH of EAST.D. 24.1 kg m/s at an angle of 24.2 degrees SOUTH of EAST.
E. 24.1 kg m/s at an angles of 48.4 degrees NORTH of EAST.
Answer:
The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EASTExplanation:
Momentum = mass*velocity of a body
For a 4.00 kg ball is moving at 4.00 m/s to the EAST, its momentum = 4*4 = 16kgm/s
For a 6.00 kg ball is moving at 3.00 m/s to the NORTH;
its momentum = 6*3 = 18kgm/s
Total momentum = The resultant of both momentum
Total momentum = √16²+18²
Total momentum = √580
total momentum = 24.1kgm/s
For the direction:
[tex]\theta = tan^{-1} \frac{y}{x}\\\theta = tan^{-1} \frac{18}{16}\\ \theta = tan^{-1} 1.125\\\theta = 48.4^{0}[/tex]
The total momentum is 24.1 kg m/s at an angle of 48.4 degrees NORTH of EAST
Ocean waves of wavelength 30m are moving directly toward a concrete barrier wall at 4.8m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.A) How far from the wall is she?B) What is the period of her up and down motion?
Answer:
a)15m
b)6.25s
Explanation:
A) She is ½ a wavelength away, or
d = λ/2 = 30/2 = 15 m
B)Speed of the wave:
V=fλ = λ/T
so,
T=λ/V= 30/4.8
T=6.25s
a) The distance from the wall is 15m
b) The period of her up and down motion is 6.25s
Calculation of the distance and period is:a.
Since Ocean waves of wavelength 30m are moving directly toward a concrete barrier wall at 4.8m/s .
Also,
She is ½ a wavelength away, or
d = λ/2
= 30/2
= 15 m
b)
Here the speed of wave should be used
T=λ/V
= 30/4.8
T=6.25s
Learn more about wavelength here: https://brainly.com/question/13524696
A 50-kg block is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 150 N. Fp is parallel to the displacement of the block. The coefficient of kinetic friction is 0.25.
a) What is the total work done on the block?
b) If the box started from rest, what is the final speed of the block?
Answer:
a) WT = 137.5 J
b) v2 = 2.34 m/s
Explanation:
a) The total work done on the block is given by the following formula:
[tex]W_T=F_pd-F_fd=(F_p-F_f)d[/tex] (1)
Fp: force parallel to the displacement of the block = 150N
Ff: friction force
d: distance = 5.0 m
Then, you first calculate the friction force by using the following relation:
[tex]F_f=\mu_k N=\mu_k Mg[/tex] (2)
μk: coefficient of kinetic friction = 0.25
M: mass of the block = 50kg
g: gravitational constant = 9.8 m/s^2
Next, you replace the equation (2) into the equation (1) and solve for WT:
[tex]W_T=(F_p-\mu_kMg)d=(150N-(0.25)(50kg)(9.8m/s^2))(5.0m)\\\\W_T=137.5J[/tex]
The work done over the block is 137.5 J
b) If the block started from rest, you can use the following equation to calculate the final speed of the block:
[tex]W_T=\Delta K=\frac{1}{2}M(v_2^2-v_1^2)[/tex] (3)
WT: total work = 137.5 J
v2: final speed = ?
v1: initial speed of the block = 0m/s
You solve the equation (3) for v2:
[tex]v_2=\sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(137.5J)}{50kg}}=2.34\frac{m}{s}[/tex]
The final speed of the block is 2.34 m/s
A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 8.1 rad/s. The drawing shows the position of the block when the spring is unstrained. This position is labeled x= 0 m. The drawing also shows a small bottle located 0.080 m to the right of this position. The block is pulled to the right, stretching the spring by 0.050 m, and is then thrown to the left. In order for the block to knock over the bottle,it must be thrown with a speed exceeding v0. Ignoring the width of the block, find v0.
Answer:
v₀ = 0.5058 m/s
Explanation:
From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m
Now, the potential energy of the block at x = 0.08 m is ½kx²
where;
k is the spring constant given by; k = ω²m
ω is the angular velocity of the oscillation
m is the mass of the block.
Thus, potential energy of the spring at the bottle(x = 0.08 m) is;
U = ½ω²m(0.08m)²
Also, potential energy of the spring at the bottle(x = 0.05 m) is;
U = ½ω²m(0.05m)²
and the kinetic energy of the block at x = 0.05 m is;
K = ½mv₀²
Thus;
½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²
Inspecting this, ½m will cancel out to give;
ω²(0.08)² = ω²(0.05)² + v₀²
Making v₀ the subject, we have;
v₀ = ω√((0.08)² - (0.05)²)
So,
v₀ = 8.1√((0.08)² - (0.05)²)
v₀ = 0.5058 m/s
Which one of the following is closely related to the law of conservation of
energy, which states that energy can be transformed in different ways but can
never be created or destroyed?
O A. Charles's Law
B. Boyle's Law
C. Second law of thermodynamics
O D. First law of thermodynamics
Answer:
D
Explanation:
Answer:
It is D
Explanation: No cap
A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. Determine the angle θ for which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the bag.
Answer:
Dear user,
Answer to your query is provided below
The angle for which the rope will break θ = 41.8°
Explanation:
Explanation of the same is attached in image
A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. The angle θ for which the rope will break, is 41.81°
What is tension?The tension is a kind of force which acts on linear objects when subjected to pull.
The maximum tension Tmax =2W
From the work energy principle,
T₂ = 1/2 mv²
Total energy before and after pushing off
0+mglsinθ = 1/2 mv²
v² = 2gflsinθ..............(1)
From the equilibrium of forces, we have
T= ma +mgsinθ = mv²/l +mgsinθ
2mg = mv²/l +mgsinθ
2g = v²/l +gsinθ
Substitute the value of v² ,we get the expression for θ
θ = sin⁻¹(2/3)
θ =41.81°
Hence, the angle θ for which the rope will break, is 41.81°
Learn more about tension.
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The uniform dresser has a weight of 91 lb and rests on a tile floor for which μs = 0.25. If the man pushes on it in the horizontal direction θ = 0∘, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight of 151 lb , determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip.
Answer:
F = 22.75 lb
μ₁ = 0.15
Explanation:
The smallest force required to move the dresser must be equal to the force of friction between the man and the dresser. Therefore,
F = μR
F = μW
where,
F = Smallest force needed to move dresser = ?
μ = coefficient of static friction = 0.25
W = Weight of dresser = 91 lb
Therefore,
F = (0.25)(91 lb)
F = 22.75 lb
Now, for the coefficient of static friction between shoes and floor, we use the same formula but with the mas of the man:
F = μ₁W₁
where,
μ₁ = coefficient of static friction between shoes and floor
W₁ = Weight of man = 151 lb
Therefore,
22.75 lb = μ₁ (151 lb)
μ₁ = 22.75 lb/151 lb
μ₁ = 0.15
How have physicists played a role in history?
A. Physics has changed the course of the world.
B. History books are written by physicists.
C. Physicists have controlled most governments.
D. Most decisions about wars are made by physicists.
Answer:
A. Physics has changed the course of the world.
Explanation:
Identify the following as combination, decomposition, replacement, or ion exchange reactions NaBr(aq) + Cl2(g) → 2 NaCl(aq) + Br2(g)
Answer:
Replacement
Explanation:
in replacements, like ions replace like. in this equation, we can see that Bromine replaced Chlorine. so, the answer is replacement.
Answer:
Single-replacement or replacement
Explanation:
The single-replacement reaction is a + bc -> ac + b, compare them.
NaBr + Cl2 -> 2 NACl + Br.
AB + C -> AC + B
As you can see they are the same ( even though the b is with the a and not with the c. The formula can be switched around a little with the order of b and c ) ((also like ions replace like ions in replacements, which they are in this))
when a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. when the paper is turned 25 degree with respect to the field the flux through it is:
Answer:
22.66Nm²/C
Explanation:
Flux through an electric field is expressed as ϕ = EAcosθ
When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. If the paper is turned 25 degree with respect to the field the flux through it can be calculated using the formula.
From the formula above where:
EA = 25N.m^2/C
θ = 25°
ϕ = 25cos 25°
ϕ = 22.66Nm²/C
Your roommate is working on his bicycle and has the bike upside down. He spins the 68.0 cm -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. A. What is the pebble's speed? B. What is the pebble's acceleration?
Answer:
a. 6.41 m/s
b. 120.85 m/s^2
Explanation:
The computation is shown below:
a. Pebble speed is
As we know that according to the tangential speed,
[tex]v = r \times \omega[/tex]
[tex]= \frac{0.68}{2} \times 18.84[/tex]
= 6.41 m/s
The 18.84 come from
[tex]= 2 \times 3.14 \times 3[/tex]
= 18.84
b. The pebble acceleration is
[tex]a = \frac{v^2}{r}[/tex]
[tex]= \frac{6.41^2}{0.34}[/tex]
= 120.85 m/s^2
We simply applied the above formulas so that the pebble speed and the pebble acceleration could come and the same is to be considered
what are the strengths and weaknesses of the four methods of waste management?
Answer & Explanation: Waste management are all activities and actions required to manage waste from its inception to its final disposal. There are several methods of managing waste with its strengths and weaknesses. The strengths include;
* It creates employment
* It keeps the environment clean
* The practice is highly lucrative
* It saves the earth and conserves energy
The weaknesses of the methods of waste management includes;
* The sites are often dangerous
* The process is mostly
* There is a need for global buy-in
* The resultant product had a short life
An ideal, or Carnot, heat pump is used to heat a house to a temperature of 294 K (21 oC). How much work must the pump do to deliver 3000 J of heat into the house (a) on a day when the outdoor temperature is 273 K (0 oC) and (b) on another day when the outdoor temperature is 252 K (-21 oC)
Answer:
a) [tex]W_{in} = 214.286\,J[/tex], b) [tex]W_{in} = 428.571\,J[/tex]
Explanation:
a) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.
[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.
Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 273\,K[/tex]. The Coefficient of Performance is:
[tex]COP_{HP} = \frac{294\,K}{294\,K-273\,K}[/tex]
[tex]COP_{HP} = 14[/tex]
Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.
[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]
The input work to deliver a determined amount of heat to the house:
[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]
If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 14[/tex], the input work that is needed is:
[tex]W_{in} = \frac{3000\,J}{14}[/tex]
[tex]W_{in} = 214.286\,J[/tex]
b) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.
[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.
Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 252\,K[/tex]. The Coefficient of Performance is:
[tex]COP_{HP} = \frac{294\,K}{294\,K-252\,K}[/tex]
[tex]COP_{HP} = 7[/tex]
Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.
[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]
The input work to deliver a determined amount of heat to the house:
[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]
If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 7[/tex], the input work that is needed is:
[tex]W_{in} = \frac{3000\,J}{7}[/tex]
[tex]W_{in} = 428.571\,J[/tex]
A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flies an additional distance in an unknown direction, only to find himself at a small airstrip that his map shows to be 70.0 km directly north of his starting point.
a. What was the length of the third leg of his trip?b. What was the direction of the third leg of his trip?
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
Consider the double slit experiment for light. Complete each statement as it would apply to Young's experiment (for each statement select "Increases", "Decreases", or "Cannot be Predicted"). If a variable is not mentioned, consider it to remain unchanged.Required:a. If the distance to the screen decreases, fringe separation:_______?b. If the frequency of the light used increases, fringe separation:_______?c. If the wavelength of the light used decreases, fringe separation:_______?d. For the fringe separation to remain unchanged, wavelength__________ while the distance to the screen decreases.e. If slit separation decreases, fringe separation :_______?f. If slit separation decreases and the distance to the screen decreases, fringe separation :_______?g. If the distance to the screen triples and slit separation doubles, fringe separation :_______?
Answer:
a) DECREASE , b) Decreases , c) DECREASE , d) the wavelength must increase , e) increasses,
Explanation:
Young's double-slit experience is explained for constructive interference by the expression
d sin θ = m λ
as in this case, the measured angles are very small,
tan θ = y / L
tan θ = sin θ / cos θ = sin θ
sin θ= y L
d y / L = m Lam
we can now examine the statements given
a) if the distance to the screen decreases
y = m λ / d L
if L decreases and decreases.
The answer is DECREASE
b) if the frequency increases
the wave speed is
c = λ f
λ = c / f
we substitute
y = (m / d l) c / f
in this case if if the frequency is increased the separation decreases
Decreases
c) If the wavelength decreases
separation decreases
DECREASE
d) if it is desired that the separation does not change while the separation to the Panamanian decreases the wavelength must increase
y = (m / d) lam / L
e) if the parcionero between the slits (d) decreases the separation increases
INCREASES
f) t he gap separation decreases and the distance to the screen decreases so well.
Pattern separation remains constant
A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at 1.0 m/s. The mass of the high-jumper is 60.0 kg. What is the magnitude and direction of the impulse that the air mattress exerts on her
-- As she lands on the air mattress, her momentum is (m v)
Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down
-- As she leaves it after the bounce,
Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up
-- The impulse (change in momentum) is
Change = (60 kg-m/s up) - (300 kg-m/s down)
Magnitude of the change = 360 km-m/s
The direction of the change is up /\ .
The direction of a body or object's movement is defined by its velocity.In its basic form, speed is a scalar quantity.In essence, velocity is a vector quantity.It is the speed at which distance changes.It is the displacement change rate.
Solve the problem ?
Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is travelling along a path.In other words, velocity is a vector, whereas speed is a scalar value. We discuss the conceptive impulse in this puzzle.A high jumper weighing 60.0 kg sprints on the matrix at minus 6 meters per second in the downhill direction before falling to the mattress.her admirer.Speed drops to 0 meters/second.We must determine the impulse's size and presumed direction, which is upward and positive.The change in momentum is then equal to the impulse.The impulse therefore equals m times.the end velocity less the starting velocity.60.0kg times 0 minus minus 6 meters per second is the impulse, therefore.The impulse is 360 kilogram meters per second, or 360 newtons, to put it another way.The second is upward, and the direction.To learn more about magnitude refer
https://brainly.com/question/24468862
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An electron has a kinetic energy of 10.1 eV. The electron is incident upon a rectangular barrier of height 18.2 eV and width 1.00 nm. If the electron absorbed all the energy of a photon of green light (with wavelength 546 nm) at the instant it reached the barrier, by what factor would the electron's probability of tunneling through the barrier increase
Answer:
factor that the electron's probability of tunneling through the barrier increase 2.02029
Explanation:
given data
kinetic energy = 10.1 eV
height = 18.2 eV
width = 1.00 nm
wavelength = 546 nm
solution
we know that probability of tunneling is express as
probability of tunneling = [tex]e^{-2CL}[/tex] .................1
here C is = [tex]\frac{\sqrt{2m(U-E}}{h}[/tex]
here h is Planck's constant
c = [tex]\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-10.1) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}[/tex]
c = 2319130863.06
and proton have hf = [tex]\frac{hc}{\lambda } = {1240}{546}[/tex] = 2.27 ev
so electron K.E = 10.1 + 2.27
KE = 12.37 eV
so decay coefficient inside barrier is
c' = [tex]\frac{\sqrt{2m(U-E}}{h}[/tex]
c' = [tex]\frac{\sqrt{2\times 9.11 \times 10^{-31} (18.2-12.37) \times (1.60 \times 10^{-19}}}{6.626\times 10^{-34}}[/tex]
c' = 1967510340
so
the factor of incerease in transmisson probability is
probability = [tex]e^{2L(c-c')}[/tex]
probability = [tex]e^{2\times 1\times 10^{-9} \times (351620523.06)}[/tex]
factor probability = 2.02029