Answer:
the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N
Explanation:
Given that;
Length L = 10 m
mass of beam m_b = 150 kg; weight W_beam = 150×9.8
mass of sign m = 75 kg
distance of sign hung from the beam from the wall d = 2.50 m
angle ∅ = 60°
g = 9.8 m/s²
Now,
Torque acting at one end of the beam will be;
[tex]T_{net}[/tex] = Tsin∅ × L - mg(d)-W × (L/2)
for equilibrium, [tex]T_{net}[/tex] = 0
therefore, 0 = Tsin∅ × L - mg(d)-W × (L/2)
so we substitute
Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0
Tsin(60°) × 10 - 1837.5 - 7350 = 0
Tsin(60°) × 10 - 9187.5 = 0
Tsin(60°) × 10 = 9187.5
divide both side by 10
Tsin(60°) = 918.75
T × 0.8660 = 918.75
T = 918.75 / 0.8660
T = 1060.9 N
Therefore, the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N
The magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.
Torque acting at one end of the beam,
= Tsin∅ × L - mg(d)-W × (L/2)
When equilibrium = 0
Tsin∅ × L - mg(d)-W × (L/2) = 0
Where,
L - Length = 10 m
m - mass of sign bord= 75 kg
g- gravitational accelaration = 9.8 m/s²
W - weight of beam = 150×9.8 = 1470 kg
Put the values in the formula,
Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0
Tsin(60°) × 10 - 1837.5 - 7350 = 0
Tsin(60°) × 10 - 9187.5 = 0
Tsin(60°) × 10 = 9187.5
Tsin(60°) = 918.75
T × 0.8660 = 918.75
T = 918.75 / 0.8660
T = 1060.9 N
Therefore, the magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.
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A 0.38 kg drinking glass is filled with a hot liquid. The liquid transfers 7032 J of energy to the glass. If the
temperature of the glass increases by 22 K, what is the specific heat of the glass?
Answer:
841 J/kg.K
Explanation:
The computation of the specific hear of the glass is shown below:
As we know that
E= cmΔt
where
c denotes specific heat
m denotes 0.38 kg
Δt = temperature = 22k
E denotes energy = 7032 J
Now
7032 J = (0.38) (22) (c)
7032 J = 8.36 (c)
So C = 7032 J ÷ 8.36
= 841 J/kg.K
a ball is thrown upward with a beginning speed of 40m/s. The graph below shows how the speed of the ball changes until it reaches its maximum height.
use the graph to find
a) the time when the ball reaches its maximum height
b) the acceleration of the ball
c) the maximum height the ball went
Answer:
a) 4.0816s
b) -9.8 ms^-1
c) 81.63265m
A bird lands on a bird feeder which is connected to a spring. The mass of the bird is exactly the same as the mass of the bird feeder. How does the added mass affect the period of oscillation of the bird feeder?
Answer:
The added mass will mean a longer period of oscillation.
Explanation:
The period of oscillation here is given by the formula;
T = 2π√(m/k)
Where m is mass and k is spring constant
From the equation of oscillation period above, it's obvious that when we increase the mass, the oscillation period will also increase.
Thus, the added mass will mean a longer period of oscillation.
Connective Tissue in a tendon is
According to Coulomb's Law, if the distance between two charged particles is doubled, the electric force will be _________. *
Answer: reduced by 1/4
Explanation:
The force will be reduced by 1/4. Try plugging in 2r, then squaring it. You will get 4r^2, which is essentially dividing the force by 4
A train 350 m long is moving on a straight track with a speed of 84.1 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 15.8 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.
Answer:
t = 25.0 s
Explanation:
Assuming that the engineer applies the brakes just over the crossing, the train moves exactly 350 m at a constant acceleration, with a final speed (when the last car of the train leaves the crossing) of 15.8km/h.Since we know the initial and final speeds, and the horizontal distance traveled (the length of the train) we can use the following kinematic equation to get the acceleration:[tex]v_{f}^{2} - v_{o}^{2} = 2*a* \Delta x (1)[/tex]
Since we need to find the time in seconds, it is advisable to convert vf and vo to m/s first, as follows:[tex]v_{o} = 84.1 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 23.4 m/s (2)[/tex]
[tex]v_{f} = 15.8 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 4.4 m/s (3)[/tex]
Replacing (2) and (3) in (1), since Δx =350m, we can solving for a:[tex]a = \frac{(4.4m/s)^{2} - (23.4m/s)^{2}}{2*350m} = -0.76 m/s2 (4)[/tex]
In order to get the time, we can simply use the definition of acceleration, and rearrange terms:[tex]t =\frac{v_{f}-v_{o}}{a} = \frac{(4.4m/s)-(23.4m/s)}{-0.76m/s2} = 25.0 s (5)[/tex]
Suppose two skiers (A and B) are racing. Assume a frictionless surface! They start from the top of a mountain at the same time, and glide down to the flat area below. Just before the finish line there is a ditch. The skiers can either go down into the ditch or take a flat bridge over the ditch. Both the bridge and the ditch are covered with frictionless snow. Skier A decides to go down into the ditch. Skier B decides to go over the bridge. . Which skier gets to the finish line first, or do they arrive at the same time?
a. Skier A (ditch) arrives first
b. Skier B (bridge) arrives first
c. The skiers arrive at the same time
d. Neither skier arrives at the finish line
Answer:
b. Skier B (bridge) arrives first
Explanation:
This is because, skier B continues along the bring with the same velocity he started with before moving over the bridge and since the bridge is frictionless, he losses no kinetic energy and his speed is constant.
Whereas, skier A losses kinetic energy as he goes into the ditch. This is due to his change in potential energy. He thus emerges from the ditch with lesser kinetic energy than skier B and thus a slower speed.
Therefore, skier B arrives first since he moves at a constant speed.
Donuts are a simple or complex carbohydrate? *
Ok
Answer:
simple carbohydrate
Explanation:
Simple carbs are like cakes, donuts, and candy, etc.
Hope this helps :D
A car traveling 85 km/h is 250 m behind a truck
traveling 73 km/h.
Time needed = t = 20.83 s
Further explanationGiven
car speed = 85 km/h
truck speed = 73 km/h
Required
the time it takes for the car to reach the truck
Solution
When the car reaches the truck, the distance between them will be the same
x car - 250 m = x truck
General formula for distance (d) :
d = v.t
So the equation becomes :
85t-250 = 73t
12t=250
t = 20.83 s
What are 3 things you could you do this week to help you connect better with kids in
your classes?
Answer:
In my physics class, something that helps connect better with kids is keeping connected with them always make sure to ask if they understand what you're teaching if they are following because sometimes most kids are to afraid to admit that they are lost, another way to connect with kids is maybe posting surveys to be able to check in with each student especially during this hard times :)
Explanation:
1. start with fun activities.
2. Encourage single-tasking.
3. Designate a learning playing field.
A baseball player hits a 0.15 kg 0.15kg0, point, 15, start text, k, g, end text baseball that is initially at rest, changing its momentum by 11 kg ⋅ m s 11 s kg⋅m 11, start fraction, start text, k, g, end text, dot, start text, m, end text, divided by, start text, s, end text, end fraction.
Answer:
73.3m/s
Explanation:
We can find the velocity of the player.
Momentum = mass * velocity
Given
Mass = 0.15kg
Momentum = 11kgm/s
Get the velocity
Velocity = Momentum/Mass
Velocity = 11/0.15
Velocity = 73.3m/s
Hence the velocity of the player is 73.3m/s
Choose the best explanation from among the following:_________.
1. Charge is conserved, and therefore the mass of the object will remain the same.
2. A positive charge increases an object's mass; a negative charge decreases its mass.
3. To give the object a negative charge we must give it more electrons, and this will increase its mass.
Answer: 3. To give the object a negative charge we must give it more electrons, and this will increase its mass.
Explanation:
Suppose we have an object and we negatively charge it.
Then we are "adding" N electrons to the object.
Remember that the mass of an electron is:
m = 9.11*10^(-31) kg
Then if we add N electrons to an object of mass M, the new mass of the object will be:
Mass = M + N*9.11*10^(-31) kg
So we will have an (almost negligible) increase of the mass of the object.
(Something similar can happen if the object is positively charged, where we remove electrons, then the mass of the object decreases)
Then the correct option is:
3. To give the object a negative charge we must give it more electrons, and this will increase its mass.
A runner completes the 200-meter dash with a time of 19.80 seconds. What was the runner's average speed in miles per hour?
Answer:
v = 22.54 mph.
Explanation:
Given that,
Distance moved, d = 200 m
Time, t = 19.8 s
We need to find the runner's average speed.
We know that,
1 mile = 1609.34 m
200 m = 0.124 miles
19.8 seconds = 0.0055 h
So,
Speed = distance/time
[tex]v=\dfrac{0.124}{0.0055}\\\\v=22.54\ mph[/tex]
So, the runner's average speed is 22.54 mph.
The "problem of perception" is best characterized as?
Answer:
making sense of a 3-d world from 2-d data
Explanation:
What is measurement ?
A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the $y$-axis, then hit and bounce off the $x$-axis, then hit the point $(7,5)$. What is the total distance the beam will travel along this path
Answer:
4units
Explanation:
To calculate the total distance the beam will travel along this path, you will use the formula for calculating the distance between two coordinates expressed as;
D = √(x2-x1)²+(y2-y1)²
Given the coordinate points
(3,5) and (7,5)
Substitute
D = √(7-3)²+(5-5)²
D = √(7-3)²+0²
D = √4²
D = √16
D = 4
Hence the total distance the beam will travel along this path is 4units
Carousel conveyors are used for storage and order picking for small parts. The conveyorsrotate clockwise or counterclockwise, as necessary, to position storage bins at the storageand retrieval point. The conveyors are closely spaced, such that the operators travel timebetween conveyors is negligible. The conveyor rotation time for each item equals 1 minute;the time required for the operator to retrieve an item after the conveyor stops rotatingequals 0.25 minute. How many carousel conveyors can one operator tend without creatingidle time on the part of the conveyors
Answer:
the number of carousel conveyors that an operator can operate without any idle time is 5
Explanation:
Given the data in the question;
first we express the equation for number of carousel conveyors that can be operated by an operator;
n' = [tex]\frac{(a + t)}{( a + b)}[/tex]
where a is the concurrent activity time ( 0.25 minute )
b is the independent operator activity time
t is the independent machine activity time( 1 )
Now independent activity time is zero as the operator is not performing any inspection or packaging tasks.
So time taken for the operator to retrieve the finished item at the end of the process is the concurrent activity and independent machine activity time, the conveyor rotation time of each item
so
we substitute
0.25min for a, 1 for t and 0min for b
n' = [tex]\frac{(0.25min + 1min)}{( 0.25min+ 0 min)}[/tex]
n' = 1.25 min / 0.25
n' - 5
Therefore, the number of carousel conveyors that an operator can operate without any idle time is 5
For each scenario below, choose the best graph.
(a) Maria bikes from home to work.
A line graph that plots either speed or distance against time.
Really, the type of graph depends on what quantity you want the graph to show.
What is the motion and arrangement of molecules in a liquid
Answer:
particles
Explanation:
in liquids, particle are close together
A car travels 100 km due East in 2 hours. It then travels 50 km South in 1 hour. What is its average velocity?
The average velocity of the car is 37.27 km/h.
The given parameters;
Initial displacement of the car, x = 100 kmTime of motion, t = 2 hoursFinal displacement of the car, y = 50 kmtime of motion, t = 1 hourThe average velocity of the car is calculated as follows;
[tex]average \ velocity = \frac{Total \ displacement}{Total \ time}[/tex]
The total displacement of the car is calculated as follows;
[tex]d = \sqrt{x^2 + y^2} \\\\d = \sqrt{100^2 \ + \ 50^2} \\\\d = 111.803 \ km[/tex]
The average velocity of the car is calculated as follows;
[tex]v = \frac{111.803}{3} \\\\v = 37.27 \ km/h[/tex]
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An electron has a mass of 9.1x10-31 kg. What is its momentum if it is travelling at a speed of 3.5x106 m/s?
Answer:
3.19*10^-24
Explanation:
the equation to find momentum is p=mv so you just multiply the mass times velocity
An electron has a mass of 9.1x10⁻³¹ kilograms. if it is traveling at a speed of 3.5x10⁶ meters/seconds then its momentum would be 3.185 ×10⁻²⁴ kgm/s.
What is momentum?It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.
As given in the problem an electron has a mass of 9.1x10⁻³¹ kilograms. if it is traveling at a speed of 3.5x10⁶ meters/seconds then its momentum would be
the momentum of the electrons = mass of the electron×velocity of the electron
= 9.1x10⁻³¹ × 3.5x10⁶
= 3.185 ×10⁻²⁴ kgm/s
Thus, the momentum of the electron would be 3.185 ×10⁻²⁴ kgm/s.
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Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N/C. What is the kinetic energy of the proton at the end of the motion
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, [tex]v_p_i[/tex] = 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
[tex]W =K.E_f - K.E_i[/tex]
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
[tex]K.E_f =EQd + \frac{1}{2}m_pv_p_i^2[/tex]
[tex]m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C[/tex]
[tex]K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5 \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\[/tex]
[tex]K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J[/tex]
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
A 45.0-kg girl stands on a 13.0-kg wagon holding two 18.0-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 6.5 m/s relative to herself . Assuming that the wagon was at rest initially, what is the speed of the girl relative to the ground after she throws both weights at the same time
Answer:
v = 4.0 m/s
Explanation:
Assuming no external forces acting during the instant that the girl throws the weights, total momentum must be conserved.Since all the masses at rest initially, the initial momentum must be zero.So, due to momentum must keep constant, final momentum must be zero too, as follows:[tex]p_{f} = m_{w} * v_{w} + m_{g+w} *v_{g+w} = 0 (1)[/tex]
Assuming the direction towards the back of the wagon as positive, and replacing the masses in (1), we can solve for vg, as follows:[tex]v_{g+w} =- \frac{m_{w} *v_{w}}{m_{g+w} } = - \frac{36.0kg *6.5m/s}{58.0kg } = -4.0m/s (2)[/tex]
This means that the girl (along with the wagon on she is standing) will move at a speed of 4.0 m/s in an opposite direction to the one she threw the weights.At what speed, in m/s, would a moving clock lose 1.3ns in 1.0 day according to experimenters on the ground?
Answer:
v=0.14c
Explanation:
A bottle of water at a room temperature of 21.0 C is placed into a refrigerator
with an air temperature of 4.5C. The thermal energy will move — *
A. from the cooler air to lower the temperature of the water to 4.5 C
B. in both directions until the temperature is equal in the water and the air
C. from the water to the air until the water temperature is zero degrees Celsius
O D. from the water to the air until the temperature is equal in both
Answer:
B. in both directions until the temperature is equal in the water and the air
Explanation:
When a warm body is in contact with a cool body , there is exchange of heat energy in both sides until there is attainment of equilibrium temperature . At this temperature both the body attains equal temperature . Initially rate of heat radiated by warm body is more than that from cool body , but after attainment of equilibrium , the rate becomes equal to each other . This is called dynamic equilibrium .
Hence option B is correct .
What genetic test would you get if there was a specific genetic disease in your family
A friend comments to you that there was a beautiful, thin sliver of a Moon visible in the early morning just before sunrise. Which phase of the Moon would this be, and in what direction would you look to see the Moon (in the southern sky, on the eastern horizon, on the western horizon, high in the sky, etc.)?
Answer: Waning Crescent
Explanation:
A plastic rod 1.6 m long is rubbed all over with wool, and acquires a charge of -9e-08 coulombs. We choose the center of the rod to be the origin of our coordinate system, with the x-axis extending to the right, the y-axis extending up, and the z-axis out of the page. In order to calculate the electric field at location A = < 0.7, 0, 0 > m, we divide the rod into 8 pieces, and approximate each piece as a point charge located at the center of the piece.
Solution :
Length of the plastic rod , L = 1.6 m
Total charge on the plastic rod , Q = [tex]$-9 \times 10^{-8}$[/tex] C
The rod is divided into 8 pieces.
a). The length of the 8 pieces is , [tex]$l=\frac{L}{8}$[/tex]
[tex]$=\frac{1.6}{8}$[/tex]
= 0.2 m
b). Location of the center of the piece number 5 is given as : 0 m, -0.09375 m, 0 m.
c). The charge q on the piece number 5 is given as
[tex]$q=\frac{Q}{L}\times l$[/tex]
[tex]$q=\frac{-9 \times 10^{-8}}{1.6}\times0.2$[/tex]
= [tex]$-1.125 \times 10^{-8}$[/tex] C
d). WE approximate that piece 5 as a point charge and we need to find out the field at point A(0.7 m, 0, 0) only due to the charge.
We know, the Coulombs force constant, k = [tex]$8.99 \times 10^9 \ N.m^2/C^2$[/tex]
So the X component of the electric field at the point A is given as
[tex]$E_x = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \cos \frac{187.628}{0.70625}$[/tex]
= -126.15 N/C
The Y component of the electric field at the point A is
[tex]$E_y = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \sin \frac{187.628}{0.70625}$[/tex]
= -16.93 N/C
Now since the rod and the point A is in the x - y plane, the z component of the field at point A due to the piece 5 will be zero.
∴ [tex]$E_z=0$[/tex]
Thus, [tex]$E= <-126.15,-16.93,0>$[/tex]
Elizabeth has always believed that people's thoughts can help heal them. She wants to help people use positive thinking to positively affect their
illnesses. What type of psychology would be MOST appropriate for Elizabeth to study?
Answer: Family
Explanation:
A rock thrown vertically upward from the surface of the moon at a velocity of 32m/sec reaches a height of sequals32 t minus 0.8 t squaredmeters in t sec.a. Find the rock's velocity and acceleration at time t.b. How long does it take the rock to reach its highest point?c. How high does the rock go?d. How long does it take the rock to reach half its maximum height?e. How long is the rock aloft?
Answer:
A) v(t) = (32 - 1.6t) m/s
a(t) = -1.6 m/s²
B) t = 20 seconds
C) 320 m
D) t = 5.85 seconds going up and 34.14 seconds going down
E) 40 seconds
Explanation:
Height is given by the equation;
S(t) = 32t - 0.8t²
A) Velocity after time t is gotten from first derivative of the distance.
Thus;
v(t) = dS/dt = 32 - 1.6t
Acceleration at time t is gotten from derivative of the velocity.
Thus;
a(t) = d²S/dt² = -1.6 m/s²
B) At highest point, velocity is zero.
Thus;
32 - 1.6t = 0
1.6t = 32
t = 32/1.6
t = 20 seconds
C) To find how high the rock goes, it means we are looking for maximum height.
This will be at t = 20 seconds.
Thus;
S(20) = 32(20) - 0.8(20)²
S(20) = 640 - 320
S(20) = 320 m
D) we want to find the time it will take to reach half its maximum height.
since maximum height is 320 m, then half the maximum height is; S_½ = 320/2 = 160
Thus;
160 = 32t - 0.8t²
0.8t² - 32t + 160 = 0
Using quadratic formula, we will get;
t = 5.85 seconds going up and 34.14 seconds going down
E) time the rock is aloft = twice the time it took to reach maximum height.
Thus; t_aloft = 2 × 20 = 40 seconds