A block is attached to one end of a spring so that it can bounce back and forth on a horizontal frictionless surface. The block is pulled a distance of 0.150 m away from equilibrium and released. When the block is 9.00 cm away from equilibrium, it is traveling with a speed of 0.800 m/s. If the spring constant of the spring is 60.0 N/m, what is the mass of the block?
Answer:
m = 1.35 kg
Explanation:
In absence of friction, total mechanical energy must be conserved, so the sum of the initial elastic potential energy plus the initial kinetic energy, must be equal to the sum of the final kinetic energy plus the final elastic potential energy, as follows:[tex]U_{i} + K_{i} = U_{f} + K_{f} (1)[/tex]
As the block starts from rest, this means that Ki =0.We can express the elastic potential energy at any point , as follows:[tex]U = \frac{1}{2} * k * \Delta x^{2} (2)[/tex]
where k is the spring constant = 60.0 N/m, and Δx is the distance from
the equilibrium position.
Replacing Ui, Uf and Kf in (1), we have:[tex]\frac{1}{2} * k *\Delta x_{i} ^{2} =\frac{1}{2} * k *\Delta x_{f} ^{2} +\frac{1}{2} * m *v ^{2} (3)[/tex]
Replacing by the givens, and solving for m, we get:m =1.35 kg
what is the advantage of increasing the length of the handles of a wheelbarrow?
1. A stone of mass 0.8 kg is attached to a 0.9 m long string. The string will break if the tension exceeds 60 N. The stone is whirled in a horizontal circle on a frictionless tabletop while the other end of the string remains fixed. What is the maximum speed the stone can attain without breaking the string?
A. 8.22 m/s
B. 7.30 m/s
C. 9.34 m/s
D. 7.76 m/s
2. A highway curve with radius 900 ft is banked so that a car traveling at 55 mph will not skid sideways even in the absence of friction. At what angle should the curve be banked to prevent skidding?
A. 14.6°
B. 18.9°
C. 10.9°
D. 12.7°
3. A button will remain on a horizontal platform rotating at 40 rev/min as long as it is no more than 0.15 m from the axis. How far from the axis can the button be placed without slipping if the platform rotates at 60 rev/min?
A. 0.365 m
B. 0.338 m
C. 0.225 m
D. 0.294 m
4. What is the tension in a cord 10 m long if a mass of 5 kg is attached to it and is being spun around in a circle at a speed of 8 m/s?
A. 67 N
B. 28 N
C. 32 N
D. 50 N
5. A 0.5 kg mass attached to a string 2 m long is whirled around in a horizontal circle at a speed of 5 m/s. What is the centripetal acceleration of the mass?
A. 11.3 m/s2
B. 12.5 m/s2
C. 5.9 m/s2
D. 10.2 m/s2
6. Find the maximum speed with which a car can round a curve that has a radius of 80 m without slipping if the road is unbanked and the coefficient of friction between the road and the tires is 0.81.
A. 44.3 m/s
B. 20.8 m/s
C. 25.2 m/s
D. 30.6 m/s
Answer:
1. A. 8.22. m/s
Explanation:
Ms. Reitman's scooter starts from rest and the final velocity is
14 m/s in 7 seconds. What is the average acceleration?
Answer:
2 m/s 2
Explanation: Just is....
As the skiers travel down the slope a portion of their total energy is lost. This means that when they perform their tricks, they will never go as high as they were when they first pushed off from the gate. Describe how this energy is lost.
Answer:
Explained below
Explanation:
In skying down a hill, usually the skiers start at an elevated position and this means that they possess a large quantity of potential energy since they are in vertical position.
Now, as the skiers start to descend down the hill, they will lose potential energy while they gain kinetic energy since they are in motion. This is because there is reduction in height which results in a loss of potential energy and there is an increase in their speed which results in an increase in kinetic energy.
Now, immediately the skiers reaches the bottom of the hill, it means they are now at zero level height which means potential energy is now zero and it implies they have completely depleted the potential they had at the beginning at the top of the hill.
In contrast, at this zero level height, their speed and kinetic energy would have reached a maximum and this kinetic energy state will be maintained until they encounter a section of unpacked snow where they have to skid to a stop under force of friction. This friction force will carry out work on the skiers which will make their total mechanical energy to decrease. This means that as the force of friction keeps acting over an increasing distance, the quantity of work will therefore increase while the mechanical energy of the skiers will gradually be dissipated.
Eventually, the skiers will run out of energy and comes to a rest position and therefore they wouldn't be able to go as high as they first were before pushing off from the gate.
9)A skier starts from rest from the top of a 40 m high slope which makes 40 degrees with the ground. Coefficient of friction is 0.1 What is the velocity of the skier at the bottom of the ramp?
Answer:
The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.
Explanation:
We can determine the final velocity of the skier at the bottom of the ramp by Principle of Energy Conservation and Work-Energy Theorem, whose model is:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+W_{disp}[/tex] (1)
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.
[tex]W_{disp}[/tex] - Work dissipated by friction, measured in joules.
By definitions of gravitational potential and translational kinetic energy and work, we expand and simplify the model:
[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2}) =\mu_{k}\cdot N\cdot \Delta s[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the skier, measured in meters.
[tex]N[/tex] - Normal force from the incline on the skier, measured in newtons.
[tex]\Delta s[/tex] - Distance covered by the skier, measured in meters.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
The normal force exerted on the skier and the covered distance are, respectively:
[tex]N = m\cdot g\cdot \cos \theta[/tex] (3)
[tex]\Delta s = \frac{z_{1}-z_{2}}{\sin \theta}[/tex] (4)
Where [tex]\theta[/tex] is the angle of the incline above the horizontal, measured in sexagesimal degrees.
By applying (3) and (4) in (2), we get that:
[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot m\cdot g \cdot \cos \theta \cdot \left(\frac{z_{1}-z_{2}}{\sin \theta} \right)[/tex]
[tex]g\cdot (z_{1}-z_{2}) +\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})= \mu_{k}\cdot g \cdot \left(\frac{z_{1}-z_{2}}{\tan \theta} \right)[/tex] (5)
Then, we clear the velocity of the skier at the bottom of the ramp is: ([tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.1[/tex], [tex]\theta = 40^{\circ}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}-z_{2} = 40\,m[/tex])
[tex]\left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = \frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex]
[tex]2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = v_{1}^{2}-v_{2}^{2}[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}-2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2})}[/tex] (6)
[tex]v_{2} = \sqrt{\left(0\,\frac{m}{s} \right)^{2}-2\cdot \left(\frac{0.1}{\tan 40^{\circ}} -1\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m)}[/tex]
[tex]v_{2} \approx 26.288\,\frac{m}{s}[/tex]
The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.
A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. How far away is the cliff? The speed of sound in air a ordinary temperature is 343 m/s.
Answer:
480.2 m
Explanation:
The following data were obtained from the question:
Speed of sound (v) = 343 m/s.
Time (t) = 2.8 s
Distance (x) of the cliff =?
The distance of the cliff from the woman can be obtained as follow:
v = 2x /t
343 = 2x /2.8
Cross multiply
2x = 343 × 2.8
2x = 960.4
Divide both side by the coefficient of x i.e 2
x = 960.4/2
x = 480.2 m
Therefore, the cliff is 480.2 m away from the woman.
The distance should be 480.2 m
The calculation is as follows:Since A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. And, there is the velocity of 343 m/s
[tex]v = 2x \div t\\\\343 = 2x \div 2.8\\\\2x = 343 \times 2.8[/tex]
2x = 960.4
x = 480.2 m
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A submarine sends out a sonar signal (sound waves) in a direction directly downward it take 2.3 s for the sound waves to travel from the submarine to the ocean bottom and back to the submarine how high (approx) up from the ocean floor is the submarine?speed of sound in water is 1490 m/s
Explanation:
Using the formula;
2x = vt
x is the distance up from the ocean floor the submarine is
v is the speed of sound in water
t is the time
Given
t = 2.3s
v = 1490m/s
Required
how high (approx) up from the ocean floor is the submarine x
From the formula;
x = vt/2
x = 1490(2.3)/2
x = 745(2.3)
x = 1,713.5m
Hence the submarine is 1713.5m high up from the ocean floor
A radio wave transmits 38.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagation of the wave. Assuming the surface is a perfect absorber, calculate the radiation pressure on it.
Answer:
[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]
Explanation:
Given that,
A radio wave transmits 38.5 W/m² of power per unit area.
A flat surface of area A is perpendicular to the direction of propagation of the wave.
We need to find the radiation pressure on it. It is given by the formula as follows :
[tex]P=\dfrac{2I}{c}[/tex]
Where
c is speed of light
Putting all the values, we get :
[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]
So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].
An elastic conducting material is stretched into a circular loop of 13.6 cm radius. It is placed with its plane perpendicular to a uniform 0.871 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 73.9 cm/s. What emf is induced in volts in the loop at that instant?
Answer: The Answer is attached to the image below
A Ferris wheel has a diameter of 60 m and a period of rotation of 70 s. A passenger weighs 500 N. What is her apparent weight at the highest point of the Ferris wheel?
A. 298 N
B. 350 N
C. 487 N
D. 513 N
Answer:
A
Explanation:
Hopefully this helps.
You were chosen to co-pilot a mission to Mars. After successfully reaching a stable orbit, a piece of space-junk hits the Shuttle and you are sent out on a space walk with out a rope and only with a large roll of duct tape to repair the damage. You lose your grip during the walk and start to float away from the shuttle and realize you don’t have a safety line to grab. Should Mr. Wright call your parents to tell them you floated out into space or is it possible you can get back to the ship? Explain your answer. If possible, include a force diagram in your explanation. Hint: Think about newton’s laws.
Answer:
im just so focused on the fact that im going to mars :O
A ball is kicked horizontally at 4.6 m/s off of a cliff 13.4 m high. How far from the cliff will it land.
7.53 m
Explanation:We are given:
Initial Horizontal Velocity of the Ball = 4.6 m/s
Initial Vertical Velocity of the Ball = 0 m/s
Height from which ball is kicked = 13.4 m
Time taken by the ball to reach the ground:
The ball has an initial vertical velocity of 0 m/s
it also has a downward acceleration of 10 m/s² due to gravity
Solving for the time taken:
s = ut + 1/2(at²) [second equation of motion]
replacing the values
13.4 = (0)(t) + 1/2 (10)(t²)
13.4 = 5t²
t² = 13.4/5 [dividing both sides by 5]
t² = 2.68
t = 1.637 seconds [taking the square root of both sides]
Horizontal distance covered by the ball:
Since there are no horizontal opposing forces on the ball,
the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground
We calculated that the ball will hit the ground in 1.637 seconds
Distance covered:
s = ut + 1/2 (at²) [seconds equation of motion]
s = ut [since a = 0m/s² in the horizontal plane]
replacing the values
s = 4.6 * 1.637
s = 7.53 m
Hence, the ball landed 7.53 m from the cliff
A 4.0 kg mess kit sliding on a frictionless surface explodes into two 2.0 kg parts: 3.0 m/s, due north, and 5.0 m/s, 30° north of east.What is the original speed of the mess kit?
Answer:
3.49m/s
Explanation:
We have mess of kit = 4kg
Then pieces are 2kg
3m/s due north
5m/s, 30⁰ due south
MiVi = m1v1 + m2v2
We break velocities to have x and y components
MiVi = m1v1(cosθ+sin θ1j) + m2v2(v2cos θ2i+sin2 θj)
θ1 = 90⁰north
θ2 = 30⁰ north east
Vi = 2/4x3m/s(cos90i + sin90j)+2/4x5m/s(cos30i + sin30j)
= 2.16i + 2.75j
Vi = √Vx²+Vy²
Vi = √ (2.16)²+(2.75)²
Vi = 3.49m/s
This is the original speed of the kit
A 350 g mass on a 45-cm-long string is released at an angle of 4.5° from vertical. It has a damping constant of 0.010 kg/s. After 25 s, how many oscillations has it completed?
Answer:
The value is [tex]n = 18.5 \ oscillations[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 350 \ g = 0.350 \ kg[/tex]
The length is [tex]L = 45 \ cm = 0.45 \ m[/tex]
The angle is [tex]\theta = 4.5^o[/tex]
The damping constant is [tex]b = 0.010 \ kg/s[/tex]
The time taken is [tex]t = 25 \ s[/tex]
Generally the angular frequency of this damped oscillation is mathematically evaluated as
[tex]w = \sqrt{ \frac{ g}{L} + \frac{b^2}{4m^2} }[/tex]
=> [tex]w = \sqrt{ \frac{9.80 }{ 0.45} + \frac{0.010 ^2}{4* 0.350^2} }[/tex]
=> [tex]w = 4.667 \ s^{-1}[/tex]
Generally the period of the oscillation is mathematically represented as
[tex]T = \frac{2 \pi }{w}[/tex]
=> [tex]T = \frac{2 * 3.142 }{ 4.667 }[/tex]
=> [tex]T = 1.35 \ s[/tex]
Generally the number of oscillation is mathematically represented as
[tex]n = \frac{t}{T}[/tex]
=> [tex]n = \frac{25}{1.35}[/tex]
=> [tex]n = 18.5 \ oscillations[/tex]
In the days before scuba gear, some divers descended to underwater depths in diving bells, which are basically just upside-down containers whose open ends face down. The bell allows the person inside to breathe the air trapped inside it, observe underwater objects and marine life, and work under the water.
Required:
a. If the bell is submerged to a depth of 30m below sea level, what is the water pressure at the air-water interface inside the bell?
b. If the air pressure inside the bell before submersion into the water was 1 atm (101.825 kPa), what air pressure does the person experience at that depth?
Answer:
a) P = 4.03 10⁵ Pa, b) P = 4.03 10⁵ Pa
Explanation:
a) The pressure as a function of the depth of a fluid is
P =[tex]P_{atm}[/tex] + ρ g y
where Patm is the atmospheric pressure, the sea density of about 1025 kg / m³
let's calculate
P = 1.01825 10⁵ + 1025 9.8 30
P = 4.03 10⁵ Pa
b) When the hood is submerged, the water exerts a perpendicular force on the entire surface, in the equilibrium position, the air is compressed by this force until the pressure it exerts is equal to the external pressure (open at the lower), therefore the air pressure is
P = 4.03 105 Pa
"[tex]3.958\times 10^5 \ Pa[/tex]" would be the pressure outside the bell .as well as the air pressure inside it. A further explanation is below.
According to the question,
Depth,
[tex]d = 20 \ m[/tex]Pressure,
[tex]P_o = 1 \ atm[/tex][tex]= 1.018\times 10^{5} \ Pa[/tex]
Now,
→ The water pressure outside the bell will be:
= [tex]P_o +p\times g\times d[/tex]
By putting the values, we get
= [tex]1.018\times 10^5+1000\times 9.8\times 30[/tex]
= [tex]3.958\times 10^5 \ Pa[/tex]
And inside the air pressure will be same as the water pressure.
Thus the response above is right.
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Ball 1 (1.5 kg) moves to the right at 2 m/s and ball 2
(2.5 kg) moves to the left at 1.5 m/s. The balls stick together after collision. What is the speed and direction of ball 2 after the collision?
Answer:
0.1875 m/s leftward
Explanation:
Taking rightwards as positive
We are given:
Ball 1:
Mass (m1) = 1.5 kg
velocity (u1) = 2 m/s
Ball 2:
Mass (m2) = 2.5 kg
velocity (u2) = -1.5 m/s [negative because it is in the opposite direction]
Speed and Direction of Ball 2:
We are told that the balls stick together after the collision
We can say that the balls have the same velocity since they are sticking together
So, Final velocity of Ball 1 (v1) = Final velocity of Ball 2 (v2) = V m/s
According to the law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
replacing the variables
1.5(2) + (2.5)(-1.5) = V (1.5 + 2.5) [v1 = v2 = V]
3 + (-3.75) = 4V
-0.75 = 4V
V = -0.75/4 [dividing both sides by 4]
V = -0.1875 m/s
Hence, the balls will move at a velocity of 0.1875 m/s in the Leftward direction
The x and y coordinates of a particle at any time t are x = 5t - 3t2 and y = 5t respectively, where x and y are in meter and t in second. The speed of the particle at t = 1 second is
Answer:
[tex]v=\sqrt{26}~m/s[/tex]
Explanation:
Parametric Equation of the Velocity
Given the position of the particle at any time t as
[tex]r(t) = (x(t),y(t))[/tex]
The instantaneous velocity is the first derivative of the position:
[tex]v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))[/tex]
The speed can be calculated as the magnitude of the velocity:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
We are given the coordinates of the position of a particle as:
[tex]x=5t-3t^2[/tex]
[tex]y=5t[/tex]
The coordinates of the velocity are:
[tex]v_x(t)=(5t-3t^2)'=5-6t[/tex]
[tex]v_y(t)=(5t)'=5[/tex]
Evaluating at t=1 s:
[tex]v_x(1)=5-6(1)=-1[/tex]
[tex]v_y(1)=5[/tex]
The velocity is:
[tex]v=\sqrt{(-1)^2+5^2}[/tex]
[tex]v=\sqrt{1+25}[/tex]
[tex]\mathbf{v=\sqrt{26}~m/s}[/tex]
Please help me out, I need someone helpful enough.
How many sig figs are in 0.00357
How many sig figs are in 0.0009239
How many sig figs are in 0.0037930
How many sig figs are in 0.93938900
Answers:
Question 1: 3 sig figs
Question 2: 4 sig figs
Question 3: 5 sig figs
Question 4: 8 sig figs
The rule for sig figs:
Any number after a zero in a decimal is one sig fig.
Example: 0.00359, it has three sig figs.
Also, remember if there are zeros after a number greater than zero, it is also a sig fig as well.
Hope this helps you!
Answer:
1) 3 sf 2) 3 sf 3) 5 sf 4) 6 sf
Explanation:
00s at the start dont count towards the sig figs and 00s at the end dont either but 00s in the middle do
your welcome :)
A speaker creates uniformly spherical sounds w/ 500 watts of power
a)What is the intensity, I, of the sound at a distance of 20 meters from the speaker? What is the sound intensity level, β, of the sound at a distance of 20 meters from the speaker?
b) What is the intensity, I, of the sound at a distance of 10 meters from the speaker? What is the sound intensity level, β, of the sound at a distance of 10 meters from the speaker?
c) How many deciBels do you increase walking from 20 meters away from the speaker to 10 meters away from the speaker?
c) What power output would the speaker have to put out in order to create 100 dB at a distance 20 meters from the speaker?
Answer:
A) I = 0.09947 W , β = 109 db , B) β = 116 db , β = 116 db , c) Δβ = 7 dB,
D) P = 50.27 W
Explanation:
A) The intensity of a spherical sound wave is
I = P / A
where A is the area of the sphere where the sound is distributed
A = 4π R²
we substitute
I = P / 4πR²
let's calculate
I = 500 / (4π 20²)
I = 0.09947 W
to express this quantity in decibels we use relate
β = 10 log (I / I₀)
The detectivity threshold is I₀ = 1 10⁻¹² W / m²
β = 10 lob (0.09947 / 10⁻¹²)
β = 10 (10.9976)
β = 109 db
B) intensity at r = 10m
I = 500 / (4π 10²)
I = 0.3979 W / m²
β = 10 log (0.3979 / 10⁻¹²)
β = 10 (11.5997)
β = 116 db
C) the change in intensity in decibles is
Δβ = β₁ - β₂
Δβ = 116 - 109
Δβ = 7 dB
D) let's find the intensity for 100 db
I = I₀ 10 (β / 10)
I = 10⁻¹² 10 (100/10)
I = 10⁻² W / m²
Thus
P = I A
P = I 4π R²
P = 10⁻² 4π 20²
P = 50.27 W
Question 2 of 10
What is the current in the 10.0 2 resistor?
10.00
w
120.0 V
20.00
w
30.00
A. 10.0 A
B. 12.0 A
C. 0.0833 A
O D. 2.00 A
Answer:
2.00 A
Explanation:
Total resistance = 10 + 20 + 30 = 60 ohms
Potential difference (V) = 120volts
Current (I) =?
from Ohms law V = IR
==> I = V/R = 120/60 = 2 A
Note: for resistors in series equal amount of electric current flows through the circuit
7. Answer each of the following questions if the student applied a 550N force to a 100kg box on the same surface. What would be the magnitude of the box's acceleration?
Acceleration of an object is the force divided by its mass. Hence, acceleration of the 100 Kg box with an applied force of 550 N is 5.5 m/s².
What is acceleration?Acceleration is a physical quantity having both magnitude and direction. This vector measures the rate of change of velocity of a moving body. Thus, acceleration has the unit of m/s².
According to Newton's second law of motion, force acting on a body is the product of its mass and acceleration. Thus, force is directly proportional to the mass and acceleration of the body.
Given the mass of the box = 100 Kg
Force applied on it = 550 N.
Acceleration = 550 N / 100 Kg
= 5.5 m/s² ( 1 N = 1 Kg m/s²)
Therefore, the acceleration of the box is 5.5 m/s².
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3. The force exerted by gravity on kg = 98N
Answer:
960.4N
Explanation:
Given parameter:
Mass of the body = 98kg
Unknown:
Force exerted by gravity = ?
Solution:
Force exerted by gravity on the body is the weight
Weight = force x acceleration due to gravity
Acceleration due to gravity = 9.8m/s²
Weight = 98kg x 9.8m/s²
Weight = 960.4N
2. Which bicyclist was traveling the fastest at the end of the race?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, to determine the bicyclist that traveled the fastest at the end of the race, the speed of the bicyclists at the end of the race will determine this (not the bicyclist that came first nor there overall speed). The speed of the bicyclist at the end of the race can be determined by using the formula below
s = d ÷ t
Where s is the speed of each bicyclist at the end of the race
d is the specific distance covered by the bicyclist at the end of the race
t is the time taken for the bicyclist to complete that distance
It should be noted that to get an accurate result, the distance covered at the end of the race must be the same for all the bicyclists.
A 20.0-kg uniform plank (10.0 m long) is supported by the floor at one end and by a vertical rope at the other as shown in the figure. A 50.0-kg mass person stands on the plank a distance three-fourths of the length plank from the end on the floor.
Answer:
Tension= 475N
Force= 225N
Explanation:
The question is not complete, here is the complete question
Also, see attached a free body diagram for your reference
"A 20.0-kg uniform plank is supported by the floor at one end by a vertical rope at the other as shown in the figure. A 50.0-kg mass person stands on the plank a distance three-fourths of the length plank from the end on the floor.
a. What is the tension in the rope?
b. What is the magnitude of the force that the floor exerts on the plank?"
given data
mass of man=50kg
mass of plank=20kg
length of plank=10m
let us make the lenght of the rope be d
The torque about the floor
That is taking moment about the floor
[tex]N*0+T*d=20*10*d/2 + 50*10*3d/4\\\\T=100+375=475N\\\\[/tex]
Force will be also zero
[tex]N+T=20*10+50*10\\\\N+T=700 \\\\N=700-475=225 newtons\\\\N+T=20*10+50*10\\\\N+T=700\\\\N=700-475=225newtons[/tex]
Tornadoes are accompanied by spinning downdrafts and updrafts that form a funnel cloud. True or False
Answer: True
Explanation: I took the quiz and got it right! have a great day.
True, tornadoes are accompanied by spinning downdrafts and updrafts that form a funnel cloud.
What is a Tornado?
A tornado can be defined as a spinning air column, usually violet resulting from a thunderstorm in the cloud and extends to the ground.
These tornadoes are accompanied by spinning downdrafts and updrafts that form a funnel cloud.
The major constituent of tornadoes are:
Air, which are invisibleWater droplets DustDebrisThus, we can conclude that the given statement about tornadoes is true.
Learn more here:https://brainly.com/question/18935303
The 600-N ball shown is suspended on a string AB and rests against the frictionless vertical wall. The string makes an angle of 30° with the wall. The line AB goes through the center of the ball, and the contact point with the wall is at the same vertical height as the center of the ball. The ball presses against the wall with a force of magnitude:
Answer: T = 692.82 and 346.4 N
Explanation:
Given that;
w = 600 N
∅ = 30°
ΣFy = ma
a = 0 m/s²
ΣF = T(cos30°) - W = 0
T(cos30°) = W
we Divide both sides by cos30°
T = W / cos30o
T= 600N / cos30°
T = 692.82
and ∑fx
F = T sin∅
F = 692.82 × (sin30°)
F = 346.4 N
The equilibrium condition allows finding the result for the force of the ball against the wall is:
The force of the ball directed towards the wall is 346.4 N
Newton's second law gives a relationship between force, mass and acceleration of bodies. In the case where the acceleration is zero, it is called the equilibrium condition.
∑ F = 0
A free-body diagram is a diagram of the forces without the details of the bodies. In the attached we have a free-body diagram of the system.
Let's use trigonometry to break down stress.
sin 30 = [tex]\frac{T_x}{T}[/tex]
cos 30 = [tex]\frac{T_y}{T}[/tex]
T_y = T cos 30
Tₓ = T sin 30
Let's write the equilibrium condition for the system.
y-axis.
T_y -W = 0
T cos 30 = W
[tex]T = \frac{W}{cos 30}[/tex]
x-axis.
R - Tₓ = 0
R = T sin 30
We substitute
[tex]R = \frac{W}{cos 30} \ sin 30 \\R = W \ tan 30[/tex]
Let's calculate.
R = 600 tan 30
R = 346.4 N
This force is directed from the wall towards the ball, by Newton's third law the force of the ball is of equal magnitude and opposite direction, that is, directed towards the wall.
In conclusion with the equilibrium condition we can find the result for the force of the ball against the wall is:
The force of the ball directed towards the wall is 346.4 N
Learn more about the equilibrium condition here: brainly.com/question/18117041
waves disturb ____, but do not transmit it
a. energy
b. matter
c. sound
d. none of the above
Answer:
energy
Explanation:
im sure
Calculate the gauge pressure at a depth of 295 m in seawater
Answer:
P = 2893.95 [kPa]
Explanation:
The manometric pressure can be calculated by means of the following equation.
[tex]P=Ro*g*h[/tex]
where:
Ro = density of the water = 1000 [kg/m³]
g = gravity acceleration = 9.81 [m/s²]
h = deep = 295 [m]
Now replacing:
[tex]P = 1000*9.81*295\\P = 2893950 [Pa]\\P = 2893.95 [kPa][/tex]
What is the ratio of thicknesses of crown glass and water that would contain the same number of wavelengths of light?
Answer:
the thickness of the glass divided by thickness of water is going to be 1.333 divided by 1.52, which is 0.877. So, the height of this glass, in order to have the same number of wavelengths as in water, the height of the glass will be 0.877 times the height of the water, and so it will be smaller.