The magnitude of the tension T in the cable is approximately 468.164 N.
To find the tension T in the cable supporting the beam, we can analyze the forces acting on the beam in equilibrium.
First, let's consider the forces acting on the beam:
1. The weight of the beam, acting downward at its center (2.0 m from the wall).
2. The tension T in the cable, acting upward and at an angle of 53 degrees with the horizontal.
3. The reaction force at the pin connection, acting horizontally to the left.
4. The vertical force exerted by the person standing on the beam, which contributes to the vertical component of the tension in the cable.
The beam is in equilibrium, so the sum of the torques about any point must be zero. Let's take moments about the pin connection at the wall.
Clockwise torques:
- Weight of the beam: [tex]\(200 \, \text{N} \times 2.0 \, \text{m}\)[/tex]
- Vertical component of the tension: [tex]\(T \times \cos(53^\circ) \times 1.5 \, \text{m}\)[/tex]
Counterclockwise torques:
- Tension in the cable: [tex]\(T \times \sin(53^\circ) \times 1.5 \, \text{m}\)[/tex]
- Vertical force exerted by the person: [tex]\(350 \, \text{N} \times 1.5 \, \text{m}\)[/tex]
Since the beam is in equilibrium, these torques must balance:
[tex]\[ 200 \, \text{N} \times 2.0 \, \text{m} + T \times \cos(53^\circ) \times 1.5 \, \text{m} = T \times \sin(53^\circ) \times 1.5 \, \text{m} + 350 \, \text{N} \times 1.5 \, \text{m} \][/tex]
Now, we can solve for T:
[tex]\[ 400 \, \text{N} + T \times \cos(53^\circ) \times 1.5 \, \text{m} = T \times \sin(53^\circ) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]
[tex]\[ 400 \, \text{N} = T \times \sin(53^\circ) \times 1.5 \, \text{m} - T \times \cos(53^\circ) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]
[tex]\[ 400 \, \text{N} = T \times (\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m} + 525 \, \text{N} \][/tex]
[tex]\[ T \times (\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m} = 125 \, \text{N} \][/tex]
[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]
Now, we can calculate T:
[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T = \frac{125 \, \text{N}}{(\sin(53^\circ) - \cos(53^\circ)) \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T \approx \frac{125 \, \text{N}}{(0.7986 - 0.6206) \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T \approx \frac{125 \, \text{N}}{0.178 \times 1.5 \, \text{m}} \][/tex]
[tex]\[ T \approx \frac{125 \, \text{N}}{0.267 \, \text{m}} \][/tex]
[tex]\[ T \approx 468.164 \, \text{N} \][/tex]
So, the magnitude of the tension T in the cable is approximately 468.164 N.
Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series with the galvanometer is 1680 Ω for the 20.0-V scale and 2930 Ω for the 30.0-V scale. Determine the coil resistance.
Answer:
Resistance of the circuit is 820 Ω
Explanation:
Given:
Two galvanometer resistance are given along with its voltages.
Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.
⇒ [tex]V=20\ \Omega,\ V_1=30\ \Omega[/tex]
⇒ [tex]G=1680\ \Omega,\ G_1=2930\ \Omega[/tex]
Concept to be used:
Conversion of galvanometer into voltmeter.
Let [tex]G[/tex] be the resistance of the galvanometer and [tex]I_g[/tex] the maximum deflection in the galvanometer.
To measure maximum voltage resistance [tex]R[/tex] is connected in series .
So,
⇒ [tex]V=I_g(R+G)[/tex]
We have to find the value of [tex]R[/tex] we know that in series circuit current are same.
For [tex]G=1680[/tex] For [tex]G_1=2930[/tex]
⇒ [tex]I_g=\frac{V}{R+G}[/tex] equation (i) ⇒ [tex]I_g=\frac{V_1}{R+G_1}[/tex] equation (ii)
Equating both the above equations:
⇒ [tex]\frac{V}{R+G} = \frac{V_1}{R+G_1}[/tex]
⇒ [tex]V(R+ G_1) = V_1 (R+G)[/tex]
⇒ [tex]VR+VG_1 = V_1R+V_1G[/tex]
⇒ [tex]VR-V_1R = V_1G-VG_1[/tex]
⇒ [tex]R(V-V_1) = V_1G-VG_1[/tex]
⇒ [tex]R =\frac{V_1G-VG_1}{(V-V_1)}[/tex]
⇒ Plugging the values.
⇒ [tex]R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}[/tex]
⇒ [tex]R =\frac{(50400 - 58600)}{(-10)}[/tex]
⇒ [tex]R=\frac{-8200}{-10}[/tex]
⇒ [tex]R=820\ \Omega[/tex]
The coil resistance of the circuit is 820 Ω .
A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/10. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her
Answer:
(a) 1058.4 J
(b) -10584 J
Explanation:
Parameters given:
Mass of astronaut, m = 72 kg
Distance moved by astronaut, d = 15 m
(a) WORK DONE BY FORCE FROM THE HELICOPTER
Work done is given as the product of Force applied to a body and the distance moved by the body:
W = F * d
The force from the helicopter is given as:
F = m * a
where a = acceleration of the astronaut due to the helicopter
Therefore, the work done is given as:
W = m * a * d
W = 72 * g/10 * 15
W = [tex]\frac{72 * 9.8 * 15}{10}[/tex]
W = 1058.4 J
(b) WORK DONE BY FORCE OF GRAVITY
W = F * d
The force of gravity is given as:
F = -m * g
where g = acceleration due to gravity
The negative sign is due to the fact that the astronaut moves in an opposite direction (upwards) to the force of gravity (Gravity acts downwards)
Therefore, the work done is given as:
W = -m * g * d
W = -72 * 9.8 * 15
W = -10584 J
A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m.What is the magnitude ofthe normal force exerted on the car by the walls of the circle at pointA?
Answer:
61.76 N.
Explanation:
Given the mass of the car, m = 1.60 kg.
The speed of the car, v = 12.0 m/s.
The radius of the circle, r = 5 m.
As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.
Let N is the normal force.
So, [tex]N - mg = F_c[/tex]
[tex]N-mg=\frac{mv^2}{r}[/tex]
Now substitute the given values, we get
[tex]N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}[/tex]
[tex]N=15.68+46.08[/tex]
N = 61.76 N.
Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.
The magnitude of the normal force exerted on the car by the walls of the circle at point A can be calculated using the concept of centripetal force. In this case, the magnitude of the normal force is equal to the weight of the car, which is the product of its mass and the acceleration due to gravity.
Explanation:The magnitude of the normal force exerted on the car by the walls of the circle at point A can be calculated using the concept of centripetal force. In a vertical circle, the net external force equals the necessary centripetal force. The only two external forces acting on the car are its weight and the normal force of the road. Since the car is not leaving the surface and the net vertical force must be zero, the vertical components of the two external forces must balance each other.
The vertical component of the normal force is equal to the car's weight, which is mg. The weight can be calculated by multiplying the mass of the car by the acceleration due to gravity. In this case, the weight is 1.60 kg multiplied by 9.8 m/s². Therefore, the magnitude of the normal force exerted on the car at point A is also 1.60 kg multiplied by 9.8 m/s².
Learn more about Centripetal Force here:https://brainly.com/question/11324711
#SPJ3
Which element is a metal? A.lithium B.Nitrogen C.sulfur D.chlorine
Answer:
the answer is b
Explanation:
hope it helps ...
Lithium (Li) is a metal located in Group 1 of the periodic table and is known as an alkali metal. It stands out for its low density and capacity to conduct electricity, along with exhibiting the typical properties of metals.
The element lithium (Li) is a metal, and notably, it is the first metal in the periodic table located in Group 1. It has the lowest atomic number (3) among the metals. The defining characteristics of metals include their ability to conduct electricity, possess malleability, and often exhibit a shiny luster. Lithium, being part of the alkali metals category, shares these properties. Furthermore, lithium is the least dense metal and has a unique electronic structure where the third electron occupies an outer shell, differentiating it from nonmetals.
Lithium is used in various applications including batteries and the aerospace industry due to its low density and high reactivity. The chemically similar elements to lithium include other alkali metals such as sodium and potassium, which also share similar chemical properties.
when a cup of hot chocolate cools from 90c to 80c which of the following is happening to the molecules of the liquid
Answer:
I think that the liquids molecules are slowing down. Hope this helps!
Explanation:
Please vote me Brainliest
The gravitational potential energy of an object is equal to its weight multiplied by its
Answer:
Height above a surface
Explanation:
Gravitational potential energy is the energy which an object possesses due to its position above a surface.
It is also the amount of work a force has to do in order to bring an object from a particular position to a point of reference.
It is given mathematically as:
P. E. = m*g*h
where m = mass of the body
g = acceleration due to gravity
h = height above a surface
m*g represents the weight of the object.
Hence, Gravitational potential energy is the product of an object's weight and its height above a surface/reference point.
1) what is a calorie?
You are coasting on your 12-kg bicycle at 13 m/s and a 5.0-g bug splatters on your helmet. The bug was initially moving at 1.5 m/s in the same direction as you. If your mass is 70 kg, answer the following questions:a.What is the initial momentum of you plus your bicycle? b.What is the intial momentum of the bug? c.What is your change in velocity due to the collision the bug? d.What would the change in velocity have been if the bug were traveling in the opposite direction?
Answer:
a) Pi,c = 1066 kgm/s
b) Pi,b = 0.0075 kgm/s
c) ΔV = - 0.0007 m/s
d) ΔV = - 0.0008 m/s
Explanation:
Given:-
- The mass of the bicycle, mc = 12 kg
- The mass of passenger, mp = 70 kg
- The mass of the bug, mb = 5.0 g
- The initial speed of the bicycle, vpi = 13 m/s
- The initial speed of the bug, vbi = 1.5 m/s
Find:-
a.What is the initial momentum of you plus your bicycle?
b.What is the initial momentum of the bug?
c.What is your change in velocity due to the collision the bug?
d.What would the change in velocity have been if the bug were traveling in the opposite direction?
Solution:-
- First we will set our one dimensional coordinate system, taking right to be positive in the direction of bicycle.
- The initial linear momentum (Pi,c) of the passenger and the bicycle would be:
Pi,c = vpi* ( mc + mp)
Pi,c = 13* ( 12+ 70 )
Pi,c = 1066 kgm/s
- The initial linear momentum (Pi,b) of the bug would be:
Pi,b = vbi*mb
Pi,b = 0.005*1.5
Pi,b = 0.0075 kgm/s
- We will consider the bicycle, the passenger and the bug as a system in isolation on which no external unbalanced forces are acting. This validates the use of linear conservation of momentum.
- The bicycle, passenger and bug all travel in the (+x) direction after the bug splatters on the helmet.
Pi = Pf
Pi,c + Pi,b = V*(mb + mc + mp)
Where, V : The velocity of the (bicycle, passenger and bug) after collision.
1066 + 0.0075 = V*( 0.005 + 12 + 70 )
V = 1066.0075 / 82.005
V = 12.9993 m/s
- The change in velocity is Δv = 13 - 12.9993 = - 0.00070 m/s
- If the bug travels in the opposite direction then the sign of the initial momentum of the bug changes from (+) to (-).
- We will apply the linear conservation of momentum similarly.
Pi = Pf
Pi,c + Pi,b = V*(mb + mc + mp)
1066 - 0.0075 = V*( 0.005 + 12 + 70 )
V = 1065.9925 / 82.005
V = 12.99911 m/s
- The change in velocity is Δv = 13 - 12.99911 = -0.00088 m/s
Answer:
a. The initial momentum of you and your bicycle is 1066 kgm/s.
b. The initial momentum of the bug is 0.0075 kgm/s.
c. The change in velocity due to the collision with the bug is -0.0008 m/s.
d. If the bug were travelling in the opposite direction, the change in velocity due to the collision would have been -0.0009 m/s.
Explanation:
The initial momentum of you and your bicycle can be easily calculated using the definition of momentum:
[tex]p=mv\\\\p=(m_{you}+m_{bicycle})v\\\\p=(70kg+12kg)(13m/s)\\\\p=1066kgm/s[/tex]
So the initial momentum of you plus your bicycle is 1066 kgm/s (a).
The initial momentum of the bug can be obtained in the same way:
[tex]p=mv\\\\p=(0.005kg)(1.5m/s)\\\\p=0.0075kgm/s[/tex]
Then the initial momentum of the bug is 0.0075 kgm/s (b).
Now, since the mass of the bug is much less than your mass, we can think of this as a perfectly inelastic collision. This means that, after the collision, the velocity of you, the bicycle and the bug is the same. From the conservation of linear momentum, we have:
[tex]p_0=p_f\\\\(m_{you}+m_{bicycle})v_{you}+m_{bug}v_{bug}=(m_{you}+m_{bicycle}+m_{bug})v_f\\\\v_f=\frac{(m_{you}+m_{bicycle})v_{you}+m_{bug}v_{bug}}{m_{you}+m_{bicycle}+m_{bug}}\\\\v_f=\frac{(70kg+12kg)(13m/s)+(0.005kg)(1.5m/s)}{70kg+12kg+0.005kg}\\ \\v_f=12.9992m/s[/tex]
As your initial velocity was 13m/s, the change in velocity is of -0.0008 m/s (c).
If the bug were travelling in the opposite direction, its initial velocity would have been negative. So:
[tex]v_f=\frac{(70kg+12kg)(13m/s)-(0.005kg)(1.5m/s)}{70kg+12kg+0.005kg}\\ \\v_f= 12.9991m/s[/tex]
So, in this case the change in velocity is of -0.0009 m/s (d).
Note that the bug is so small that the change in velocity is negligible in most cases. That's why we don't notice when we hit a bug when riding bicycle.
Trace the path of a ray emitted from the tip of the object toward the focal point of the mirror and then the reflected ray that results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.
When a ray is emitted from the tip of an object towards the focal point of a mirror, it follows the law of reflection. The location and orientation of the reflected ray depend on the shape of the mirror. Additional rays can be traced from the base of the object to locate the extended image.
Explanation:When a ray is emitted from the tip of an object towards the focal point of a mirror, it follows the law of reflection. For a concave mirror, the reflected ray passes through the focal point, while for a convex mirror, the reflected ray extends backward through the focal point. The location and orientation of the reflected ray depend on the shape of the mirror. To locate the extended image, additional rays can be traced from the base of the object along the optical axis. All four principal rays run parallel to the optical axis, reflect from the mirror, and then run back along the optical axis. The image of the base of the object is located directly above the image of the tip, as the mirror is symmetrical from top to bottom.
Learn more about Ray tracing in mirrors here:https://brainly.com/question/15506761
#SPJ12
Ray tracing involves drawing rays from an object through points of interest like the focal point of a mirror and then determining where the reflected rays intersect to locate the image. It is a crucial method in optics to determine the properties of images formed by mirrors and lenses.
Explanation:Understanding Ray Tracing in MirrorsTo trace the path of a ray emitted from the tip of an object toward the focal point of a mirror, as well as the reflected ray, one must understand the basic principles of ray tracing with mirrors. With a concave mirror, an incident ray that travels parallel to the optical axis will be reflected through the focal point on the same side of the mirror. Conversely, with a convex mirror, a ray that travels parallel to the optical axis will reflect as if it originates from the focal point behind the mirror, forming a virtual focus. To construct the reflected ray, you draw the ray until it reaches the mirror’s surface and then redirect it according to the mirror type's ray tracing rules.
Furthermore, the intersection of the reflected rays (real or virtual) determines the location of the image. If the rays meet in real space, the image is real; if they only appear to intersect upon extension in virtual space, the image is virtual.
To achieve a complete picture of how an image is formed, one must also trace rays from another point on the object. For instance, tracing rays from the base of the object can help determine the orientation of the image. In the case of rays that are collinear with the optical axis, the image will maintain the object's vertical orientation. By tracing at least two different rays following the simple ray tracing rules, one can locate the image formed by mirrors and lenses.
Learn more about Ray Tracing here:https://brainly.com/question/35884592
#SPJ3
. A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00m above the ground. At the same instant the balloon is released; a second balloon is thrown straight down at 2.00 m/s from the same height. Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon
Answer:
Explanation:
Height of building
H = 6m
Horizontal speed of first balloon
U1x = 2m/s
Second ballot is thrown straight downward at a speed of
U2y = 2m/s
Time each gallon hits the ground
Balloon 1.
Using equation of free fall
H = Uoy•t + ½gt²
Uox = 0 since the body does not have vertical component of velocity
6 = ½ × 9.8t²
6 = 4.9t²
t² = 6 / 4.9
t² = 1.224
t = √1.224
t = 1.11 seconds
For second balloon
H = Uoy•t + ½gt²
6 = 2t + ½ × 9.8t²
6 = 2t + 4.9t²
4.9t² + 2t —6 = 0
Using formula method to solve the quadratic equation
Check attachment
From the solution we see that,
t = 0.9211 and t = -1.329
We will discard the negative value of time since time can't be negative here
So the second balloon get to the ground after t ≈ 0.92 seconds
Conclusion
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
Answer:
Second balloon hits ground Δt = 0.185 seconds sooner than first balloon
Explanation:
Given:-
- The first balloon is thrown horizontally with speed, u1 = 2.0 m/s
- The second balloon is thrown down with speed, u2 = 2.0 m/s
- The height from which balloon are thrown, si = 6.0 m (above ground)
Find:-
Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon
Solution:-
- We will first determine the time taken (t1) for the first balloon thrown horizontally with speed u1 = 2.0 m/s from top of building from a height of s = 6.0 m from ground to it the ground.
- Using the second kinematic equation of motion in vertical direction:
si = 0.5*g*t1^2
Where, g: The gravitational constant = 9.81 m/s^2
6.0 = 4.905*t1^2
4.905*t1^2 - 6.0 = 0
- Solve the quadratic equation:
t 1 = 1.106 s
- Similarly, the time taken (t2) for the second balloon thrown down with speed u2 = 2.0 m/s from top of building from a height of s = 6.0 m from ground to it the ground.
- Using the second kinematic equation of motion in vertical direction:
si = u2*t2 + 0.5*g*t1^2
Where, g: The gravitational constant = 9.81 m/s^2
6.0 = 4.905*t1^2 + 2*t2
4.905*t1^2 + 2*t2 - 6.0 = 0
- Solve the quadratic equation:
t 2 = 0.9208 s
- We see that the second balloon thrown down vertically hits the ground first. The second balloon reaches ground, t1 - t2 = 0.185 seconds, sooner than first balloon.
By what factor will the electrostatic force between two charged objects change when the amount of charge on one object doubles?
Answer:
The electrostatic force between two charge doubles when when the amount of charge on one object doubles.
Explanation:
Force between two charge particle [tex]F=\frac{1}{4\pi\epsilon } \frac{q_1\cdot\ q_2}{r^{2} }[/tex]
Where, [tex]\frac{1}{4\pi\epsilon } = constant[/tex]
[tex]q_1,q_2\ magnitude\ of\ charges.\\r\ is\ the\ distance\ between\ the\ particles.[/tex]
Now suppose as per the question
[tex]q_1= 2 q_1\ with\ q_2\ and\ r\ with\ same\ value.\\Hence\ force\ between\ the\ particle\ F_2 =2 F[/tex]
Hence, electrostatic force between two charge doubles when when the amount of charge on one object doubles.
En un planeta distante,un objeto en caida libre cambia su velocidad de 10m/s a 90m/s en 2 segundos ¿Cual es la aceleración gravitacional de dicho planeta?
Answer:
a = 40 m / s²
Explanation:
For this exercise we can use the vertical launch equations to which these are valid in any system with inertial
v = v₀ + a t
a = (v-v₀) / t
We calculate
a = (90 - 10) / 2
a = 80/2
a = 40 m / s²
This is the value of the planet's gravity acceleration
Kopnięta poziomo piłka o masie 1,5 kg 1,5 kg w momencie kopnięcia uzyskała przyspieszenie 5 m s 2 5ms2 . Oblicz wartość siły kopnięcia. Po nasiąknięciu wodą masa piłki wzrosła do 2 kg 2 kg . O ile większej siły musi użyć zawodnik, aby nadać jej takie samo przyspieszenie jak poprzednio?
1) 7.5 N
2) 10 N
Explanation:
1)
We can solve this first part of the problem by using Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:
[tex]F=ma[/tex]
where
F is the net force
m is the mass of the object
a is its acceleration
In this problem we have:
m = 1.5 kg is the mass of the ball
[tex]a=5 m/s^2[/tex] is the acceleration
So, the kick force on it was:
[tex]F=(1.5)(5)=7.5 N[/tex]
2)
In this case, the mass of the ball has increased to
m' = 2 kg
We can also solve this part by using again Newton's second law of motion:
[tex]F'=m'a[/tex]
where
F' is the new kick force
m' = 2 kg is the new mass of the ball
a is the acceleration
The acceleration is the same as before,
[tex]a=5 m/s^2[/tex]
Therefore, the new kick force is:
[tex]F'=(2)(5)=10 N[/tex]
List down different layers of the sun. Rank these layers based on their distance from the sun’s center
Answer:
Core
Radiative zone
Convective zone
Photosphere
Chromosphere
Transient region
Corona
Ranks of layers based on their distance from the sun’s center
1st-corona
2nd-Transient region
3rd-chromosphere
4th-Photosphere
5th-convective zone
6th-radiative zone
7th-core
A double-concave (thickest at the edges, thinnest in the middle) thin lens is made of glass with an index of refraction of 1.6. The radius of curvature of the left and right faces of the lens are 65 cm and 75 cm, respectively. The object lies to the left of the lens. What is the focal length of the lens
Answer:
- 58 cm
Explanation:
refractive index, n = 1.6
radius of curvature of left face, R1 = - 65 cm
Radius of curvature of the right face, R2 = 75 cm
Use the lens maker's formula
[tex]\frac{1}{f}=\left ( n-1 \right )\times \left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]
[tex]\frac{1}{f}=\left ( 1.6-1 \right )\times \left ( -\frac{1}{65}-\frac{1}{75} \right )[/tex]
[tex]\frac{1}{f}=\left ( 0.6 \right )\times \left ( \frac{-75-65}{75\times 65}\right )[/tex]
f = - 58 cm
Thus, the focal length of the lens is - 58 cm.
Why is a shadow formed
Answer:
Shadows are made by blocking light. Light rays travel from a source in straight lines. If an opaque (solid) object gets in the way, it stops light rays from traveling through it. The size and shape of a shadow depend on the position and size of the light source compared to the object.
Explanation:
Answer:
shadows are made by blocking light
You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it at position y0. Imagine that you slowly pull down with a spring scale so the block is now at position y bottom, below the equilibrium position y0 where it was hanging at rest. The scale reading when you let go of the block is 3.0 N.
a. Where was the block when you let go? Assume y0 is the equilibrium position of the block and that "down" is a positive direction.
b. Determine the work you did stretching the spring.
Express your answer to two significant figures and include the appropriate units.
c. What was the energy of the spring-Earth system when you let go (assume that zero potential energy corresponds to the equilibrium position of the block)?
Express your answer to two significant figures and include the appropriate units.
d. How far will the block rise after you release it?
Express your answer to two significant figures and include the appropriate units.
Answer:
a) Δy = 0.144 m
b) W = 0.145 J
c) Us = 0.32 J
d) ymax = 0.144 m
Explanation:
a) First let's find the spring constant using Hooke's Law
F = k*Δy ⇒ k = F/Δy
where
F = m*g = 0.1 kg*9.81 m/s² = 0.981 N
and Δy = 0.07 m. Hence
k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m
In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:
f = 3 N - F = 3 N - 0.981 N = 2.019 N
Now that we have found the force we can use Hooke's Law in order to find the position of the block
f = k*Δy ⇒ Δy = f/k
⇒ Δy = 2.019 N/14 N/m
⇒ Δy = 0.144 m
b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:
W = ΔU
W = ΔUs + ΔUg
W = (Usf - Usi) + (Ugf - Ugi)
Notice that
Us = 0.5*k*y²
where
yf = 0.07 m + 0.144 m = 0.214 m and
yi = 0.07 m
and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence
W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)
⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)
⇒ W = 0.145 J
c) When we let the block go the spring was stretched by
y = 0.07 m + 0.144 m = 0.214 m
Therefore:
Us = 0.5*k*y²
⇒ Us = 0.5*14 N/m*(0.214 m)²
⇒ Us = 0.32 J
d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is
ymax = Δy = 0.144 m
An object is thrown with a horizontal velocity of 20 m/s from a cliff that is 125 m above level ground. If air resistance is negligible, the time that it takes the object to fall to the ground from the cliff is most nearly
Given that,
Horizontal velocity of the object, v = 20 m/s
Height of the cliff, h = 125 m
We need to find the time that it takes the object to fall to the ground from the cliff is most nearly. It can be calculated using second equation of motion. Let us consider that the initial speed of the object is 0. So,
[tex]h=ut+\dfrac{1}{2}at^2[/tex]
Here, a = g and u = 0
[tex]h=\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 125}{10}} \\\\t=5\ s[/tex]
So, the object will take 5 seconds to fall to the ground from the cliff.
The time taken for the object to fall to the ground is 5 s
From the question given above, the following data were obtained:
Horizontal velocity (u) = 20 m/sHeight (h) = 125 mAcceleration due to gravity (g) = 10 m/s²Time (t) =?The time taken for the object to fall to the ground can be obtained as follow:
[tex]t = \sqrt{ \frac{2h}{g}} \\ \\ t = \sqrt{ \frac{2 \times 125}{10}} \\ \\ t = 5 \: s \\ \\ [/tex]
Therefore, the time taken for the object to get to the ground is 5 s
Learn more about projectile motion: https://brainly.com/question/15206806
In an experiment, a large number of electrons are fired at a sample of neutral hydrogen atoms and observations are made of how the incident particles scatter. The electron in the ground state of a hydrogen atom is found to be momentarily at a distance a0/2 from the nucleus in 1 300 of the observations. In this set of trials, how many times is the atomic electron observed at a distance 2a0 from the nucleus?
Answer:
N = 1036 times
Explanation:
The radial probability density of the hydrogen ground state is given by:
[tex]p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }[/tex]
[tex]p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }[/tex]
[tex]p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }[/tex]
[tex]N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}[/tex]
[tex]N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } } }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}[/tex]
[tex]N = 1300(16) e^{-3}[/tex]
N = 1035.57
N = 1036 times
What is the kinetic energy of a bike with a mass of 16 kg traveling at 4 m/s?
Answer: 128J
Explanation:
[tex]Formula: E_k=\frac{1}{2}mv^2[/tex]
[tex]E_k=\frac{1}{2}(16kg)(4m/s)^2 \\E_k=\frac{1}{2}(16kg)(16m^2/s^2)\\ E_k=\frac{1}{2}(256kg*m^2/s^2)\\ E_k=128kg*m^2/s^2\\or\\E_k=128J[/tex]
Answer:
That is correct
Explanation:
128 J
what is the wavelength of microwaves
Answer:
1 mm PLS mark brainliest
Explanation:
Microwaves are electromagnetic waves with wavelengths longer than those of terahertz (THz) wavelengths, but relatively short for radio waves. Microwaves have wavelengths approximately in the range of 30 cm (frequency = 1 GHz) to 1 mm (300 GHz).
Use the following information for determining sound intensity. The number of decibels β of a sound with an intensity of I watts per square meter is given by β = 10 log(I/I0), where I0 is an intensity of 10−12 watt per square meter, corresponding roughly to the faintest sound that can be heard by the human ear. Find the number of decibels β of the sound.
Answer:
120 decibels
Explanation:
we know sound in decibels is given by.
[tex]\beta =10log(\frac{I_{} }{I_{0} } )[/tex]
Where I is intensity of the sound and [tex]I_{0}=10^-12W/m^{2 }[/tex] is reference intensity.
substituting All of this in our decibel formula with I =1W/[tex]m^2[/tex]
gives us 120 decibels.
The sound intensity level of a sound is measured in decibels (dB) using a logarithmic scale that uses a reference intensity of 10^-12 W/m², which is the threshold of human hearing. The decibel level of a sound is determined by the ratio of its intensity to this reference intensity, with each increment of 10 dB corresponding to a tenfold increase in intensity.
Explanation:The sound intensity level ß, measured in decibels (dB), provides a logarithmic measure of sound intensity I in watts per meter squared (W/m²), with the formula
ß = 10 log(I/I0)
where I0 = 10⁻¹² W/m² is the reference intensity, corresponding to the threshold of human hearing. The decibel level of a sound is thus derived from the ratio of its intensity to this reference intensity. A sound with the same intensity as the reference intensity (I = I0) has a decibel level of 0 dB, because log 10(1) = 0.
Interpreting decibels involves understanding that because the formula uses a logarithm, each increment of 10 dB corresponds to a tenfold increase in intensity. Therefore, a sound with an intensity of 10^-11 W/m² would be 10 dB, a sound with an intensity of 10⁻¹⁰ W/m² would be 20 dB, and so on. The decibel scale helps to compress the vast range of sound intensities that human ears can perceive into a more manageable range of numbers.
Learn more about Sound Intensity Level here:https://brainly.com/question/30101270
#SPJ11
An open container holds ice of mass 0.550 kg at a temperature of -15.3 âC . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 900 J/minute .The specific heat of ice to is 2100 J/kgâK and the heat of fusion for ice is 334Ã103J/kg.a) How much time tmelts passes before the ice starts to melt?tmelts=___minutesb)From the time when the heating begins, how much time trise does it take before the temperature begins to rise above 0âC?trise=____minutes
The ice in the open container begins to melt after 19.82 minutes of heating, and the temperature of the system starts to rise above 0°C after a total heating time of 223.93 minutes.
Explanation:To answer both parts of this question, we need to calculate the time for two processes: the heating of the ice to 0°C (melting point), and the melting of the ice into water at 0°C.
Firstly, we find the heat (Q) required to raise the temperature of the ice to 0°C using Q=mcΔT, which is 0.55kg * 2100 J/kg°C * (0 - (-15.3°C)) = 17842.5 J. As the heater adds 900J per minute, the time for this is 17842.5J ÷ 900J/min = 19.82 minutes.
Next, we find the heat required to melt ice into water at 0°C using Q=mLf, where Lf = 334000 J/kg. This is 0.55kg * 334000 J/kg = 183700J. The time for this can be found by 183700J ÷ 900J/min = 204.11 minutes.
tmelts (the time before ice begins to melt) is 19.82 minutes, and trise (the total time before the temperature begins to rise above 0°C) is 19.82 min + 204.11 min = 223.93 minutes.
Learn more about Heat Transfer Calculations here:https://brainly.com/question/31080599
#SPJ12
The time before the ice starts to melt (tmelts) is 19.63 minutes, and the total time before the temperature begins to rise above 0 °C (trise) is 223.74 minutes.
Explanation:Calculating Time Until Ice Starts to MeltTo calculate the time tmelts before the ice starts to melt, you firstly need to determine how much heat is necessary to raise the ice from -15.3 °C to 0 °C using the formula Q = m * c * ΔT, where Q is the heat, m is the mass of the ice, c is the specific heat of ice, and ΔT is the change in temperature. In this case, Q would be:
Q = (0.550 kg) * (2,100 J/kg·K) * (15.3 K) = 17,661.5 J.
Since heat is supplied at 900 J/minute, the time tmelts can be calculated as:
tmelts = Q / (heat rate) = 17,661.5 J / (900 J/min) = 19.63 minutes.
Calculating Time Until Temperature Rises Above 0 °CTo calculate the time trise until the temperature begins to rise above 0 °C, we must add the time it takes to melt the ice completely at 0 °C. The heat of fusion (∆Hfus) formula is used here: Q = m * Lf where Lf is the heat of fusion for ice. Considering that the heat of fusion for ice is 334,000 J/kg:
Q = (0.550 kg) * (334,000 J/kg) = 183,700 J.
For melting the ice at 0 °C, it will take:
trise = Q / (heat rate) = 183,700 J / (900 J/min) = 204.11 minutes.
Therefore, the total time before the temperature begins to rise above 0 °C is the sum of the time to warm the ice to 0 °C and the time to melt it completely, which will be 19.63 minutes + 204.11 minutes = 223.74 minutes.
Learn more about Heat Transfer in Ice here:https://brainly.com/question/18940221
#SPJ3
Suppose that 3.00 g of hydrogen is separated into electrons and protons. Suppose also that the protons are placed at the Earth's North Pole and the electrons are placed at the South Pole. What is the resulting compressional force on the Earth?
Answer:
Explanation:
Atomic mass of hydrogen is 1 . so 1 g of hydrogen will have 6.02 x 10²³ atoms
3 g of hydrogen will have 3 x 6.02 x 10²³ atoms . This will give 3 x 6.02 x 10²³ protons and same number of electrons
So number of protons at north pole = 3 x 6.02 x 10²³
charge on these protons = 1.6 x 10⁻¹⁹ x 3 x 6.02 x 10²³ C
= 28.9 x 10⁴ C
similarly total charge on electrons at south pole = 28.9 x 10⁴ C
distance between them = diameter of the earth = 2 x 6356 x 10³ m
= 12712 x 10³ m
Attractive force between these charges
= k q₁q₂ / r² , q₁ ,q₂ are charges and r is distance between charges.
= 9 x 10⁹ x (28.9 x 10⁴)² / (12712 x 10³ )²
= 4.6517 x 10⁻⁵ x 10¹¹
= 4.6517 x 10⁶ N
A student's life was saved in an automobile accident because an airbag expanded in front of his head. If the car had not been equipped with an airbag, the windshield would have stopped the motion of his head in a much shorter time. Compared to the windshield, the airbag:
Answer:
Compared to windshield the airbag exerts much lesser force
Explanation:
Impulse is defined as change in momentum of the object when it is acted upon by a force during interval of time
Impulse = Impulsive force *time
I = F*Δt
If the object should be bought to rest from certain velocity there should be change in momentum. If the duration in which the momentum is increased then there would be less force applied and hence less damage.
Airbags are used to reduce the force experience by the people when they are met with accident by extending the time required to stop the momentum.
During the collision, the passenger is carried towards the windshield and if they are stopped by collision with wind shield the force will be larger and more damage.But if they are hit with airbag then the force will be less due to increased time.
The change is momentum will be the same with or without momentum but its the time that decides the impact of force.By making it longer the force become less.
Thus compared to the windshield the airbag exerts much lesser force.
A kettle is rated at 1 kW, 220 V. Calculate the working resistance of the kettle.
Resistance = ...........
When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point. Calculate how much energy has been supplied?
Explanation:
Power of electric kettle, P = 1 kW
Voltage, V = 220 V
(a) Electric power is given by the formula as follows :
[tex]P=\dfrac{V^2}{R}[/tex]
R is resistance
[tex]R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{10^3}\\\\R=48.4\ \Omega[/tex]
(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.
Energy supplied is given by :
[tex]E=P\times t[/tex]
P is power, [tex]P=\dfrac{V^2}{R}[/tex]
[tex]E=\dfrac{V^2}{R}t\\\\E=\dfrac{(220)^2}{48.4}\times 180\\\\E=180000\ J\\\\E=180\ kJ[/tex]
The working resistance of the kettle is 48.35 ohms. The amount of energy supplied is 180,000 J.
Explanation:To calculate the working resistance of the kettle, we can use Ohm's law, which states that resistance (R) is equal to the voltage (V) divided by the current (I).
In this case, we know that the kettle is rated at 1 kW, which is equal to 1000 W. Since power (P) is equal to voltage multiplied by current, we can rearrange the equation to solve for current: I = P/V. Thus, I = 1000 W / 220 V = 4.55 A. Now, we can use Ohm's law to determine the resistance: R = V/I = 220 V / 4.55 A = 48.35 ohms.To calculate the amount of energy supplied, we need to use the formula E = P * t, where E is energy, P is power, and t is time.
Since the kettle is rated at 1 kW or 1000 W, we can calculate the energy supplied over 3 minutes as follows: E = 1000 W * 3 min * (60 s / 1 min) = 180,000 J.Therefore, The working resistance of the kettle is 48.35 ohms. The amount of energy supplied is 180,000 J.
All organisms require a set of instructions that specify its traits.The instructions that are responsible for all the inherited traits of an organisms
Answer:
That's DNA
Explanation:
DNA(Deoxyribonucleic acid) is the building blocks of life, the Legos for genetically complex life if you will. All living things have DNA, it is the instruction that that organism must follow to specify what exact traits it'll have, For example, I have Blue eyes because Blue eyes is apart of my DNA. The DNA told my body that I was gonna have blue eyes before I was born.
Which factors are most likely to affect whether a galaxy will evolve by merging with a second galaxy? Select the two correct answers.(1 point)
its proportion of gas compared to that of the other galaxy
its distance from the other galaxy
its size compared to that of the other galaxy
its age compared to that of the other galaxy
Answer:
ii. its distance from the other galaxy.
iii. its size compared to that of the other galaxy
Explanation:
A galaxy is a vast constituent of interstellar medium of gas, dust and billions of stars bounded by the gravitational pull. There are tens of billions of galaxies existing in the whole universe. Various numbers of galaxies can merge to form an inter-cluster, which is a more massive and bigger galaxy.
The constituent of an individual galaxy are held together by mutual gravitational pull, but may interact with other galaxies depending on; its distance from other galaxies and its size compared to that of the other galaxy. This process would result into the formation of a giant elliptical galaxy and a number of stars.
Answer:
1.Gravity acts on existing material, shaping it into a new form.
2.a much larger galaxy, a nearby galaxy
3.low gas content, little star formation
Explanation:
i just did the test and its 100%. dont believe me? just do the test and see for your self
A rockets initially at rest , it steady gains speed at a rate of 10m/s^2 for 4.6s during take off .
What was the rocket’s top speed ?
Final answer:
The rocket's top speed is 46 m/s.
Explanation:
The rocket's top speed can be found using the equation v = vo + at, where v is the final velocity, vo is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity (vo) is 0 m/s since the rocket starts at rest, the acceleration (a) is 10 m/s², and the time (t) is 4.6 s.
Plugging these values into the equation, we get:
v = 0 m/s + (10 m/s²)(4.6 s) = 46 m/s.
Therefore, the rocket's top speed is 46 m/s.
Star A appears brighter than star B, as seen from Earth. Therefore, star A must be closer to Earth than star B. Star A appears brighter than star B, as seen from Earth. Therefore, star A must be closer to Earth than star B. a.True b.False
Answer:
The statement is not true always.
Explanation:
Given:
Star A appears brighter than Star B.
The apparent brightness of a star depends on two factors:
i) the distance of the star from the earth
ii) the limunosity of the star
If two stars are at same distance from the earth and one has more luminosity than other then ir appears to be more luminous. Also if two stars are at a different distance from the earth, then the star having more luminosity appears to be more luminous.
Thus, the statement is not true.