A spring has a spring constant of 450 N/m. How much must this spring be stretched to store 49 J of potential energy?
Answer:
W = 1/2 K x^2
x^2 = 2 * W / K = 2 * 49 J / (N/m) = .218 / m^2
x = .467 m
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward and ideal spring that is attached to a wall , After the block collides with the spring, the spring is compressed a maximum distance of 0.68m . what is the speed of the block when the spring is compressed to only one-half of the maximum distance?
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall, the maximum speed of the block when the spring is compressed to one-half of the maximum distance is 4.33 m/s
From the conservation of energy; the kinetic energy of the mass is equal to the work done on the spring.
i.e.
[tex]\mathbf{\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2_{max}}[/tex]
Given that:
the mass of the block = 4.0 kg the speed at which it is moving = 5.0 m/scompression of the spring = 0.68 m∴
From the equation above, multiplying both sides with 2, we have:
[tex]\mathbf{mv^2 =kx^2_{max}}[/tex]
Making (k) the subject of the formula;
[tex]\mathbf{k = \dfrac{mv^2}{x^2_{max}}}[/tex]
[tex]\mathbf{k = \dfrac{4 \times 5^2}{0.68^2}}[/tex]
k = 216.26 N/m
However, when compressed to one-half of the maximum distance; the speed is computed as follows:
x = 0.68/2 = 0.34 m
∴
[tex]\mathbf{\dfrac{1}{2}mv_o^2 - \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{m(v_o^2 -v^2) =kx^2}[/tex]
[tex]\mathbf{(v_o^2 -v^2) =\dfrac{kx^2}{m}}[/tex]
[tex]\mathbf{(5^2 -v^2) =\dfrac{216.26 \times 0.34^2}{4.0}}[/tex]
25 - v² = 6.25
25 -6.25 = v²
v² = 18.75
[tex]\mathbf{ v= \sqrt{18.75 }}[/tex]
v = 4.33 m/s
Therefore, we can conclude that the speed of the block when the spring is compressed to only one-half of the maximum distance is 4.33 m/s
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The ear drum vibrates when struck by sound waves and directly sends a message to the brain that is then recognized as sound
True or False
Answer:
true
Explanation:
A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft/s while connected to the end of a rope. If the rope is pulled inward with a constant speed of vr = 4 ft/s, determine the speed of the box at the instant rB = 1 ft. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box
Answer:
W = 1.875 J
Explanation:
For this exercise let's use the relationship between work and kinetic energy
W = ΔK
The kinetic energy of rotational motion is
K₀ = ½ I w²
we can assume that the box is small, so it can be treated as a point object, with moment of inertia
I = m rₐ²
angular and linear velocity are related
v = w r
w = v / r
we substitute in the equation, for point A
K₀ = ½ (m rₐ²) (v / rₐ)²
K₀ = ½ m v²
For the final point B, as the system is isolated the angular momentum is conserved
initial L₀ = Io wo
final L_f = I_f w_f
L₀ = L_f
I₀ w₀ = I_f w_f
(m rₐ²) w₀ = (m [tex]r_{b} ^2[/tex]) w_f
w_f = (rₐ/r_b)² w₀
with this value we find the final kinetic energy
K_f = ½ I_f w_f²
K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)² w₀) ²
K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]
we substitute in the realcion of work
W = K_f - K₀
W = ½ m [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²
W = ½ m [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²
W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2
W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]
let's calculate
W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)
W = 0.625 (3)
W = 1.875 J
Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. Bifocal lenses are used to correct his vision. A diverging lens is used in the top part of the lens to allow the person to clearly see distant objects.
1. What power lens (in diopters) should be used in the top half of the lens to allow her to clearly see distant objects?
2. What power lens (in diopters) should be used in the bottom half of the lens to allow him to clearly see objects 25 cm away?
Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
[tex]\frac{1}{f_1} = \frac{1}{q}[/tex]
q = f₁
2) for an object located at p = 25 cm
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]
[tex]\frac{1}{f_2}[/tex] = 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P = [tex]\frac{1}{f}[/tex]
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
. Calculate
the length of this wire required to make a 10 Ω coil.
Answer:
the length of the wire is 134.62 m.
Explanation:
Given;
resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm
cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²
resistance of the wire, R = 10Ω
The length of the wire is calculated as follows;
[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]
Therefore, the length of the wire is 134.62 m.
In a movie production, a stunt person must leap from a balcony of one building to a balcony 3.0 m lower on another building. If the buildings are 2.0 m apart, what is the minimum horizontal velocity the stunt person must have to accomplish the jump? Assume no air resistance and that ay = −g = −9.81 m/s2 . (Ans. 2.6m/s) PLS SHOW WORK
This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The minimum horizontal velocity required is "2.6 m/s".
First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.
[tex]h=v_it+\frac{1}{2}gt^2[/tex]
where,
h = height = 3 m
vi = initial vertical speed = 0 m/s
t = time interval = ?
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]
t = 0.78 s
Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:
[tex]s = vt\\\\v = \frac{s}{t}[/tex]
where,
s = horizontal distance = 2 m
t =0.78 s
v = minimum horizontal velocity = ?
Therefore,
[tex]v = \frac{2\ m}{0.78\ s}[/tex]
v = 2.6 m/s
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The attached picture shows the equations of motion in the horizontal and vertical directions.
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500
Answer:
8 kV
Explanation:
Here is the complete question
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?
Solution
Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V
So, Q = CV
= 500 × 10⁻⁶ F × 800 V
= 400000 × 10⁻⁶ C
= 0.4 C
Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV
V = Q/C
= 0.4 C/500 × 10⁻⁶ F
= 0.0008 × 10⁶ V
= 800 V
The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)
= 10 × 800 V
= 8000 V
= 8 kV
A wave has a frequency of 67 Hz and a wavelength of 7.1 meters. What is the speed of this
wave?
Answer:
475.7 m/s
Explanation:
Given,
Frequency ( f ) = 67 Hz
Wavelength ( λ ) = 7.1 m
To find : Speed ( v ) = ?
Formula : -
v = f λ
v
= 67 x 7.1
= 475.7 m/s
Therefore,
the speed of the wave is 475.7 m/s.
A go-cart is traveling at a rate of 25 m/sec for 20 seconds. How far will the go cart travel?
Answer:
Distance travel by go-cart = 500 meter
Explanation:
Given:
Speed of go cart = 25 m/s
Time travel = 20 seconds
Find:
Distance travel by go-cart
Computation:
Distance = Speed x time
Distance travel by go-cart = Speed of go cart x Time travel
Distance travel by go-cart = 25 x 20
Distance travel by go-cart = 500 meter
Help me please with both questions?
Answer:
question #1 is A
Question #2 is C
Explanation:
Protons, neutrons, electrons, and a nucleus are
NO LINKS PLEASE
At what speed do a bicycle and its rider, with a combined mass of 90 kg
k
g
, have the same momentum as a 1500 kg
k
g
car traveling at 4.0 m/s
m
/
s
?
Answer:
2
Explanation:
A scientist measuring the resistivity of a new metal alloy left her ammeter in another lab, but she does have a magnetic field probe. So she creates a 4.5-m-long, 2.0-mm-diameter wire of the material, connects it to a 1.5 V battery, and measures a 3.0 mT magnetic field 1.0 mm from the surface of the wire. What is the material's resistivity
Answer:
[tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex]
Explanation:
r = Radius = [tex]\dfrac{2}{2}=1\ \text{mm}[/tex]
B = Magnetic field = 3 mT
1 mm = Distance from the surface of the wire
V = Voltage
x = Distance from the probe = [tex]r+1=1+1=2\ \text{mm}[/tex]
R = Resistance
L = Length of wire = 4.5 m
Magnetic field is given by
[tex]B=\dfrac{\mu_0I}{2\pi x}\\\Rightarrow I=\dfrac{B2\pi x}{\mu_0}\\\Rightarrow I=\dfrac{3\times 10^{-3}\times 2\times \pi 2\times 10^{-3}}{4\pi 10^{-7}}\\\Rightarrow I=30\ \text{A}[/tex]
Voltage is given by
[tex]V=IR\\\Rightarrow R=\dfrac{V}{I}\\\Rightarrow R=\dfrac{1.5}{30}\\\Rightarrow R=0.05\ \Omega[/tex]
Resistivity is given by
[tex]\rho=\dfrac{RA}{L}\\\Rightarrow \rho=\dfrac{0.05\times \pi (1\times 10^{-3})^2}{4.5}\\\Rightarrow \rho=3.49\times 10^{-8}\ \Omega\text{m}[/tex]
The resistivity of the material is [tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex].
Which two chemical equations show double-replacement reactions?
A. C+02 - CO2
B. 2Li + CaCl2 - 2LiCl + Ca
I C. Ca(OH)2 + H2S04 - CaSO4 + 2H20
D. Na2CO3 + H2S - H2CO3 + Na2S
The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.
What is double replacement reaction?A double replacement reaction have two ionic compounds that are exchanging anions or cations.
From the given options, we can choose the following based on their exchange of anions or cations.
Ca(OH)2 + H2S04 - CaSO4 + 2H20Na2CO3 + H2S - H2CO3 + Na2SThus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.
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#SPJ2
. Estimate the buoyant force that air exerts on you. (To do this, you can estimate your volume by knowing your weight and by assuming that your weight density is a bit less than that of water.)
Answer:
[tex]0.886[/tex] N buoyant force is exerted by air
Explanation:
My weight is [tex]75[/tex] Kg
Weight = mass * gravity
As we know
Buoyant Force is equal to the product of density * acceleration due to gravity and volume of the body
Assuming weight density is a bit less than that of water or equal to water i.e [tex]997.77[/tex] kg/m3
Volume is equal to mass / density
[tex]= 75[/tex] Kg * g/[tex]997.777[/tex]
[tex]= 0.0751[/tex] * g
Buoyant Force
= Volume * g * density
[tex]= 0.0751 * 9.8 * 1.2041[/tex]kg/m3
[tex]= 0.886[/tex] N
You are testing a new amusement park roller coaster with an empty car with a mass of 130 kg . One part of the track is a vertical loop with a radius of 12.0 m . At the bottom of the loop (point A) the car has a speed of 25.0 m/s and at the top of the loop (point B) it has speed of 8.00 m/s . Part A As the car rolls from point A to point B, how much work is done by friction
Answer:
work done by friction = 5889 J
Explanation:
We are given;
Mass of car; m = 130 kg
Speed at point A; v1 = 25 m/s
Speed at point B: v2 = 8 m/s
Since radius is 12 m
At point A, distance is; y1 = 12 m
At point B, distance is; y2 = -12 m
Now, formula for work done by all the forces is given by the equation;
Total work;
W_gravity + W_others = K2 - K1
Where W_others is work done by other forces which is equal to work done by friction
Where K2 - K1 is change in kinetic energy.
W_grav is also change in potential energy and is expressed as;
W_grav = mgy1 - mgy2
K2 - K1 = ½m(v1)² - ½m(v2)²
Thus;
mgy1 - mgy2 + W_others = ½m(v1)² - ½m(v2)²
Making W_others the subject;
W_others = ½m(v1)² - ½m(v2)² + mgy2 - mgy1
Plugging in the relevant values;
W_others = (½ × 130 × 25²) - (½ × 130 × 8²) + (130 × 9.8 × -12) - (130 × 9.8 × 12)
W_others = 5889 J
Recall that I earlier said W_others = work done by friction.
Thus, work done by friction = 5889 J
35 POINTSS!!! PLSSSS HELLPPP!!!
Work is the transfer of power from one object to another.
Please select the best answer from the choices provided
T
F
Answer:
T
beacuse:
Energy can be transferred from one object to another by doing work. ... When work is done, energy is transferred from the agent to the object, which results in a change in the object's motion (more specifically, a change in the object's kinetic energy).
The water pressure to an apartment is increased by the water company. The water enters the apartment through an entrance valve at the front of the apartment. Where will the increase in the static water pressure be greatest when no water is flowing in the system
Answer:
Option C
Explanation:
Options for the question are as follows -
A. At a faucet close to entrance valve
B. At a faucet away from the entrance valve
C. It will be the same at all faucets
D. There will be no increase in the pressure at the faucets
Solution -
The static force will be the same at all faucets and also the area of the faucets be same.
Thus, the pressure created at all faucets will be the same.
Thus, option C is correct
In a nuclear fusion reaction, atoms:
split apart.
combine.
explode.
cool down.
A uniform magnetic field is in the positive z direction. A positively charged particle is moving in the positive x direction through the field. The net force on the particle can be made zero by applying an electric field in what direction
Answer:
We apply an electric field in the negative y direction
Explanation:
Since A uniform magnetic field is in the positive z direction and A positively charged particle is moving in the positive x direction through the field, the magnetic force acting on the positively charged particle is in the positive y direction according to Fleming's right-hand rule.
For the net force on the particle to be zero, we apply an electric field in the negative y direction to create an electric force on the positively charged particle, so as to cancel out the magnetic force.
Tony ran 600 meters in 60 seconds. What was Tony's speed during the
race?
A woman shouts at a boy who is underwater what happens to the speed of the sound wave as it moves from the air into the water
Answer:
B. it increases
Explanation:
As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).
Answer:
B is the correct answer.
Explanation:
A carnival ride starts at rest and is accelerated from an initial angle of zero to a final angle of 6.3 rad by a rad counterclockwise angular acceleration of 2.0 s2 What is the angular velocity at 6.3 rad?
The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Final angular velocity of the carnival ride
The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;
ωf = ωi + 2αθ
where;
ωf is the final angular velocity of the carnival ride = ?ωi is the initial angular velocity of the carnival ride = 0α is the angular acceleration = 2.0 rad/s²θ is the angular displacement of the carnival ride = 6.3 radωf = 0 + 2(2.0) x 6.3
ωf = 25.2 rad/s
Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
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Answer: 5.0 rad/s
Explanation: Because that’s what khan said so try it out.
Which of the following is NOT an example of Potential Energy? *
Chemical bonds
Gravitational energy
Electricity
Answer:
chemical bonds
Explanation:
because no energy is required
Which one the answer to this question
Tameika makes a table about sensory organs
Eye
skin
brain
tongue
Which organ should be removed from the table?
A. eye
B. skin
C. brain
D. tongue
Answer:
I think its d
Explanation:
I'm not sure I'm sorry if I'm wrong
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction
acting on the box is a constant 20 N. How much work is done by the gravitational force?
A. 03
OB. 10 J
C. 100
D. 50 J
Answer:
B i think
Explanation:
...
Which of the following best defines
weather?
A. the expanding or contracting of the atmosphere
B. the measurement of the amount of water vapor in the
atmosphere
C. the condition of the atmosphere at a certain time and
place
Help Resources
D. the average air temperature of a specific region
Answer:
I'd say D
Explanation:
because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)
If a virtual image is formed 10.0 cm along the principle axis from a convex mirror of focal length-15.0 cm, how far is the object from the mirror
Answer:
U=30cm
Explanation:
All you have to do is to put
Mirror formula , 1/f=1/u + 1/v
You should be careful in sign convention .
Virtual image is negative
we take focal length of convex lens negative even if its not given and so on...
