M = mass of the boy = 55kg
V = initial velocity of the boy = 12 m/s
m= mass of stationary skateboard = 15kg
v= velocity os stationary sketeboard= 0 m/s
V' = velocity of the boy on the skateboard after collision
Conservation of momentum:
MV + mv = (M + m) V'
Replacing:
55 kg * 12 m/s + 15 kg *0 = (55 kg+ 15 ) V'
Solve for V´'
660 = 70 V´'
660/70 = V'
V'= 9.42 m/s
What is the difference in the path of a ball tossed straight up in the air by a passenger on a a bus from the view point of the passenger, and the person on the street?
The difference in the path of a ball tossed straight up in the air by a passenger on a a bus from the view point of the passenger, and the person on the street is that the ball would appear to the passenger to be making an up and down movement when the ball is thrown up, while the stationary observer will actually see the ball moving along a parabolic path.
What is a parabolic path?A parabolic path is described as a Kepler orbit with the eccentricity equal to 1 and is an unbound orbit that is exactly on the border between elliptical and hyperbolic.
The Parabolic path is also defined as the angle of trajectory of a projectile.
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A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 4.0cm. If a timer is started when it's displacement is a maximum (hence x=4cm when t=0), what is the speed of the mass when t=3s ?
Since the frequency is 4 Hz, it completes one cycle in 1/4 seconds (its period, T, is 1/4 sec).
f = 1/T
Let's draw a graph of the displacement of the mass along time.
This pattern repeats, with each crest (high point) at every 1/4 second. This means that at t = 3, which is a multiple of 1/4, the mass is at its high point.
Notice that when the spring reaches its high or low point, it changes direction.
At this instant when it changes direction, its speed momentarily becomes 0.
That means that each multiple of 1/8 seconds, the speed is momentarily 0.
The speed of the mass is 0 at t = 3.
Using an electric current, you can split liquid water to form two new substances, hydrogen and oxygen gases. Is this a change in state. Explain your answer.
This process (called electrolysis) is not have a change of state. This comes from the fact that the substance (in this case water) does not keep its identity during the change, this means that we don't end up with the same substance; rather we have two new ones, hydrogen and oxygen.
An unbanked asphalt highway has turns of 40m radii. How fast should the speed limit be if cars may be traveling in the rain? (Us for wet asphalt on rubber is .755)
Given data:
* The radius of the turn is r = 40 m.
* The coefficient of friction is,
[tex]\mu_s=0.755[/tex]Solution:
The centripetal force acting on the car is,
[tex]F=\frac{mv^2}{r}[/tex]where m is the mass of the car,
The frictional force acting on the car is,
[tex]F_r=\mu_smg[/tex]where g is the acceleration due to gravity,
In order to travel a car in rain, the centripetal force acting on the car must be equal to the frictional force on the same car.
Thus,
[tex]\begin{gathered} F_r=F_{} \\ \mu_smg=\frac{mv^2}{r} \\ \mu_sg=\frac{v^2}{r} \\ v^2=r\mu_sg \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} v^2=0.755\times40\times10 \\ v^2=302 \\ v=17.4\text{ m/s} \end{gathered}[/tex]Thus, the maximum speed limit of the car in rain is 17.4 m/s.
Hence, the nearest possible correct answer is option b.
The inductor in the RLC tuning circuit of an AM radio has a value of 9 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 66 kHz
Given:
The inductance is,
[tex]\begin{gathered} L=9\text{ mH} \\ =9\times10^{-3}\text{ H} \end{gathered}[/tex]The radio frequency is,
[tex]\begin{gathered} f=66\text{ kHz} \\ =66\times10^3\text{ Hz} \end{gathered}[/tex]To find:
value of the variable capacitor, in picofarads
Explanation:
The frequency of the AM is,
[tex]\begin{gathered} f=\frac{1}{2\pi\sqrt{LC}} \\ \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} 66\times10^3=\frac{1}{2\pi\sqrt{9\times10^{-3}\times C}} \\ \sqrt{9\times10^{-3}\times C}=\frac{1}{2\pi\times66\times10^3} \\ \sqrt{9\times10^{-3}\times C}=2.41\times10^{-6} \\ 9\times10^{-3}\times C=5.81\times10^{-12} \\ C=6.45\times10^{-10} \\ C=645\times10^{-12}\text{ F} \\ C=645\text{ pF} \end{gathered}[/tex]Hence, the capacitance is 645 pF.
When you eat cereal and thenlift weights, how is the energytransformed?A. The chemical energy in the cereal istransformed into mechanical energy.B. The thermal energy in the cereal istransformed into mechanical energy.C. The mechanical energy in the cereal istransformed into chemical energy.
in any food not just in cereal there are chemical energy is stored in the form of
carbon, protein, fats, etc.
when we eat the food our body transforms this energy into
energy by decomposing the food into its elementary
particles and store it in the form of chemical energy and when we need energy
to do work it (body) coverts the energy into the mechanical energy by various
process.
so the correct answer is option (A)
A man takes in 0.95 × 107 J of energy each day from consuming food, and maintains a constant weight. what power in watt supplied by the food?
Given data
*The given energy is E = 0.95 × 10^7 J
*The given time is t = 1 day = (24 × 60 × 60) s = 86400 s
The formula for the power in watt supplied by the food is given as
[tex]P=\frac{E}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} P=\frac{(0.95\times10^7)}{(86400)} \\ =1.09\times10^2\text{ W} \end{gathered}[/tex]Hence, the power in watt supplied by the food is P = 1.09 × 10^2 W
if you had only one telescope and wanted to take both visible-light and ultraviolet pictures of stars, where should you locate your telescope?
If we just had one telescope and wanted to photograph stars in both visible and ultraviolet light, we should put it in space.
While visible light is observable from Earth, ultraviolet light can only be seen from space. Indeed, Hubble's ability to observe ultraviolet light gives it a major advantage over larger ground-based observatories.
Rank the visible light hues from left to right according to the altitude in the atmosphere where they are totally absorbed. All visible light wavelengths reach the Earth's surface, which is why we can see all colors and why visible-light telescopes perform well on the ground.
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When an object falls through air, there is a drag force that depends on the product
of the surface area of the object, and the square of its velocity, i.e. F=CAv2
, where C
is a constant, determine the dimension of C and its SI unit
The dimensions and unit of C from the expression above is [ML^-3] and kgm^-3
What is dimension and units ?
Despite the fact that the solution to any engineering problem must incorporate units, dimensions and units are frequently misunderstood. While units are arbitrary names that are correlated to certain dimensions to make it relative, dimensions are tangible quantities that can be measured (e.g., a dimension is length, whereas a metre is a relative unit that describes length). Through a conversion factor, all units for the same dimension are related to one another.
given in question drag force depends upon the product of the cross sectional area of the object and the square of its velocity
and drag force can be given by
F = CAv^2 ----------- (1)
It is given that
Dimension of mass = [M]
Dimension of length = [L]
Dimension of time = [T]
So, by using above dimension we can write
the dimension of force,
F = [ MLT^-2]
dimension of cross-section area,
A = [L^2]
and dimension of velocity
v = [LT^-1]
now, by putting these values in equation (1), we will get
F = CAv^2
[ MLT^-2] = C[L^2][LT^-1]^2
C = [ML^-3T^0]
C = [ML^-3]
The dimensions and unit of C from the expression above is [ML^-3] and kgm^-3
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Hence, the dimension of constant C will be,
Below is a diagram of a 5.0 kg block being dragged to the right, along a horizontal surface. The
coefficient of dynamic friction, is 0.40. Take acceleration due to gravity, g, as 9.81 ms ².
5 kg
30. N
What is the acceleration of the block? Give your answer correct to two significant figures, in
m-s-2, without units.
The acceleration of the block when the coefficient of friction is 0.40 is 3.924 m/s².
The ratio of the frictional resistive force to the perpendicular force pushing the objects together is known as the coefficient of friction.
Acceleration is the rate at which an object's velocity with respect to time changes.
The block of mass 5 kg is dragged to the right on a horizontal surface.
The coefficient of friction is 0.40.
The acceleration due to gravity is 9.81 m/s².
The force on the block is 30 N.
Now, the force is defined as the product of the mass and acceleration of the object.
Now, using the conservation of force:
F = μmg
ma = μmg
a = μg
a = 0.40 × 9.81
a = 3.924 m/s²
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ine? Bir10. Two people are pulling on opposite ends of a rope so that ithas a tension of 150 newtons. If the rope is not moving, withwhat pulling force is each of the two people pulling?
ANSWER
150 N
EXPLANATION
The rope is not moving, so the net force on it is 0. The force that each person exerts on the rope is equal and opposite to the tension on the rope, so the sum of the forces acting on it is zero.
Hence, the force that each of the two people is exerting on the rope while pulling is 150 Newtons.
A dentist causes the bit of a high speed drill to accelerate from an angular speed of 1.76 x 10^4 rads to an angular speed of 4.61 x 10^4 rat. In the process, the bit turns through 1.97 x 10 ^4 rad. Assuming a constant angular acceleration, how long would it take the reach its maximum speed of 7.99 x 10^4 rads starting from rest?
The time taken for the bit to reach the maximum speed is 1.35 seconds.
What is the angular acceleration of the bit?The angular acceleration of the bit is determined by applying the following kinematic equation as shown below.
ωf² = ωi² + 2αθ
where;
ωf is the final angular speedωi is the initial angular speedθ is the angular displacementα is the angular accelerationα = (ωf² - ωi²)/2θ
α = (46,100² - 17,600²) / (2 x 19,700)
α = 46,077.4 rad/s²
The time taken for the bit to reach the maximum speed is calculated as follows;
ωf = ωi + αt
t = (ωf - ωi) / α
t = (79,900 - 17,600) / (46,077.4)
t = 1.35 seconds
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Example of a balanced force
A card is drawn at random from a standard deck. Determine whether the events are mutually exclusive or not mutually exclusive. Then find each probability (2 or black card).
Consider that the event are no mutually exclusive because you can obtain a black card and 2 in one card.
In a standard deck you have 52 cards. There are 26 black cards and 4 cards with number 2.
Consider that in the 26 black cards there are two black cards with number 2.
Then, to get the probability consider that the required result can be obtained for 26 + 2 = 28 cards (26 black cards and two cards with number 2).
The probability is the quotient between the number of options over the number of cards:
[tex]p=\frac{28}{52}=0.54[/tex]Hence, the probability is 0.54
A/An _____ is described as a device that is used to measure potential difference across any part of a circuit.ammeterfuseground fault interruptervoltmeter
Answer:
[tex]\text{Voltmeter}[/tex]Explanation: We need to find an instrument that measures potential difference across any part of the circuit, a potential difference is basically voltage difference across a circuit difference and the device used to measure this difference is known as:
[tex]\text{ Voltmeter}[/tex]Voltmeter used in a circuit
If Hubble's constant had a value of 95 km/s/Mpc, what would be the age of the Universe?
The age of universe will be 0.0103 million years.
Hubble's law states that the rate at which any galaxy is receding from another galaxy is proportional to its distance from the galaxy. In simple form,
v = H₀ × d
where v is the velocity of the galaxy, d is its distance, and H₀ is the Hubble's constant.
Given: Hubble's constant, H₀ = 95 km/s/Mpc.
The term 1 / H₀ is called the Hubble's time and gives the age of the universe. That is, if 't' is the age of the universe, then
t = 1 / H₀.
Substitute the given value.
⇒ t = 1 / (95 km/s/Mpc)
(1 pc = 3.086 ×[tex]10^{13}[/tex] km)
⇒ t = 1 / (95/3.086 ×[tex]10^{13}[/tex] ) s
⇒ t = (1 / 30.784) ×[tex]10^{13}[/tex] s
⇒ t = 0.0325 ×[tex]10^{13}[/tex] s
(1 year = 3.154 ×[tex]10^{7}[/tex] s)
⇒ t = 10305.68 years
⇒ t = 0.0103 million years
Therefore, the age of universe will be 0.0103 million years.
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A string with both ends fixed is vibrating in its second harmonic. The waves have a speed of 36m/s and a frequency of 60Hz. The amplitude of the standing wave at an antinode is 0.6cm. calculate the amplitude of the motion of points on the string a distance of 30cm,15cmand 7.5cm on the left hand end of the string.
The amplitude of the motion of points on the string a distance of 30cm,15cmand 7.5cm is 0.0547m, 0.027m and 0.0137m respectively.
Amplitude is the maximum range of vibration or oscillation, measured from the equilibrium position
According to the equation of the second harmonic motion
A = sin (kx)
A = Amplitude
k = [tex]\frac{2\pi f}{v}[/tex] = [tex]\frac{2*\pi * 60}{36}[/tex] = 10.467
x = distance of the point
For x = 30 cm = 0.3 m
A = sin (kx)
A = Sin (10.467 * 0.3)
A = 0.0547 m
For x = 15 cm = 0.15 m
A = sin (kx)
A = Sin (10.467 * 0.15)
A = 0.027 m
For x = 7.5 cm = 0.075 m
A = sin (kx)
A = Sin (10.467 * 0.075)
A = 0.0137 m
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Determine the speed of the Earth in its motion around the Sun using Newton's Law of Universal Gravitation and centripetal force. Look up the values of the Earth's mass, the Sun's mass, and the average distance of Earth from the Sun; other than G, nothing else is needed
In order to determine the speed of the Earth, proceed as follow:
Consider that the centripetal force must be equal to the gravitational force between the Earth and the Sun (because guarantees the stability of the system):
[tex]F_g=F_c[/tex]Fg is the gravitational force and Fc the centripetal force. The expressions for each of these forces are:
[tex]\begin{gathered} F_g=\text{G}\frac{\text{mM}}{r^2} \\ F_c=ma_c=m\frac{v^2}{r} \end{gathered}[/tex]where,
G: Cavendish's constant = 6.67*10^-11 Nm^2/kg^2
m: Earth's mass = 5.97*10^24 kg
M: Sun's mass = 1.99*10^30kg
v: speed of Earth around the Sun = ?
r: distance between the center of mass of Earth and Sun = 1.49*10^8km = 1.49*10^11 m
Equal the expressions for Fg and Fc, solve for v, replace the previous values of the parameters and simplify:
[tex]\begin{gathered} \text{G}\frac{\text{mM}}{r^2}=m\frac{v^2}{r} \\ v^{}=\sqrt[]{\frac{GM}{r}} \\ v=\sqrt[]{\frac{(6.67\cdot10^{-11}N\frac{m^2}{\operatorname{kg}^2})(1.99\cdot10^{30}kg)}{1.49\cdot10^{11}m}} \\ v\approx29846.7\frac{m}{s} \end{gathered}[/tex]Hence, the speed of the Earth around the Sun is approximately 29846.7m/s
What is troubling about the circumstance when a coil is stationary and a magnet moves as compres to when a magnet is stationary and the coil moves?
When the magnet is moved, the galvanometer needle will deflect. It shows that the current is flowing in the coil. When the magnet moves into the coil, the needle deflects into one way, and if the magnet moves out of the coil, the needle deflects into the other way.
When the magnet is held stationary near, or even inside, the coil, no current will flow through the coil.
A 4-kg ball traveling westward at 25 m/s hits a 15-kg ball at rest. The 4-kg ball bounces east at 8.0 m/s. What is the speed and direction of the 15-kg ball? What is the impulse of the second ball?
Given:
The mass of the first ball is,
[tex]m_1=4\text{ kg}[/tex]The initial velocity of the first ball towards West is,
[tex]u_1=25\text{ m/s}[/tex]The mass of thr second ball is,
[tex]m_2=15\text{ kg}[/tex]the second object is initially at rest.
The final velocity of the first ball is,
[tex]v_1=-8.0\text{ m/s}[/tex]we are taking West as positive.
Applying momentum conservation principle we can write,
[tex]m_1u_1+m_2\times0=m_1v_1+m_2v_2[/tex]Substituting the values we get,
[tex]\begin{gathered} 4\times25+0=4\times(-8.0)+15\times v_2 \\ v_2=\frac{100+32}{15} \\ v_2=8.8\text{ m/s} \end{gathered}[/tex]THe final velocity of the second ball is towards East and the magnitude is 8.8 m/s.
The impulse of the Second ball is,
[tex]\begin{gathered} I=m_2v_2-m_2\times0 \\ =15\times8.8 \\ =132\text{ kg.m/s} \end{gathered}[/tex]As the speed of an object falling toward Earth increases, the gravitational potential energy of the object with respect to EarthA. IncreasesB. DecreasesC. Remains the same
Answer:
B. Decreases
Explanation:
When an object is falling toward Earth, the height of the object decreases, and the speed increases. Then, the gravitational potential energy decreases, and the kinetic energy increase because the first one depends on the height and the second one depends on the speed. Therefore, the answer is:
B. Decreases
The role or purpose of the battery in a circuit is to ____. Choose three.Group of answer choicesa) supply electric charge so that a current can existb) supply energy to the chargec) move the charge from the - to the + terminal of the batteryd) transform energy from electrical energy into light energye) establish an electric potential difference between the + and - terminalsf) replenish the charge which is lost in the light bulbg) offer resistance to the flow of charge so that the light bulb can get hot
Given:
The battery is connected to a circuit.
To find:
The purpose of the battery in the circuit
Explanation:
A battery is a source of energy in a circuit. The battery provides a push or a voltage to get the current flowing in a circuit.
Hence, the purposes are
b) supply energy to the charge
e) establish an electric potential difference between the + and - terminals
f) replenish the charge which is lost in the light bulb
Four masses are arranged as shown. They are connected by rigid, massless rods of lengths 0.780 m and 0.500 m. What torque must be applied to cause an angular acceleration of 0.750 rad/s2 about the axis shown?
Given,
The length of the rods;
L=0.780 m
l=0.500 m
The angular acceleration, α=0.750 rad/s²
The masses;
m_A=4.00 kg
m_B=3.00 kg
m_C=5.00 kg
m_D=2.00 kg
The moment of inertia of the given system of masses is given by,
[tex]\begin{gathered} I=\Sigma mr^2 \\ =m_A(\frac{L}{2})^2+m_B(\frac{L}{2})^2+m_C(\frac{L}{2})^2+m_D(\frac{L}{2})^2 \\ =(\frac{L}{2})^2(m_A+m_B+m_C+m_D) \end{gathered}[/tex]Where r is the distance between each mass and the axis of rotation.
On substituting the known values,
[tex]\begin{gathered} I=(\frac{0.780}{2})^2(4.00+3.00+5.00+2.00) \\ =2.13\text{ kg}\cdot\text{m}^2 \end{gathered}[/tex]The torque required is given by,
[tex]\tau=I\alpha[/tex]On substituting the known values,
[tex]\begin{gathered} \tau=2.13\times0.750 \\ =1.6\text{ Nm} \end{gathered}[/tex]Thus the torque that must be applied to cause the required acceleration is 1.6 Nm
A mass of 0.520 kilograms is attached to a spring and lowered into a resting position, causing the spring to stretch 18.7 centimeters. The mass is then lifted up and released. a. What is the spring constant of the spring? Include units in your answer.b. What is the frequency of its oscillation? Include units in your answer.Answer must be in 3 significant digits.
Given data
*The given mass is m = 0.520 kg
*The spring stretches at a distance is x = 18.7 cm = 0.187 m
*The value of the acceleration due to gravity is g = 9.8 m/s^2
(a)
The formula for the spring constant of the spring is given as
[tex]\begin{gathered} F=kx \\ mg=kx \\ k=\frac{mg}{x} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} k=\frac{(0.520)(9.8)}{(0.187)} \\ =27.2\text{ N/m} \end{gathered}[/tex]Hence, the spring constant of the spring is k = 27.2 N/m
(b)
The formula for the frequency of its
Energy transformations always produce a wasteful amount of energy called ?
Energy transformation always produce a wasteful amount of energy called heat energy.
Hence, the answer is heat energy.
Sodium has a density of 1.95 g/cm3. What is the volume of 56.2 g of sodium?
Answer: couldn't type in that little 3 haha
Explanation:
g) 0.35 oz to mgh) 75 mL to gali) 54 mi to kmj) 1789 ft to km
In order to covnert the given quantities, use the correct covnersion factor, as follow:
g) 0.35 oz to mg
[tex]0.35oz\cdot\frac{28.3495g}{1oz}\cdot\frac{1000mg}{1g}=9922.325mg[/tex]h) 75 mL to gal
[tex]75mL\cdot\frac{1L}{1000mL}\cdot\frac{0.2641722gal}{1L}\approx0.02gal[/tex]i) 54 mi to km
[tex]54mi\cdot\frac{1.61\operatorname{km}}{1mi}=86.94\operatorname{km}[/tex]j) 1789 ft to km
[tex]1789ft\cdot\frac{0.3048m}{1ft}\cdot\frac{1\operatorname{km}}{1000m}\approx0.54\operatorname{km}[/tex]
5. Draw a transverse wave with two wavelengths and label amplitude, crest, trough, and
equilibrium position.
A wave is considered to be transverse if its oscillations run counterclockwise to the wave's direction of advance. A longitudinal wave, on the other hand, moves in the direction of its oscillations. Transverse waves include water waves.
A waveform signal's wavelength, which is the distance between two identical locations (adjacent crests) in the succeeding cycles, determines whether it is sent through space or via a wire. This length is typically defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm).
The largest displacement or distance made by a point on a wave or vibrating body relative to its equilibrium position is its amplitude. It is equal to one-half of the vibration path.
The crest and trough of a wave, respectively, are its highest and lowest surface portions. The wave height is the vertical distance between the peak and trough. The wavelength is the horizontal separation between two consecutive crests or troughs.
The horizontal line at the wave's centre stands in for balance. A period is the length of time it takes to complete a cycle, which includes travelling from one peak to another, from one trough to another, or from one equilibrium point to another (both equilibrium points same direction).
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A parallel plate capacitor is charged by connecting it to a battery. After reaching steady state, the electric energy stored in the capacitor is UE . Select the correct claim about the amount of work needed to charge the capacitor.1) The amount of work needed to charge the capacitor is UE, because the work done is equal to the final potential energy of the system.2) No work is needed to charge the capacitor, because the charges will naturally move to their equilibrium position.3) The amount of work needed to charge the capacitor is 2UE , because it takes UE of work to remove the charges from the battery and and additional UE to place the charges on the capacitor.4) The amount of work needed to charge the capacitor is UE , because integrating the equation W=∫q dV yields the equation for the energy stored on a capacitor, Ue= 1/2qV
4. IS correct, that is the for mula used to calculate the potential energy stored
1. Is more or lees correct, but is less explained than the 4
2. Is false, is needed some energy
3. Is missunderstand, cause you are transfering energy from the battery to the capacitor, you dont have to count it twice
The safe loadof a wooden beam supported at both ends varies jointly as the width, w, the square of the depthd, and inversely as the length A wooden beam 7 inwide, 10 indeepand 19 ft long holds up 4422 What load would a beam inwide. 5 indeep and long of the same material support? (Round off your answer to the nearest pound )
L = k(wd^2/ l)
L = load
L1= 4422 lb
w = width
w1= 7 in
d = depth
d1= 10 in
l = lenght
l1= 19 ft
L2=
w2= 5in
d2=5in
l2= 11ft
First solve the constant with values 1
L1= k [(w1) (d1)^2 / l1]
4422= k [(7) (10)^2 / 19]
k = 4422 / [(7) (10)^2 / 19]
k= 120 lb/ft in^3
Replace with values 2
L2= k [(w2) (d2)^2 / l2]
L2= 120 [(5) (5)^2 / 11]
L2= 1363.63 LB