Uncle Harry's mass is 7.5 kg.
The weight of a body is defined as the gravitational force with which a body is attracted toward the center of the earth.
If [tex]g[/tex] is the acceleration due to gravity at a place, then a body of mass [tex]m[/tex] is attracted towards the center of the earth with a force equal to [tex]mg[/tex] at the place. Whereas, the mass of a body remains the same everywhere.
Hence the weight of a body is given by
[tex]W = mg[/tex] -------------- (1)
As the value of [tex]g[/tex] varies from place to place, the weight of a body also varies from place to place.
Given that, the weight of Uncle Harry, [tex]W=[/tex] [tex]75 N.[/tex]
From equation (1), we get mass as,
[tex]m=\frac{W}{g}[/tex] --------------(2)
Taking acceleration due to gravity, [tex]g= 10 m/s^{2}[/tex],.
Substitute for [tex]W[/tex] and [tex]g[/tex] in equation (2), we have,
[tex]m=\frac{75}{10}\\ \\m=7.5 kg[/tex]
Thus, the mass of Uncle Harry is [tex]7.5 kilograms.[/tex]
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A certain violet light has a wavelength of 413 nm. Calculate the frequency of the light. Calculate the energy content of one quantum of the light.
We will have the following:
First, we transform from nm to m,t that is:
[tex]\lambda=413nm\cdot\frac{m}{1\cdot10^9nm}\Rightarrow\lambda=4.13\cdot10^{-7}m[/tex]Then:
[tex]4.13\cdot10^{-7}m=\frac{3.0\cdot10^8m/s}{f}\Rightarrow f=\frac{3.0\cdot10^8m/s}{4.13\cdot10^{-7}m}[/tex][tex]\Rightarrow f\approx7.26\cdot10^{14}Hz[/tex]So, the frequency of the ligth is approximately 7.26*10^14 Hz.
Now, the energy will be:
[tex]E=(6.63\cdot10^{-34}m^2Kg/s)(7.26\cdot10^{14}/s)\Rightarrow E\approx4.81\cdot10^{-19}m^2Kg/s^2[/tex][tex]\Rightarrow E\approx4.81\cdot10^{-19}m^2Kg/s^2\cdot(J/m^2Kg/s^2))\Rightarrow E\approx4.81\cdot10^{-19}J[/tex]And the energy is approximately 4.81*10^-19 J.
I was wondering if I could get help matching these definitions?
We are asked to match the following terms.
1. Accretion: This is the phenomenon when layers of the matter is added to a celestial body due to gravity. Therefore, this is matched with: the "Process of the steady growth of an object by a steady accumulation of material".
2. Interstellar. This is a word of Latin origin that means between the starts.
3. Nebula: These are clouds of hydrogen ions where the stars are formed, therefore, this is matched with a "cloud of gas and dust in space".
4. Nuclear fusion: When the nuclei of atoms combine they emit energy. This is a reaction that occurs inside the stars and is responsible for its emission of radiation. Therefore, this is matched with "a nuclear reaction where nuclei combine and release intense energy".
5. Solar nebula: This is the nebula from which the solar system is formed, therefore, this is matched with: the "cloud of gas and dust from which our solar system is formed"
6. Stellar evolution: These are the changes that a star undergoes during its life cycle, therefore, it is matched with the "life cycle of a star".
7. Supernova. This is a stage in the stellar evolution when a star emits light and energy through an explosion, therefore, this is matched with "an explosion of a star that emits large amounts of matter and energy".
The cores of the terrestrial worlds are made mostly of metal because ______.
A. metals sunk to the centers a long time ago when the interiors were molten throughout
B. the terrestrial worlds as a whole are made mostly of metal
C. the core contained lots of radioactive elements that decayed into metals
D. over billions of years, convection gradually brought dense metals downward to the core
True or false. Just as like magnetic poles attract, unlike poles repel.
Answer: False
Unlike poles attract each other. Think of a south pole of a magnet and the north pole of another. They attract each other, while two of the same poles would repel each other.
A set of charged plates have anarea of 5.10*10^-3 m^2 andseparation 1.42*10^-5 m. Howmuch charge must be placed onthe plates to create a potentialdifference of 125 V across them?(The answer is *10^-7 C. Just fill inthe number, not the power.)
Given data:
Area of plates:
[tex]A=5.10\times10^{-3}\text{ m}^2[/tex]Separation between the plates:
[tex]d=1.42\times10^{-5}\text{ m}[/tex]Potential difference:
[tex]V=125\text{ V}[/tex]The capacitance of the capacitor is given as,
[tex]C=\frac{A\epsilon_{\circ}}{d}[/tex]Here, ε_o is the permittivity of the free space.
Substituting all known values,
[tex]\begin{gathered} C=\frac{5.10\times10^{-3}\times8.85\times10^{-12}}{1.42\times10^{-5}} \\ =3.178\times10^{-9}\text{ F} \end{gathered}[/tex]The charge on the capacitor is given as,
[tex]Q=CV[/tex]Substituting all known values,
[tex]\begin{gathered} Q=3.178\times10^{-9}\times125 \\ =3.9725\times10^{-7}\text{ C} \end{gathered}[/tex]Therefore, the charge on the plates is 3.9725×10^-7 C.
Thursday, February 11, 202110:08 AMCSA bullet travelling 420 m/s strikesa 1.5 kg target. How fast is thetarget moving if the bullet has amass of 0.1kg and is travelling300 m/s after the collision.
To solve this problem, we have to use conservation of momentum, which states that the initial momentum is equal to the final momentum. We know that momentum is defined as
[tex]p=mv[/tex]Let's apply it to the problem
[tex]\begin{gathered} p_{i1}+p_{i2}=p_{f1}+p_{f2} \\ m_1\cdot v_{i1}+m_2\cdot v_{i2}=m_1\cdot v_{f1}+m_2\cdot v_{f2} \end{gathered}[/tex]Using all the given information, we have
[tex]0.1\cdot420+1.5\cdot0=0.1\cdot300+1.5\cdot v_{f2}[/tex]Now, we solve the equation for v_f2:
[tex]\begin{gathered} 42=30+1.5\cdot v_{f2} \\ 42-30=1.5\cdot v_{f2} \\ 12=1.5\cdot v_{f2} \\ v_{f2}=\frac{12}{1.5} \\ v_{f2}=8 \end{gathered}[/tex]Hence, the target is moving 8 m/s after the collision.b. Calculate the electric force that exists between two objects that are 0.500 m apart and carrycharges of 0.00450 C and 0.00240
Given
The two charges,
[tex]\begin{gathered} q_1=0.00450\text{ C} \\ q_2=0.00240\text{ C} \end{gathered}[/tex]Distance between them,
[tex]r=0.5\text{ m}[/tex]To find
The electric force
Explanation
We know the force is given by
[tex]F=\frac{kq_1q_2}{r^2}[/tex]Putting the values,
[tex]\begin{gathered} F=9\times10^9\frac{0.00450\times0.00240}{(0.500)^2} \\ \Rightarrow F=388,800\text{ N} \end{gathered}[/tex]Conclusion
The electric force is 388,800 N
Which one would repel a negatively charged object?+АBС
Configuration A will repel a negative charged object.
This comes from the fact that opposite charges attract each other while equal charges repel each other. Now, we notice that configuration B is neutral (it has the same amount of positive and negative charges) then it won't have an effect on a negative charged object; meanwhile configuration C has more positive charges, then it would attract a negative charged object. Configuration A has more negative charges then it will repel a negative charged object.
if a rocket travles 5600km in 3 hours, what is its speed?
Given the distance traveled by rocket is
[tex]D=5600[/tex]The time taken by the rocket is 3 hours.
[tex]T=3\text{ hours}[/tex]Calculate the speed of the rocket.
[tex]\begin{gathered} \text{Speed = }\frac{5600km}{3\text{ hours}} \\ \text{Speed = 1866.66 }\frac{km}{hr} \end{gathered}[/tex]Thus, the time taken by the rocket is 1866.66 km/hr.
Find the power output of the engine of a 1200 kg car while the car accelerates from 30 km/h to 100 km/h in 10 s.
According to work energy thereom, the work done on the car to accelerate is,
[tex]W=\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]Plug in the known values,
[tex]\begin{gathered} W=\frac{1}{2}(1200kg)(100km/h)^2(\frac{1000\text{ m}}{1\text{ km}})^2(\frac{1\text{ h}}{3600\text{ s}})^2(\frac{1\text{ J}}{1kgm^2s^{-2}})-_{}\frac{1}{2}(1200kg)(30km/h)^2(\frac{1000\text{ m}}{1\text{ km}})^2(\frac{1\text{ h}}{3600\text{ s}})^2(\frac{1\text{ J}}{1kgm^2s^{-2}}) \\ =462963\text{ J-}41667\text{ J} \\ =421296\text{ J} \end{gathered}[/tex]The power output of the engine can be given as,
[tex]P=\frac{W}{t}[/tex]Substitute the values,
[tex]\begin{gathered} P=\frac{421296\text{ J}}{10\text{ s}}(\frac{1\text{ W}}{1\text{ J/s}}) \\ =42129.6\text{ W} \end{gathered}[/tex]Therefore, the power output of the engine is 42129.6 W.
An object is dropped from the top of a building. How fast is it moving after 5s? What is the acceleration at this time?
Any object under free fall accelerates at a constant rate given by the gravitational acceleration:
[tex]g=9.81\frac{m}{s^2}[/tex]On the other hand, the speed v of an object under free fall after t seconds, if it starts from rest, is given by the formula:
[tex]v=gt[/tex]Replace g=9.81m/s^2 and t=5s to find the speed of the object 5 seconds after the object is dropped:
[tex]v=(9.81\frac{m}{s^2})(5s)=49.05\frac{m}{s}[/tex]Therefore, the speed of the object after 5 seconds is approximately 49 meters per second. Its acceleration is always the same and it is equal to 9.81 meters per second squared.
What is the potential difference across a curling iron if there is a current of 2.5 A transferring 9360 J in 32s?
Power is expressed as energy/time
Thus means that
power = energy/time
From the information given,
energy = 9360 J
time = 32 s
Thus,
power = 9360/32 = 292.5 J/s = 292.5 watts
The formula relating potential difference which is the same as voltage, current and power is expressed as
P = IV
where
P = power = 292.5
I = current = 2.5
V = potential difference = ?
Thus, we have
292.5 = 2.5V
V = 292.5/2.5
V = 117 V
The potential difference is 117 V
A person is pushing on a car with a force of <-14,4> and another person is pushing on a car with a force of -6,1> What is the net force in the vertical direction?
We are given the following vector forces:
[tex]\begin{gathered} F_1=(-14,4) \\ F_2=(-6,1) \end{gathered}[/tex]These vector forces are expressed as:
[tex]F=(x,y)[/tex]Where the first component "x" is the horizontal component, and "y" is the vertical component. Since we are asked about the net force in the vertical component we need to add the components "y" of both vectors, we get:
[tex]\Sigma F_y=4+1=5[/tex]Therefore, the net force in the vertical direction is 5.
Which of the following choices correctly ranks the colors of visible light from the lowest to the highest frequency?Select one:a. Ab. Bc. Cd. D
d.D
Explanation
Visible light is the small part of the electromagnetic spectrum that we can see.
Colors exist at different wavelengths from lowest energy to highest energy:
The more energy a wave has, the higher its frequency, and vice versa. When it comes to visible light, the highest frequency color, which is violet, also has the most energy. The lowest frequency of visible light, which is red, has the least energy
therefore, the visible colors from the lowest to the highest frequency are
therefore, the answer is
d.D
I hope this helps you
If you drop a 2.3kg ball from the top of a 44 m high building, how much potential energy will it have just before it hits the ground? Round your answer to the nearest whole number and include an appropriate unit
Answer: Potential energy = 0
Explanation:
The formula for calculating potential energy is expressed as
Potential energy = mgh
where
m = mass of object
g = acceleration due to gravity = 9.8m/s^2
h = height of object above the ground
From the information given,
m = 2.3
h = 44
Just before the object hits the ground, all potential energy is converted to kinetic energy. Thus, the answer is
Potential energy = 0
The amount of energy in a photon is directly proportional to the ________________ of the electromagnetic wave, according to this equation:a) volumeb) none of these answers are correctc) directiond) motione) frequencyf) wavelength
Given:
[tex]E=h\nu=hf[/tex]Explanation:
The amount of energy in a photon is given as:
[tex]E=h\nu=hf[/tex]Here, E is the energy, h is Planck's constant, and ν (or f) is the frequency of the electromagnetic wave.
As h is a constant, we see that the energy E has a direct dependence on the frequency ν (or f) of the electromagnetic wave. In other words, the energy E of the photon is directly proportional to the frequency ν (or f) of the electromagnetic wave.
Final answer:
The amount of energy in a photon is directly proportional to the frequency of the electromagnetic wave.
Thus, the correct option is (e) frequency.
A 5 kg mass, hung onto a spring, causes the spring to stretch 7.0 cm. What is the spring constant? What is the potential energy of the spring?
Given,
The mass is m=5 kg.
The extensionis d=7.0 cm
The force is
F=mg
F=5x0.07=0.35N
Thus the spring constant is:
[tex]k=\frac{F}{d}=\frac{0.35}{0.07}=\frac{5N}{m}[/tex]The potential energy is:
[tex]U=\frac{1}{2}kd^2=\frac{1}{2}\times5\times(0.07)^2=0.012Nm^2[/tex]A block weighing 200N is pushed along a surface. If it takes 80N toget the block moving and 40N to keep the block moving at a constantvelocity, what are the coefficient of friction us and uk ?
Answer:
see below
Explanation:
coefficient of static F = 80 / 200 = .4
kinetic F = 40 /200 = .2
A yo‑yo with a mass of 0.0600 kg and a rolling radius of =1.60 cm rolls down a string with a linear acceleration of 5.20 m/s2.
Calculate the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.
What is the moment of inertia of this yo‑yo?
The tension magnitude in the string is 0.312 N.
The angular acceleration magnitude of the yo‑yo is 325 rad/s².
The moment of inertia of the yo-yo is 7.68 x 10⁻⁶ kgm².
What is the moment of inertia of the yo-yo?
The moment of inertia of the yo-yo is calculated by applying the following equation as shown below;
I = ¹/₂MR²
where;
M is the mass of the yo-yoR is the radius of the yo-yoI = ¹/₂(0.06 kg)(0.016)²
I = 7.68 x 10⁻⁶ kgm²
The angular acceleration of the yo-yo is calculated as follows;
α = a/R
where;
a is the linear accelerationα = 5.2/0.016
α = 325 rad/s²
The tension in the string is calculated as follows;
T = ma
where;
m is the mass of the yo-yoa is the linear acceleration of the yo-yoT = 0.06 kg x 5.2 m/s²
T = 0.312 N
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relationship between force of failure and diameter if ultimate tensile strength is same for all material tested i need equation
The relationship between the force and area of the material is,
[tex]F_f=\frac{P}{A}[/tex]where P is the load and A is the area.
As the tensile strength of the material is same for all the material tested.
As the area of the material is directly proportional to the diameter.
Thus, the force of failure is inversely proportional to the diameter.
An object weighing 297 N in air is immersedin water after being tied to a string connectedto a balance. The scale now reads 263 N.Immersed in oil, the object appears to weigh270 N.Find the density of the object.Answer in units of kg/m^3Do not round.
Given
Wair = 297 N
Wwater = 263 N
Density of water = 1000 kg/m^3
Acceleration due to gravity = 9.8
Procedure
We can calculate the thrust of the water with the difference of weights
[tex]\begin{gathered} W_{\text{air}}-W_{\text{water}}=\rho_wgV \\ V=\frac{W_{\text{air}}-W_{\text{water}}}{\rho_wg} \\ V=\frac{297N-263N}{1000\cdot9.8} \\ V=0.00347\text{ m}^3 \end{gathered}[/tex]Now for the object
[tex]\begin{gathered} \rho_{\text{obj}}=\frac{W_{\text{air}}}{gV} \\ \rho_{\text{obj}}=\frac{297}{9.8\cdot0.00347} \\ \rho_{\text{obj}}=8735.29kg/m^3 \end{gathered}[/tex]
A pitcher throws a 0.140 kg baseball with a speed of 42.3 m/s the batter strikes it with an average force of 5120 N which result in the ball traveling with an initial speed of 31.0 m/s toward the pitcher for how long were the bat and ball in contact
The impulse exerted over an object is equal to the change in its linear momentum:
[tex]I=\Delta p[/tex]On the other hand, the impulse is equal to the force exerted over the object multiplied by the time during which the force was exerted:
[tex]I=F\cdot\Delta t[/tex]First, find the impulse by finding the change in linear momentum of the baseball. The linear momentum is given by the product of the speed of the baseball times its mass:
[tex]p=mv[/tex]Assume that the negative direction is towards the batter and the positive direction is towards the pitcher. Then, the initial velocity of the ball is -42.3 m/s and the final velocity of the ball is 31.0 m/s. Then:
[tex]\begin{gathered} p_i=mv_i=(0.140\operatorname{kg})(-42.3\frac{m}{s})=-5.922\operatorname{kg}\cdot\frac{m}{s} \\ \\ p_f=mv_f=(0.140\operatorname{kg})(31.0\cdot\frac{m}{s})=4.34\operatorname{kg}\cdot\frac{m}{s} \end{gathered}[/tex]Use the initial and final linear momentum to find the change in linar momentum, which is equal to the impulse:
[tex]\begin{gathered} I=\Delta p \\ =p_f-p_i \\ =(4.34\operatorname{kg}\cdot\frac{m}{s})-(-5.922\operatorname{kg}\cdot\frac{m}{s}) \\ =10.262\operatorname{kg}\cdot\frac{m}{s} \end{gathered}[/tex]Isolate Δt from the equation that relates force and impulse and substitute the corresponding values for I and F to find the time during which the bat and the ball were in contact:
[tex]\begin{gathered} \Delta t=\frac{I}{F} \\ =\frac{10.262\operatorname{kg}\cdot\frac{m}{s}}{5120N} \\ =0.00200429687\ldots s \\ \approx0.00200s=2.00ms \end{gathered}[/tex]Therefore, the bat and the ball were in contact during a time interval of 2 miliseconds.
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BRAINLIEST!!!! AND 100 POINTS!!!!!!
Answer:
Its the fourth/last answer
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it is the symbol of dot (.)
The inductor in the RLC tuning circuit of an AM radio has a value of 10 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 97 kHz
Given:
The inductance is I = 10 mH
The frequency is f = 97 kHz
Required:
Value of capacitor in picofarads.
Explanation:
In order to tune the radio, the condition is
[tex]X_L=X_c[/tex]Here, X_L is the inductive reactance and
X_C is the capacitive reactance.
The capacitance can be calculated as
[tex]\begin{gathered} 2\pi fL=\frac{1}{2\pi fC} \\ C=\frac{1}{(2\pi f)^2L} \\ =\frac{1}{(2\times3.14\times97\times10^3)^2\times10\times10^{-3}} \\ =2.69\text{ }\times10^{-10}\text{ F}\times\frac{10^{12}\text{ pF}}{1\text{ F}} \\ =269\text{ pF} \end{gathered}[/tex]Final Answer: The value of the variable capacitor is 269 picoFarad.
3. 1.819 m4. 5.291 m5. 6.321 m39. Answer: B40. Work done by the non conservative forces actingon an object is equal1.to the change in the mechanical energy of the object2. to the change in the kinetic energy of the object3. to the work done by the conservative forces4. to the change in the potential energy of the object5. to the net work done on the object41. Answer: A
We are asked to determine the distance that spring will stretch when a given mass is attached to it.
To do that we will use Hook's law:
[tex]F=kx[/tex]where:
[tex]\begin{gathered} F=\text{ force } \\ k=\text{ spring constant} \\ x=\text{ distance that the spring is stretched} \end{gathered}[/tex]We will determine the constant of the spring "k" first using the fact that the spring is stretched 0.8 meters when a mass of 3kg hangs from it.
Since the only force acting on the spring is the weight of the object we have:
[tex]F=mg[/tex]Where:
[tex]\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now, we substitute and we get:
[tex]mg=kx[/tex]Now, we divide both sides by "x":
[tex]\frac{mg}{x}=k[/tex]Now, we plug in the values:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{0.8m}=k[/tex]Solving the operations:
[tex]36.75\text{ N/m}=k[/tex]Now, we substitute the value of "k":
[tex]F=(36.75\text{ N/m\rparen}x[/tex]Now, we solve for "x":
[tex]\frac{F}{36.75\text{ N/m}}=x[/tex]Now, we substitute the value of the weight of the second object:
[tex]\frac{(14kg)(9.8\frac{m}{s^2})}{36.75\text{ N/m}}=x[/tex]Solving the operations:
[tex]3.733m=x[/tex]Therefore, the spring will stretch by 3.733 meters.
Three facts about voltage sources
The voltage source produces potential difference across two ends of a conductor.
The voltage source has two terminals - positive and negative.
Voltage sources can be alternating or direct.
If a planet was located approximately 25 thousand light-years from the center of a galaxy and orbits that center once every 247 million years, how fast is the planet traveling around the galaxy in km/hr? If needed, use 3.0 × 10^8 m/s for the speed of light.
The distance of the planet from the centre i.e. radius,
[tex]r=25000\text{ light years}[/tex]The time period is
[tex]T=247\text{ million years}[/tex]The speed of the planet is given by the formula
[tex]v=\frac{2\pi r}{T}[/tex]Substituting the values, the speed will be
[tex]\begin{gathered} v=\frac{2\times\pi\times25000\text{ }\times9.461\times10^{12}}{247\times10^6\times876} \\ =6.866\times10^5\text{ km/hr} \end{gathered}[/tex]An air-filled pipe is found to have successive harmonics at 800 Hz , 1120 Hz , and 1440 Hz . It is unknown whether harmonics below 800 Hz and above 1440 Hz exist in the pipe. The length is 53.5 cm. Identify the correct pressure variation graph for the 1120 Hz standing wave in the pipe. Note that the closed end of the pipe is on the right.
Using the concept of Harmonic-wave, we got the desired graph which is shown in the image.
It must be an open-closed pipe.
Open-closed pipe
Let the fundamental frequency of f be
The second harmonic is 3f.
3rd harmonic = 5f
Therefore, the difference between harmonics is 3f-f = 2f
So, 2f = (1120-800) = 1440-1120 = 320
Also, f = 160Hz
800Hz = 5th Harmonic
1120 Hz = 7th
1440 = 9th
For an open- or closed-pipe,
f = v/4L
where v is the speed of sound in the air = 343 M/s
So, 160 = 343/4L
Also, L = 0.5336 m = 53.6cm
Hence for the frequency of 1120Hz, we got the desired graph of a wave which is shown in the below graph.
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Find the net force on the 4th part. Given, the value of velocity remain same at every point.
In the fourth case, when the weight and the tension are acting vertically downward.
The net force acting on the bucket is,
[tex]F_{\text{net}}=\frac{mv^2}{r}[/tex]where v is the velocity, r is the radius, and m is the mass,
The net force acting on the bucket will remain the same if the magnitude of the velocity remains the same.
As the direction of the bucket changes thus, the value of net force is not equal to zero.
As the velocity of the bucket is 3 m/s.
Thus, the value of net force acting on the bucket is,
[tex]\begin{gathered} F_{\text{net}}=\frac{0.5\times3^2}{0.6} \\ F_{\text{net}}=7.5\text{ N} \end{gathered}[/tex]Thus, (at the constant velocity at every point case) the net force acting is 7.5 N.