Answer:
it would decrease
Explanation:
f=1/T
Uranus (mass = 8.68 x 1025 kg) and its moon Miranda (mass = 6.59 x 1019 kg) exert a gravitational force of 2.28 x 1019 N on each other. How far apart are they? cs [?] x 10?'m Coefficient (green) Exponent (yellow) Enter
Answer:
Explanation:
F = GMm/d²
d = √(GMm/F)
d = √(6.674e-11(8.68e25)(6.59e19) / 2.28e19)
d = 1.29398e8 = 1.29 x 10^8 m center to center
Answer:
1.29 x 10^8 m apart
Explanation:
Works in Acellus!
The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A
We have that for the Question "" it can be said that Calculate the moment which must be applied to the handle of the screw to raise the block is
M = 7.30 N.mFrom the question we are told
The vertical position of the 100-kg block is adjusted by the screw activated wedge. Calculate the moment which must be applied to the handle of the screw to raise the block. The single thread screw has square threads with a mean diameter of 30 mm and advances 10 mm each complete turn. The coefficient of friction for the screw threads is 0.24, and the coefficient of friction for all the mating surfaces of the block and the wedge is 0.40. Neglect friction at the ball joint A
Generally the equation for the Block is mathematically given as
[tex]\sum Fy=0[/tex]
[tex]981cos21.80 = R_2cos53.6\\\\R_2=1535N[/tex]
the equation for the Wedge is mathematically given as
[tex]\sum Fx=0\\\\1535cos36.4=Pcos21.8\\\\P=1331N[/tex]
the equation for the Screw is mathematically given as
[tex]\beta = tan^{-1}*\frac{L}{2*\pi*r} \\\\\beta = tan^{-1}*\frac{10}{2*\pi*(15)} \\\\\\beta = 6.06\\\\\theta = tan^{-1}*0.25 \\\\\theta = 14.04\\\\\\Therefore\\\\\theta + \beta = 20.1\\\\[/tex]
Therefore
[tex]M = Pr tan (\theta + \beta)\\\\M = 1331(0.015) tan20.09\\\\M = 7.30 N.m[/tex]
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At the molecular level, as the kinetic energy increases, what happens to the temperature?
decreases
increases
stays the same
Answer: increases
Explanation:
Temperature is a measure of the average velocity of the molecular particles. The faster they go, the higher the temperature.
A stomp rocket takes 3.1 seconds to reach its maximum height.
- What is its initial velocity? (Do not use units. If the answer is negative, please put a
negative sign in front of the answer.)
- What is its maximum height? (Do not use units. If the answer is negative, please put a
negative sign in front of the answer.)
Answer:
Explanation:
Ignoring air resistance the time to rise will equal the time to fall and initial velocity will be the same magnitude as final velocity just before impact.
v = at
v = 9.8(3.1)
v = 30.38
v = 30 m/s
max height can be found knowing the velocity is zero at the top of its flight.
v² = u² + 2as
s = (v² - u²) / 2a
s = (0.00² - 30.38²) / (2(-9.8))
s = 47.089
s = 47 m
A stomp rocket takes 3.1 seconds to reach its maximum height then the initial velocity is given as v = 30 m/s and maximum height is 47.089 m.
What is Velocity?Velocity is defined as rate of change of position with respect to time.
SI unit of velocity is m/sec. Velocity is a vector quantity.
Given that in the question time taken by rocket to reach maximum height is 3.1 sec. Ignoring air resistance the time to rise will equal the time to fall and initial velocity will be the same magnitude as final velocity just before impact.
v = at
v = 9.8(3.1)
v = 30.38
v = 30 m/s
Max height can be found knowing the velocity is zero at the top of its flight.
v² = u² + 2as
s = (v² - u²) / 2a
s = (0.00² - 30.38²) / (2(-9.8))
s = 47.089
s = 47 m
So, the initial velocity is given as v = 30 m/s and maximum height is 47.089 m.
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A 1.1 kg ball drops vertically onto a floor, hitting with a speed of 23 m/s. It rebounds with an initial speed of 5.0 m/s. (a) What impulse acts on the ball during the contact
Hi there!
We know that:
I = Δp = m(vf - vi)
Plug in the given values. Remember to take into account direction ⇒ let the rebound velocity be positive and initial be negative.
I = 1.1(5 - (-23)) = 30.8 Ns
A thin piece of semiconducting silicon will be used to fabricate an electrical device. This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. Electrical contacts are placed at opposite ends of its length. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.
Required:
a. If the application of 1.0 volt to the contacts results in a current of 0.019 amps, what is the resistivity in (ohm-cm) of the material?
b. If the material's conductivity is due to doping with aluminum to a level of [Al]= 1x10^17 atoms/cm^3, what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material (assuming that the aluminum is fully ionized - i.e. all Al atoms donated electrons).
We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"
Answer:
Resistivity = [tex]1.754 ohm-cm[/tex]Conductivity = [tex]6.25*10^{25} cm^3/V-s[/tex]
From the question we are told
This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.
A) Resistivity is given as,
[tex]p = \frac{RA}{l}[/tex]
where,
[tex]R = \frac{V}{I}[/tex]
Therefore,
[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]
B) Conductivity is given as,
[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]
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EXAM ENDS IN 30 MINS
PLSSS HELPPP ILL MAKE U BRAINLIEST
Explanation:
F = Icurrent×length×Bfieldstrength×sin(angle field to wire)
in our case
Icurrent = 10 A
length = 0.02km = 20 meters
B = 10^-6 T
angle = 30 degrees.
F = (20 A)(20m)(10^-6 T)×sin(30) = 400× 10^‐6 ×0.5 N =
= 200 × 10^-6 = 2 × 10^‐4 N
c. Boat travels north then west
A boat travels 76.0 km due north in 8.0 hours then 56.0 km due west in 5.0 hours.
Determine the direction (as a bearing) of the average velocity (to 1 decimal places) of the boat in the 8 + 5 hour period.
PLEASE HELP!!
Answer:
Explanation:
θ = arctan(56.0/76.0) = 36.4° West of North
average velocity is √(56.0² + 76.0²) / (8 + 5) = 94.4/13 = 7.26 m/s
A 2457 kg car moves with initial speed of 18 ms-l. It is stopped in 62 m by its brakes.
What is the force applied by the brakes?
Answer:
Explanation:
The work of the brakes will equal the initial kinetic energy of the car
Fd = ½mv²
F = mv²/2d
F = 2457(18²) / (2(62))
F = 6,419.903...
F = 6.4 kN
A mass vibrates back and forth from the free end of an ideal spring of spring constant 20 N/m with an amplitude of 0.30 m. What is the kinetic energy of this vibrating mass when it is 0.30 m from its equilibrium position?
Hi there!
We can begin by using the work-energy theorem in regards to an oscillating spring system.
Total Mechanical Energy = Kinetic Energy + Potential Energy
For a spring:
[tex]\text{Total ME} = \frac{1}{2}kA^2\\\\\text{KE} = \frac{1}{2}mv^2\\\\PE = \frac{1}{2}kx^2[/tex]
A = amplitude (m)
k = Spring constant (N/m)
x = displacement from equilibrium (m)
m = mass (kg)
We aren't given the mass, so we can solve for kinetic energy by rearranging the equation:
ME = KE + PE
ME - PE = KE
Thus:
[tex]KE = \frac{1}{2}kA^2 - \frac{1}{2}kx^2\\\\[/tex]
Plug in the given values:
[tex]KE = \frac{1}{2}(20)(0.3^2) - \frac{1}{2}(20)(0.3^2) = \boxed{0 \text{ J}}[/tex]
We can also justify this because when the mass is at the amplitude, the acceleration is at its maximum, but its instantaneous velocity is 0 m/s.
Thus, the object would have no kinetic energy since KE = 1/2mv².
3) A force of magnitude Fx acting in the x-direction on a 2.00 kg particle varies in time as shown
in FIGURE 2. Find
a) The impulse of the force
b) The final velocity of the particle if it is initially at rest
c) The final velocity of the particle if it is initially moving along the x-axis with velocity
of -2.00 ms -1
Answer:
a) Impuise of force =F∗?(t) = area of F-T graph area= impulse =triangle + rectangle + triangle = 0.5*4*2 + 4*1 + 0.5*4*2 = 12 N-s (b) impulse = change in momentum \(= mExplanation:
An object of mass m is hanging by a string from the ceiling of an elevator. The elevator is moving down at constant speed. What is the tension of the string?
A. Zero
B. Equal to mg
C. Less than mg
D. Greater than mg
Answer:
D. Greater than mg
Explanation:
According to Newton’s second law of motion, the net force equals mass times acceleration. We are going to use a free body diagram (force diagram) to show that the equation of the motion is given by
T – mg = – ma
Thereby,
T = mg – ma
and the answer is: (d)
D. Greater than mg
_________________________________
(hopet his helps can I pls have brainlist (crown)☺️)
calculate the surface area of a box whose mass is 200 kg and exerts a pressure of 100 Pascal on the floor.
Answer:
Explanation:
If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change
What do alcohol, drugs, and tobacco all have in common?
All have some medicinal value.
All are harmful to the body.
All are depressants.
All are stimulants.
Answer:
all are harmful to the body
Question: A NEO distance from the Sun is 1.18 AU. What is its relative speed compared to Earth (round your answer to 3 decimal places)
Its relative speed compared to Earth is 0.921
The speed of the object v = 2πr/T where r = radius of orbit and T = period of orbit.
Let v = speed of earth, r = radius of earth orbit = 1 AU and T = period of earth orbit.
So, v = 2πr/T
Also, v' = speed of NEO, r' = radius of NEO orbit = distance of NEO from sun = 1.18 AU and T' = period of NEO orbit.
So, v' = 2πr'/T'
v'/v = 2πr'/T' ÷ 2πr/T
v'/v = r'/r × T/T'
From Kepler's law, T² ∝ r³
So, T'²/T² = r'³/r³
(T'/T)² = (r'/r)³
T'/T = √[(r'/r)]³
T/T' = √[(r'/r)]⁻³
So, substituting this into the equation, we have
v'/v = r'/r × T/T'
v'/v = r'/r × √[(r'/r)]⁻³
v'/v = √[(r'/r)]⁻¹
Since r' = 1.18 AU and r = 1 AU, r'/r = 1.18
So, v'/v = √[(r'/r)]⁻¹
v'/v = √[(1.18)]⁻¹
v'/v = [1.0863]⁻¹
v'/v = 0.921
So, its relative speed compared to Earth is 0.921
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Just before it strikes the ground, what is the watermelon's kinetic energy?
Answer:
Answer: At its lowest point, the kinetic energy of a watermelon just as it touches the ground is zero if it does not touch anything on its way down.
Explanation:
This is because that upon having been dropped from a height, an object no longer has any kinetic energy at all. Kinetic energy transforms to gravitational potential energy during the fall and there's nothing left over for kinetic once you stop accelerating anymore. Fortunately, things don't stay still until they land very often! For example, if a person catches the fruit with his hands after some air resistance slows him down - making him more similar in speed to the lag of the trajectory - then he'll be able to share some of his saved up gravitational potential with that watermelon and do some
Velocity and Acceleration Quick Check
C
D
E
During which of the labeled time segments is the object moving forward but slowing down?
(1 point)
Ο Α
0 С
OD
ОВ
Answer:
Explanation:
1 Object C has an acceleration that is greater than the acceleration for D.
2 B
3 17M
4 The velocity is zero.
5 a straight line with negative slope
just took it
100 J of work was done to lift a 10-N rock and set it at Position A near the edge of a cliff.
1. If the 100 Joules of work lifted the rock to the top of the cliff, how much potential energy did the rock gain?
2. At point C, the rock's potential energy will be
3. The rock's kinetic energy at point A is
4. At point B, some of the rock's potential energy will be changed to Kinetic energy
5. What is the mass of the rock?
6. What is the rock's velocity just before it hits the ground?
The rock to the right is sitting at the top of a ramp. I wonder how much work it required to get that rock up there.
Answer:
lol
Explanation:
(f) As the pions move away from each other, the ratio of the absolute value of electric potential energy to the final total kinetic energy of the two pions changes. At some point, the potential energy becomes negligible compared to the final total kinetic energy . We can consider that the value of the ratio is about 0.01 when , where is the final total kinetic energy of the two pions. What will the distance, , between the pions be when this criterion is satisfie
The distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m
If we consider the potential energy (U) between the pions, then (U) can be expressed as:
[tex]\mathbf{U = \dfrac{kq^2}{r} ---- (1)}[/tex]
Given that at some instance, the potential energy becomes negligible compared to the final K.E.
As such the conservation of the total energy in the system can be given as:
E = U + KAgain, if we consider the ratio of the potential energy to the kinetic energy to be about 0.01, then:
[tex]\mathbf{\dfrac{U}{K}= 0.01} \\ \\ \mathbf{U = 0.01 K----(2)}[/tex]
∴
Equating both equations (1) and (2) together, we have:
[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 K}[/tex]
[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }[/tex]
[tex]\mathbf{r =\dfrac{kq^2}{ 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }}[/tex]
where:
r = distancek = Columb's constantq = charge on a protonm_o = rest mass of each pion in the previous questionc = velocity of light[tex]\mathbf{v_\pi}[/tex] = calculated velocity of proton in the previous questionReplacing their values in the above equation, the distance (r) between the pions is calculated as:
[tex]\mathbf{r =\dfrac{(9\times 10^9 \ N.m^2/C^2) (1.6022 \times 10^{-19} \ C)^2}{ 0.01 \Bigg [ (2.5 \times 10^{-28\ kg } )\times (3\times 10^8 \ m/s)^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{(2.97 \times 10^8 \ m/s)^2}{(3 \times 10^8 \ m/s)^2} }} \Big ] \Bigg] }}[/tex]
distance (r) = 1.45 × 10⁻¹⁶ m
Therefore, we can conclude that the distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m
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The car on this ramp starts from rest. When released, it
accelerates at a constant rate. It has an initial position of 12 cm
from the top of the ramp, and has an average velocity of 1.20 m/s
for a total of 1.80 s. Which is the correct final position of the car?
Answer:
Explanation:
s 0.12 + 1.20(1.80) = 2.28 m from the top.
If the voltage across a 5-F capacitor is 2*e^-3
V find the current and the power
A cyclist on a training ride records the distance she travels away from home. The data only shows the first150 minutes of the ride before her cycling computer ran out of battery.
Answer:
A) 58 km
B) 30 mins
Explanation:
In pic details
graph in pic
A boat is using echo-sounding equipment to measure the depth of the water underneath it, as illustrated in the first diagram.
The equipment in the boat sends a short pulse of sound downwards and detects the echo after a time interval of 0.80s. i Describe how an echo is caused. ii The speed of sound in water is 1500 m/s. Calculate the distance travelled (in metres) by the sound in 0.80 s.
Answer:
Explanation:
Echo is caused by sound energy reflecting off of "hard" surfaces. It could be as simple as a change in density of the material the sound is traveling through.
In 0.8 s, the sound has traveled 0.8(1500) = 1200 m.
That means the object that reflected the sound is 600 m below the boat. The sound took 0.4 s to reach the object and another 0.4 s to return the echo.
Accelerations are produced by
A. Masses
B.accelerations
C. Velocities
D.unbalanced, net forces
The oscillation of the 2.0-kg mass on a spring is described by x = 3.0 cos (4.0 t) where x is in centimeters and t is in seconds. What is the force constant k of the spring?
The force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.
What is force?Force is the influence of either pull or pushes in the body. Basically, gravitation forces, nuclear forces, and friction forces are the types of forces. For e.g. when the wall is hit by a hand then a force is exerted by the hand on the wall as well as the wall also exerts a force on the hand. There are different laws given to Newton to understand force.
Newton is a unit of force used by physicists that is part of the International System (SI). The force required to move a body weighing one kilogram one meter per second is known as a newton.
Given:
The mass of the block, m = 2 kg,
The oscillation of spring, x = 3 cos 4t,
Calculate the omega by comparing the standard equation given below,
[tex]x = A cos \omega t[/tex]
ω = 4
Calculate the spring constant by the formula given below,
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
4² = k / 2
k = 32 N / m
Therefore, the force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.
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MCQ
A body of mass 5kg is pushed for distance x with accleration a. Then workdone against static friction is
1.ma*X cosB
2.ma*X sinB
3.zero
4.ma/X
Answer:
ma*XsinB
option 2 is correct
How much distance does a car travel with a speed of 2m/s in 15 min?
Explanation:
1 minute is 60 seconds so you multiply 60 * 15 and then multiply that answer * 2
A crane is lifting a 500 lb car. If the power of the crane 1.82 hp, find the velocity of the car.
Answer:
Explanation:
550 ft•lb/s / hp•(1.82 hp) / 500 lb = 2.00 ft/s
Which is the most famous book of the philosopher Alexis karpouzos? I think it is the ''Cosmology, philosophy and physics''.
Answer: Yes, the " Cosmology, philosophy and physics" is the most famous book of the philosopher, Alexis karpouzos. But and the other books are important. For example, the " The self-criticism of science", the "Universal conscilusness" and the "Non-duality".
Explanation:
A 2 kg ball is rolling down a hill at a constant speed of 4 m/s. How much kinetic energy does the ball have?