Answer:
the current value is [tex]0.335 \mu A[/tex]
Explanation:
The computation of the value of the current is given below:
[tex]z_i = I_s e^{\frac{0.7}{ut} }= 10^{-3}\\\\Z_z = I_s e^{\frac{0.5}{ut} }\\\\\frac{Z_z}{Z_i}= \frac{Z_z}{10^{-3}} = e^{\frac{0.5\times 0.7}{0.025} }\\\\= 0.335 \mu A[/tex]
Hence, the current value is [tex]0.335 \mu A[/tex]
Which action will leave the dump trucks inertia unchanged?? PLEASE ANSWER FAST!!!
A. add gas
B. increase force applied to engine
Answer:
B.
Explanation:
Can someone please help meee .
Answer:
32 amu is the right choice because both protons and neutrons have a mass of 1 amu. Electrons have no mass so go with the last choice
In the figure, given ∆x=30cm, k=200N/cm, g=10m/s². Find the mass of the object
Answer:
600 Kg
Explanation:
From the question given above, the following data were obtained:
Extention (∆x) = 30 cm
Spring constant (K) = 200 N/cm Acceleration due to gravity (g) = 10 m/s²
Mass (m) of object =?
Next, we shall determine force exerted. This can be obtained as follow:
Extention (∆x) = 30 cm
Spring constant (K) = 200 N/cm
Force (F) =?
F = K∆x
F = 200 × 30
F = 6000 N
Finally, we shall determine the mass of the object. This can be obtained as follow:
Acceleration due to gravity (g) = 10 m/s²
Force (F) = 6000 N
Mass (m) of object =?
F = mg
6000 = m × 10
Divide both side by 10
m = 6000 / 10
m = 600 Kg
Thus, the mass of the object is 600 Kg
if we ignore air resistance the mass of an object does not affect the rate at which it accelerate why?
Answer:
See explanation
Explanation:
The acceleration due to gravity on an object is independent of the mass of the object. This is so because, the acceleration due to gravity depends only on the radius of the earth and the mass of the earth.
As a result of this, all objects are accelerated to the same extent and should reach the ground at the same time when released from a height as long as other forces other than gravity are not at work.
28. Which of the following correctly shows the order of highest amount of friction to the lowest amount of
friction?
a. Static, Rolling, Sliding
b. Sliding, Rolling, Static
c. Rolling, Static, Sliding
d. Static, Sliding, Rolling
Answer:
[tex]\mathrm{d.\:Static,\: Sliding,\:Rolling}[/tex]
Explanation:
Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.
During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.
Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.
What is the gravitational force between two students, John and Mike, if John has a mass of 81.0 kg, Mike has a mass of 93.0 kg, and their centers are separated by a distance of .620 m?
Answer:
1.31×10¯⁶ N
Explanation:
From the question given above, the following data were obtained:
Mass of John (M₁) = 81 Kg
Mass of Mike (M₂) = 93 Kg
Distance apart (r) = 0.620 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force (F) =?
The gravitational force between the two students, John and Mike, can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 81 × 93 / 0.62²
F = 6.67×10¯¹¹ × 7533 / 0.3844
F = 1.31×10¯⁶ N
Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N
Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly walk 30.0 m west to a bench where you sit and watch the sunrise. It takes you 27.0 s to walk from your house to the windmill and then 47.0 s to walk from the windmill to the bench. For the entire trip from the front door to the bench, what are your :
a. average velocity
b. average speed
Answer:
Explanation:
Total displacement for entire trip = final position - initial position
= 50 m - 30 m = 20 m
Total time = 27 + 47 = 74 s
Average velocity = Total displacement / total time
= 20 / 74 = .27 m /s
Total distance covered in entire trip = 50 + 30 = 80 m
Total time = 74 s
Average speed = Total distance covered / total time
= 80 / 74 = 1.08 m /s .
describe measurement in our daily life
In the equation for the gravitational force between two objects, which quantity must be squared?
•mi
•m2
•G
•d
Answer:
d
Explanation:
The quantity that must be squared in the equation of gravitational force is distance d.
According to the universal gravitational law, the square of the distance between two objects is inversely proportional to the force of gravity.
Therefore, the quantity to be squared is dThe formula is given as:
Fg = [tex]\frac{G m_{1} m_{2} }{d^{2} }[/tex]
So d is the quantity that must be squared
On a slope where does a marble have to most kinetic energy?
a) it is always the same
b) at the initial position
c) at the final position
d) somewhere between the initial and the final position
Answer:
C
Explanation:
Kinetic energy is the energy of motion. It has the most potential energy at the top but the most kinetic at the bottom after it's accelerated fully down the slope.
Water will expand more than ___
A. door
B. juice
C. air
Answer:
Its A. door because liquids expand better than solids
An Egyptian pyramid contains approximately 1.95 million stone blocks. The average weight of each block is 2.55 tons. What is the weight of the pyramid in pounds?
Answer:
More than 2,300,000 limestone and granite blocks were pushed, pulled, and dragged into place on the Great Pyramid. The average weight of a block is about 2.3 metric tons (2.5 tons).
Some giant ocean waves have a wavelength of 25 m and travel at 6.5 m/s with a frequency of 0.26 HZ. What is the period of such a wave ?
Answer:
3.85s
Explanation:
Given parameters:
Wavelength = 25m
Velocity = 6.5m/s
Frequency = 0.26Hz
Unknown:
Period of the wave = ?
Solution:
The period of a wave is the inverse of the frequency of the wave.
Period = [tex]\frac{1}{frequency}[/tex]
Period = [tex]\frac{1}{0.26}[/tex] = 3.85s
If a woman walks at a speed of 5 miles/hour for 3 hours, she will have walked how many miles?
The distance walked by the woman at the given speed an time, is 15 miles.
What is meant by speed?The speed of an object is defined as the rate of change of the distance travelled by the object.
Here,
Speed with which the woman is walking,
v = 5 miles/hour
Time taken by the woman for walking,
t = 3 hours
We know speed is the rate of change of distance,
v = d/t where d is the distance travelled by the woman
So, d = v x t
d = 5 x 3
d = 15 miles
Hence,
The distance walked by the woman at the given speed an time, is 15 miles.
To learn more about speed, click:
https://brainly.com/question/17661499
#SPJ3
The main idea of this paragraph is:
a. Boots is my dog.
b. Boots can sit.
c. Boots is very smart.
d. Most dogs can't dance.
find the volume of an object with a density of 3.2 g/mL and a mass of 12 g.
1. 9mA electric current is flowing through a conducting wire. Then the number of electron
passing through it in 3mimute is?
A)2x10^18
B) 1X10^18
C) 2x10^19
D) 1.01x10^19
Answer:
use the formula negative -eE/A×t
A safety plug is designed to melt when the pressure inside a metal tank becomes too high. A gas
at 51.0 atm and a temperature of 23.0°C is contained in the tank, but the plug melts when the
pressure reaches 75.0 atm. What temperature did the gas reach?
a boy of mass 40kg sits at point 2m from the pivot of a see-saw . find the weight if a girl who can balance the see-saw by sitting at a distance of 3•2m from the pivot.( take g=10NKg)
Answer:
The weight of the girl is 250 N
Explanation:
Static Equilibrium
Static equilibrium occurs when an object is at rest, i.e., neither rotating nor translating.
In the static rotational equilibrium, the total torque is zero with respect to any rotational axis.
The torque applied by a force F perpendicular to a displacement X with respect to a reference rotating point is:
T = F*X
The seesaw will be in rotational equilibrium if the torque applied by the boy of mass m1=40 Kg at x1=2 m from the pivot is equal to the torque applied by the girl of unknown mass m2 at x2=3.2 m from the pivot.
The force applied by both children is their weight:
[tex]F_1 = W_1 = m_1g[/tex]
[tex]F_2 = W_2 = m_2g[/tex]
It must be satisfied:
[tex]m_1gx_1=m_2gx_2[/tex]
Simplifying:
[tex]m_1x_1=m_2x_2[/tex]
Solving for m2:
[tex]\displaystyle m_2=\frac{m_1x_1}{x_2}[/tex]
[tex]\displaystyle m_2=\frac{40*2}{3.2}[/tex]
[tex]m_2=25\ kg[/tex]
Her weight is:
[tex]\mathbf{W_2=25*10 = 250\ N}[/tex]
The weight of the girl is 250 N
Find the momentum of a 500,000 kg train that is stopped on the tracks?
a. O kg m/s
b. 250,000 kg m/s
c. 500,000 kg m/s
d. 16,000,000 kg m/s
Answer:
The answer should be A) 0m/s
Explanation:
It is stopped on the train tracks therefore it is not moving.
Please tell me if I am wrong because I'm not 100% sure on this. Hope it's right and that it helped you.
How many miles per day can you walk at a MODERATE Intensity level and your heart rate is 170?
Answer:
Not enough detail as it is very defendant on the person and a bunch of factors in health, but overall your heart rate shouldn't reach 170 as an adult walking at a moderate intensity level, that would be closer to extreme intensity.
Explanation:
A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.
Part A
What is the speed of the rock just before it hits the street?
Express your answer with the appropriate units.
Part B
How much time elapses from when the rock is thrown until it hits the street?
Express your answer with the appropriate units.
Answer:
A) v = 28.3 m/s
B) t = 4.64 s
Explanation:
A)
Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:[tex]v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h (1)[/tex]
Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:[tex]\Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)[/tex]
So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:Δh = 26.0 m + 14. 8 m = 40.8 m (3)Replacing now in (1), we can solve for vf, as follows:[tex]v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)[/tex]
B)
In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =02) Time elapsed from this point until it hits the street, with vo=0.For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:[tex]v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)[/tex]
Replacing by the givens in (5) and solving for Δt, we get:[tex]\Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)[/tex]
For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:[tex]\Delta h = \frac{1}{2} * g * t^{2} (7)[/tex]
Replacing by the givens and solving for t in (7), we get:[tex]t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)[/tex]
So, total time is just the sum of (6) and (8):t = 2.9 s + 1.74 s = 4.64 sExplain how momentum is determined and conserved.
ASAP!!
Explanation:
Momentum is conserved in the collision. Momentum is conserved for any interaction between two objects occurring in an isolated system.
If 478 watts of power are used in 14 seconds,how much work was done
Answer:
6692J
Explanation:
Power is defined as the rate at which work is being done.
So,
Power = [tex]\frac{workdone}{time }[/tex]
Work done = Power x time
Given parameters:
Power = 478watts
Time = 14s
So;
Work done = 478 x 14 = 6692J
3. As the mass of an object increases, the force of gravity
Answer:
As the mass of an object increases, the force of gravity increases as well.
Explanation:
Objects with more mass have more gravity. They work together.
Please answer this question I don't know how to do it.
Assuming no friction, how does the initial gravitational potential energy of
the marble on a downward slope compare to the final kinetic energy?
a) they are the same
b) the initial gravitational potential energy is greater than the final kinetic energy
c) the initial gravitational potential energy is less then the final kinetic energy
Answer:
a) They are the same.
Explanation:
Assuming no friction, there should be no energy transfer and thus the Law of Conservation of Energy says:
[tex]PE=KE,\\mgh=\frac{1}{2}mv^2[/tex]
These types of problems also disregard any air resistance the surface of the object may cause. Therefore, no energy is transferred and from the Law of Conservation of Energy, [tex]100\%[/tex] of energy is preserved.
Particle A with charge q and mass ma and particle B with charge 2q and mass
mb, are accelerated from rest by a potential difference AV and subsequently
deflected by a uniform magnetic field into semicircular paths. The radii of the
trajectories by particle A and B are R and 3R, respectively. The direction of
the magnetic field is perpendicular to the velocity of the particle. Determine
their mass ratio?
calculate the average speed of talias car during the trip
Answer:
We're no strangers to love
You know the rules and so do I
A full commitment's what I'm thinking of
You wouldn't get this from any other guy
I just wanna tell you how I'm feeling
Gotta make you understand
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
We've known each other for so long
Your heart's been aching but you're too shy to say it
Inside we both know what's been going on
We know the game and we're gonna play it
And if you ask me how I'm feeling
Don't tell me you're too blind to see
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
No, I'm never gonna give you up
No, I'm never gonna let you down
No, I'll never run around and hurt you
Never, ever desert you
We've known each other for so long
Your heart's been aching but
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
No, I'm never gonna give you up
No, I'm never gonna let you down
No, I'll never run around and hurt you
I'll never, ever desert you
Explanation:
RICK ROLLED
Heather drives her Super-Beetle around a turn on a circular track which has a radius of 200 m. The Super-Beetle has a mass of 1500 kg and the coefficient of static friction between the road and tires is 0.6.
a. What is the force of static friction the road can apply batore the car starts to selon (use Ft= uFn).
b. What is the maximum speed the car can travel before it would start to slide?
Answer:
a) The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.
Explanation:
a) Let suppose that the car is on a horizontal ground and travels at constant speed. The vehicle experiments a centripetal acceleration due to friction, which can be seen in the Free Body Diagram (please see image attached for further details). By Newton's Laws, we construct the following equations of equilibrium:
[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (1)
[tex]\Sigma F_{y} = N -m\cdot g = 0[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]N[/tex] - Normal force from ground to the car, measured in newtons.
[tex]v[/tex] - Maximum speed of the car, measured in meters per second.
[tex]R[/tex] - Radius of the circular track, measured in meters.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
By applying (2) in (1):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex] (3)
The force of static friction the road can apply in the car ([tex]f[/tex]), measured in newtons, is: ([tex]\mu_{s} = 0.6[/tex], [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
[tex]f = \mu_{s}\cdot m \cdot g[/tex]
[tex]f = (0.6)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]f = 8826.3\,N[/tex]
The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) Then, we calculate the maximum speed of the car by (3):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex]
[tex]\mu_{s}\cdot g = \frac{v^{2}}{R}[/tex]
[tex]v = \sqrt{\mu_{s}\cdot g\cdot R}[/tex]
If we know that [tex]\mu_{s} = 0.6[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 200\,m[/tex], then the maximum speed of the car can travel before it would start to slide is:
[tex]v =\sqrt{(0.6)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (200\,m)}[/tex]
[tex]v \approx 34.305\,\frac{m}{s}[/tex]
The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.