Answer:
nothing
Explanation:
bocouse of darkness
Hydroplaning causes your front wheels to actually leave the pavement and the wheels are riding on a thin layer of water. This lack of traction with the pavement takes away your control of the vehicle. To regain control of your vehicle you should___________.
Answer: Remove your foot from the gas pedal and slow down
Explanation:
Hydroplaning also refers to aquaplaning and this refers to the scenario whereby there's a layer of water that builds between the tyres of a car and the surface of the road which then brings about a loss of traction which eventually leads to a situation whereby the car doesn't respond to control inputs.
This lack of traction with the pavement takes away your control of the vehicle and to regain control of the vehicle, one should remove your foot from the gas pedal and slow down.
At the end of an investigation, you must__________ ____________. Your results may or may not support your hypothesis.
Answer:
could and largejsjisj and we look like they can get to
I need help. Please explain
Answer:
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Explanation:
A scientist measures the light from a distant star
at 525 nm. The constant for Wien's
displacement law is 2.9 x 10-3 m K. What is the
approximate temperature of the star in Kelvins?
A) 1500 K
B) 180,000 K
C) 1.5 K
D) 5500 K
The approximate temperature of the star as determined is D) 5500 K.
The Wien's displacement law relates the maximum wavelength of a body to its absolute temperature. Wien's displacement law states that:
λ = [tex]\frac{b}{T}[/tex]
where λ is the maximum wavelength of the body, b is the constant of proportionality and T is the absolute temperature.
Thus from the given question, λ = 525 nm (525 x [tex]10^{-9}[/tex]), and b = 2.9 x [tex]10^{-3}[/tex] mK.
So that,
525 x [tex]10^{-9}[/tex] = [tex]\frac{2.9*10^{-3} }{T}[/tex]
Make T the subject of the formula to have;
T = [tex]\frac{2.9*10^{-3} }{525*10^{-9} }[/tex]
= 5523.81
T = 5523.81 K
T ≅ 5500.00 K
The approximate temperature of the star in Kelvin is 5500 K.
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Write any three importance of Measurement.
Stay Safe,Stay Healthy and Stay Happy
Please answer me this question fast, PLEASE
Explanation:
The three importance of measurement are as follows :-
measurement is used to compare the items when barter system takes place .measurement helps in weighing the foods , groceries etc .Measurement are important in laboratory when there experiments is being taken for different purposes .Hope it is helpful to you
✌️✌️✌️✌️✌️✌️✌️
When 24.0 V is applied to a
capacitor, it stores 3.92 x 10-4 J of
energy. What is the capacitance?
[?] x 10!? E
[tex]\boxed{\sf E=QV^2}[/tex]
[tex]\\ \sf\longmapsto Q=\dfrac{E}{V^2}[/tex]
[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{24^2}[/tex]
[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{576}[/tex]
[tex]\\ \sf\longmapsto Q=0.006\times 10^{-4}C[/tex]
[tex]\\ \sf\longmapsto Q=6\times 10^{-1}C[/tex]
[tex]\\ \sf\longmapsto Q=0.6C[/tex]
Now
[tex]\boxed{\sf Q=CV}[/tex]
[tex]\\ \sf\longmapsto C=\dfrac{Q}{V}[/tex]
[tex]\\ \sf\longmapsto C=\dfrac{0.6}{24}[/tex]
[tex]\\ \sf\longmapsto C=0.025F[/tex]
Note:-
SI unit of charge is Coulomb(C)SI unitvof Capacitance is Farad(F)How does 'g' vary from place to place?
Explanation:
The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.
Which of the following changes would not lead to changes in the efficiency of
a heat engine?
A. Doubling the work done while keeping the heat flow into the
engine the same
B. Doubling the heat flow into the engine while halving the work done
C. Doubling both the work done and the heat flow into the engine
D. Doubling the heat flow into the engine while keeping the work
done the same
Explanation:
B. Doubling the heat flow into the engine while halving the work done
hope this helps you
have a nice day
A toy car with a mass of 5.5 kg is moving horizontally over flat ground at a speed of 2.1 m/s. An unknown force then directly pushes the car for a distance of 3 meters, after which the car has a speed of 7.3 m/s. You may assume that air resistance and friction are both negligible. What was the magnitude of the unknown force
Answer:
The magnitude of the unknown force is 44.8 N.
Explanation:
The force can be found with Newton's second law:
[tex] F = ma [/tex]
Where:
m: is the mass of the toy car = 5.5 kg
a: is the acceleration
F: is the force =?
We can calculate the acceleration with the following kinematic equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex] v_{f} [/tex]: is the final speed = 7.3 m/s
[tex] v_{0} [/tex]: is the initial speed = 2.1 m/s
d: is the distance traveled = 3 m
Hence, the acceleration is:
[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{(7.3 m/s)^{2} - (2.1 m/s)^{2}}{2*3 m} = 8.15 m/s^{2} [/tex]
Finally, the magnitude of the force is:
[tex]F = ma = 5.5 kg*8.15 m/s^{2} = 44.8 N[/tex]
Therefore, the magnitude of the unknown force is 44.8 N.
I hope it helps you!
standard unit definition
Answer:
Standard units are the units we usually use to measure the weight, length or capacity of objects.
Answer:
THE SYSTEM OF UNITS WHICH IS AGREED BY THE INTERNATIONAL CONVENTION OF SCIENTISTS HELD IN FRANCE IN 1960 IS CALLED SI SYSTEM.
How does electric force affect an atom?
O A. It keeps the neutrons in the atom.
B. It pushes the neutrons away from the protons.
O c. It pushes the electron away from the protons.
D. It keeps the electrons in the atom.
why a person feel weightlessness in a spacecraft orbiting around a heavenly body
Answer:
The orbital velocity an aircraft orbiting around a heavenly body is found as follows;
At the orbital velocity, [tex]F_G[/tex] = [tex]F_C[/tex]
Where;
[tex]F_G[/tex] = The gravitational force = [tex]\dfrac{G \cdot M \cdot m}{R_E^2}[/tex]
[tex]F_C[/tex] = The centripetal force = [tex]\dfrac{m \cdot v_0^2}{R_E}[/tex]
Therefore
[tex]v_0 = \sqrt{\dfrac{G \cdot M}{R_E} }[/tex]
Therefore, at the orbital velocity of the spacecraft, the centripetal force attracting the person away from the central region heavenly body is equal to the gravitational force pulling the person towards the center of the heavenly body (which was felt as her or his weight), and the person feels weightless while inside the orbiting spacecraft
Explanation:
1. How much heat energy ( Q ) is required to heat 2.0 kg of copper from 30.0 oC to 80.0 oC?
Answer:
38500
Explanation:
I looked it up so it may be wrong
Which of the following is NOT a type of wave?
a.
deep-water wave
c.
shallow-water wave
b.
open wave
d.
storm surge
Please select the best answer from the choices provided
A
B
C
D
Answer:
I think deep-water wave is not a wave
Explanation:
but it will be better if u wait for the second answer as I am not sure
Shallow water wave is not a type of wave. Thus, the correct option is C.
What is a Wave?A wave is a disturbance in a medium which carries energy without a net movement of particles in the medium. A wave may take the form of elastic deformation, it is a variation of pressure, electric or the magnetic intensity, electric potential, or the temperature of the medium.
Shallow water waves take place in the shallow water, this means that the waves which occur at depths that is shallower than the wavelength of the wave divided by 20. This would mean that a wave with a wavelength of about 60 meters is a shallow water wave at depth range of less than 60 divided by 20, or 3 meters.
Therefore, the correct option is C.
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The mass of a body:
(a) decreases when accelerated
(b) increases when accelerated
(c)decreases when moving with high velocity
(d)none of above
1. Car A is heading East at 25 m/s and car B is also heading East at 25 m/s. What is the relative velocity between car A
and car B? How would the passenger in Car A describe Car B's motion
The relative velocity between Car A and Car B is 0 m/s
Since Car A is heading East at 25 m/s, its velocity is u = + 25 m/s. (since it is moving in the positive x- direction)
Also, car B is also heading East at 25 m/s, its velocity is v = + 25 m/s. (since it is moving in the positive x- direction)
So, the relative velocity between car A and Car B is V = velocity of car A - velocity of car B.
So, V = u - v
V = + 25 m/s - (+ 25 m/s)
V = + 25 m/s - 25 m/s
V = 0 m/s
So, the relative velocity between Car A and Car B is 0 m/s
The passenger in Car A would describe Car B's motion as stationary.
Since Car A is heading East at 25 m/s and car B is also heading East at 25 m/s, both cars are moving at a velocity of + 25 m/s.
Since their relative velocity is 0 m/s, it would appear to a passenger in Car A that car B is stationary according to motion from car A's reference frame.
So, the passenger in Car A would describe Car B's motion as stationary.
Câu 1. Con lắc lò xo treo thẳng đứng, dao động điều hòa với biên độ 2cm và tần số góc 20 rad/s. Chiều dài tự nhiên của lò xo là 30cm. Chiều dài nhỏ nhất và lớn nhất của lò xo trong quá trình dao động là bao nhiêu? Lấy g = 10m/s2.
Answer:
28 cm and 32 cm
Explanation:
1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of
a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.
As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.
So, the minimum length is 28 cm and the maximum length is 32 cm.
Una onda se propaga en un medio P, de tal forma que recorre una distancia 8D en un tiempo 4T. La misma onda cuando se propaga por un medio Q recorre una distancia 4D en el mismo tiempo anterior, es decir, en 4T. Respecto a esta onda es correcto afirmar que *
Sabemos todo material tiene asociado un coeficiente de difracción n.
De tal forma que la velocidad de una onda electromagnetica que viaja a travez de dicho material, será:
v = c/n
donde c es la velocidad de la luz.
Con la información dada, podremos concluir que el coeficiente de difracción del medio Q es dos veces el del medio P.
Ahora veamos como llegamos a esto:
Sabemos que en el medio P, la onda recorre una distancia 8*D en un tiempo 4*T
entonces la velocidad de la onda en el medio P es:
[tex]V_p = \frac{8*D}{4*T} = 2*\frac{D}{T}[/tex]
Mientras que en el medio Q, recorre una distancia 4*D en un tiempo 4*T, por lo que la velocidad en el medio Q será:
[tex]V_Q = \frac{4*D}{4*T} = \frac{D}{T}[/tex]
Podemos ver que:
[tex]V_P = 2*V_Q[/tex]
Reescribiendo las velocidades como el cociente entre la velocidad de la luz y el correspondiente coeficiente de difracción obtenemos:
[tex]\frac{c}{n_P} = 2*\frac{c}{n_Q} \\\\n_Q = 2*n_P[/tex]
Es decir, podemos concluir que el coeficiente de difracción del medio Q es dos veces el del medio P.
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b. The role of the moon is greater than that of the sun in the occurrence of tides. ???
Our sun is 27 million times larger than our moon. Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth. If tidal forces were based solely on comparative masses, the sun should have a tide-generating force that is 27 million times greater than that of the moon. However, the sun is 390 times further from the Earth than is the moon. Thus, its tide-generating force is reduced by 3903, or about 59 million times less than the moon. Because of these conditions, the sun’s tide-generating force is about half that of the moon.
Straight wire of indefinite length (transient) passed by an electric current of 5.0 A. The magnetic field generated around this conductor at point M is 50 μT. Thus the distance from the conductor to point M is:
A. 2π cm
B. 0.2π cm
C. 2.0 cm
D. 0.20 cm
Answer:
C. 2.0 cm
Explanation:
The magnetic field around the wire at point M is given by Biot-Savart Law:
[tex]B = \frac{\mu_o I}{2\pi R}[/tex]
where,
B = Magnetic field = 50 μT = 5 x 10⁻⁵ T
I = current = 5 A
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
R = distance of point M from wire = ?
Therefore,
[tex]5\ x\ 10^{-5}\ T = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(5\ A)}{2\pi R}\\\\R = \frac{(2\ x\ 10^{-7}\ N/A^2)(5\ A)}{5\ x\ 10^{-5}\ T}\\[/tex]
R = 0.02 m = 2 cm
Hence, the correct option is:
C. 2.0 cm
when a .... cause an object to move through a ...., the .... on the object
Explanation:
velocity - the speed with a direction. Thus, inertia could be redefined as follows: Inertia: tendency of an object to resist changes in its velocity. ... Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.
(This is for other people with this Question i hope you find this when you need help) Need answer for this Help!!!
Question 7 of 20 You plan to use a slingshot to launch a ball that has a mass of 0.025 kg. You want the ball to accelerate straight toward your target at 19 m/s2. How much force do you need to apply to the ball? O A. 19.03 N OB. 0.48 N O C. 4.51 N D. 760.00 N
Answer:
0.48N
Explanation:
according to the second law of motion
force=mass×acceleration
the mass in this question is 0.025,the acceleration 19
therefore f=0.025×19
=0.48N
I hope this helps
what will happen to the gravitational force between two bodies if the distance between them is halved keeping their masses constant ?..
I need help asap please
Answer:
I dont know answer Sorry For that thank u
A force of 10 N is making an angle of 30° with the horizontal. Its horizontal component will be:
A. 4N
B. 5N
C. 7N
D. 8.7 N
Answer:The answer is A
Explanation:just did it on a test
An astronaut on Pluto attaches a small brass ball to a 1.00-m length of string and makes a simple pendulum. She times 10 complete swings in a time of 257 seconds. From this measurement she calculates the acceleration due to gravity on Pluto. What is her result
Answer:
The acceleration due to gravity at Pluto is 0.0597 m/s^2.
Explanation:
Length, L = 1 m
10 oscillations in 257 seconds
Time period, T = 257/10 = 25.7 s
Let the acceleration due to gravity is g.
Use the formula of time period of simple pendulum
[tex]T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2[/tex]
How many meters are in 10 miles?
Answer:
Explanation:
16093.4
5. A child rides a pony on a circular track with a radius of 5 m Find the distance traveled and the placement
after the child has gone halfway around the track (8) Does the distance traveled increase, decrease, or stay the
same when the child completes one circuit of the track? Does the displacement increase, decrease, or stay the
same? Explain. (C) Find the distance and displacement ter a complete circuit of the track
The answer are :
A. Distance = 15.71 m , The displacement = 10 m
B. distance traveled will increase, displacement will decrease
C. The distance = 31.42 m , Displacement = 0
Difference between Distance and DisplacementDistance is a scalar quantity, while displacement is a vector quantity. Displacement is the distance travelled in a specific direction.
Given that a child rides a pony on a circular track with a radius of 5 m, after the child has gone halfway around the track
(A)
The distance traveled will be = 2[tex]\pi[/tex]r / 2
The distance = πr
The distance = 22/7 x 5
The distance = 15.71 m
The displacement = 2r
The displacement = 2 x 5
The displacement = 10 m
(B) The distance traveled will increase as the child completes one circle of the track but the displacement will decrease because the displacement in one cycle of a circular motion is zero since it is a vector quantity.
(C) The distance after a complete circuit of the track = 2πr
The distance = 2 × 22/7 × 5
The distance = 31.42 m
The displacement after a complete circuit of the track will be zero.
That is;
Displacement = 0
Therefore, the distance traveled and the displacement after the child has gone halfway around the track are 15.7 m and 10 m respectively. While the distance and displacement after a complete circle of the track are 31.42 m and 0 respectively.
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Maya and Kenzie are discussing oil spills and how they impact the environment. How can humans help reduce the impact of oil spills?
Answer:
Using physical and chemical clean up methods to remove the oil
Explanation:
If the wave is detected 12.5 minutes after the earthquake, estimate the distance from the detector to the site of the quake
Answer:
Remember the relation:
Speed*Time = Distance.
We can estimate that the speed at which an earthquake "moves", in the surface, is:
S = 6km/s (this is a low estimation actually)
Then if the wave is detected 12.5 minutes after the earthquake, we know that it traveled for 12.5 minutes before reaching the detector.
So we know the speed of the wave and the time it took to reach the detector, then we can use the equation:
Speed*Time = Distance.
to find the distance.
First, we should write the time in seconds
we know that:
1 min = 60 s
then:
12.5 min = 12.5*(60 s) = 750 s
Then, the wave traveled with a speed of 6 km/s for 750 seconds until it reached the detector, then the distance that it traveled is:
(6km/s)*750s = 4500 km
The distance between the detector and the site of the quake is around 4500 km.
