A satellite orbiting Earth at a velocity of 3700 m/s collides with a piece of
space debris traveling at 5000 m/s. If the objects have the same mass and
the space debris has a velocity after collision of 3700 m/s, what is the
velocity of the satellite after the collision?
Answer:
Explanation:
they trade velocities... so the Satellite now is going 5000 m/s
What voltage indicates the voltmeter connected to the ends of a conductor, if the ammeter connected in series with this conductor indicates a current of 400 mA. The resistance of the conductor is equal to 5kΩ.
Explanation:
V= Current × Resistance
V= 400mA × 5kohms
V= 2000V
A+10 u charge and a -10 4C (1 HC - 106 C), at a distance of 0.3 m,
A. Repel each with a force of 10 N
B. Attract each other with a force of 10 N
C. Repel each other with a force of 20 N
D. Attract each other with a force of 20 N
If anyone knows what’s going on here please help
Answer:
B. Attract each other with a force of 10 newtons.
Explanation:
Statement is incorrectly written. The correct form is: A [tex]+10\,\mu C[/tex] charge and a [tex]-10\,\mu C[/tex] at a distance of 0.3 meters.
The two particles have charges opposite to each other, so they attract each other due to electrostatic force, described by Coulomb's Law, whose formula is described below:
[tex]F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}[/tex] (1)
Where:
[tex]F[/tex] - Electrostatic force, in newtons.
[tex]\kappa[/tex] - Electrostatic constant, in newton-square meters per square coulomb.
[tex]|q_{A}|,|q_{B}|[/tex] - Magnitudes of electric charges, in coulombs.
[tex]r[/tex] - Distance between charges, in meters.
If we know that [tex]\kappa = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}[/tex], [tex]|q_{A}| = |q_{B}| = 10\times 10^{-6}\,C[/tex] and [tex]r = 0.3\,m[/tex], then the magnitude of the electrostatic force is:
[tex]F = \frac{\kappa \cdot |q_{A}|\cdot |q_{B}|}{r^{2}}[/tex]
[tex]F = 9.987\,N[/tex]
In consequence, correct answer is B.
The energy of motion is called...?
Which of the following is an example of a healthy behavior?
OA
Binge drinking of alcohol
OB.
Smoking cigarettes
OC.
Driving a car recklessly
OD.
None of the above
PE Not Physics
Answer moderate alcohol can be good for you health
Explanation: but I wouldn't do none of those things to be honest
SMARTEST
Answer:
none of the above
Explanation:
its an easy question haha
what is the wavelength of a wave with the frequency of 330 Hz and a speed of 343 m/s
Answer:
The wavelength of a wave with the frequency of 330hz and a speed of 343m/s would be 1.04m
Explanation:
You can get the wavelength of a wave by dividing the speed of the wave by its frequency, which in this case would be:
343/300, which as a decimal number, it'd be 1.04.
I hope I helped you, and a "Brainliest" is always appreciated! ☺
Question 7 of 10
Which of these is a product of beta decay?
O A. An alpha particle
OB. A helium nucleus
C. An electron or a positron
O D. A beam of electromagnetic radiation
Answer:
c.an electron or positron
Answer:
your answer would be c
Explanation:
What is the wavelength associated with an electron with a velocity of 4.8x10*5 m/s?
(mass of the electron is 9.1 X 10-*31kg)
Answer:
Explanation:
Given the following data;
Mass = 9.1 X 10^{-31} kg
Velocity = 4.8x10⁵ m/s
Planck’s constant ( h ) = 6.6262 * 10^{-34}
To find the wavelength;
Applying de broglie's equation;
Wavelength = h/mv
Wavelength = ) = 6.6262 * 10^{-34}/(9.1 X 10^{-31} * 4.8x10⁵)
Wavelength = 1.52 * 10^{-71} meters
a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertical cliff, what is the the hight of the cliff
Answer:
Approximately [tex]281.25\; \rm m[/tex]. (Assuming that the drag on this ball is negligible, and that [tex]g = 10\; \rm m \cdot s^{-2}[/tex].)
Explanation:
Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:
Horizontal: no acceleration, velocity is constant (at [tex]v(\text{horizontal})[/tex] is constant throughout the descent.)Vertical: constant downward acceleration at [tex]g = 10\; \rm m \cdot s^{-2}[/tex], starting at [tex]0\; \rm m \cdot s^{-1}[/tex].The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: [tex]x(\text{horizontal}) = 30\; \rm m[/tex]. Combine these two quantities to find the duration of this descent:
[tex]\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}[/tex].
In other words, the ball in this question start at a vertical velocity of [tex]u = 0\; \rm m \cdot s^{-1}[/tex], accelerated downwards at [tex]g = 10\; \rm m \cdot s^{-2}[/tex], and reached the ground after [tex]t = 7.5\; \rm s[/tex].
Apply the SUVAT equation [tex]\displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t[/tex] to find the vertical displacement of this ball.
[tex]\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}[/tex].
In other words, the ball is [tex]281.25\; \rm m[/tex] below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be [tex]281.25\; \rm m\![/tex].
A flag is waved 3.2 m above the surface of a flat pool of water. When viewed from under the water, what is the magnification of the flag? Let the indices of refraction nwater = 1.33 and nair = 1.00.
1.33
0.754
1
1.67
Answer:
1.33
Explanation:
For an optical instrument, the magnification ratio of the apparent diameter of the image to that of the object.
Mathematically, from the given information;
Magnification[tex]= \dfrac{n_{water}}{n_{air}}[/tex]
where;
[tex]n_{water} =1.33\\ \\ n_{air} = 1.00[/tex]
[tex]= \dfrac{1.33}{1.00} \\ \\= \mathbf{1.33}[/tex]
At what rate is work done if the 250 Newton object from number six is moved into a hot at 4 m in four seconds
Answer:
250w
Explanation:
W=Fd
W= 250 x 4= 1000J
P=E/t=1000/4s=250w
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that the pendulum completes 98.0 full swing cycles in a time of 135s.
Required:
What is the value of g on this planet?
Answer:
g = 11.2 m/s²
Explanation:
First, we will calculate the time period of the pendulum:
[tex]T = \frac{t}{n}[/tex]
where,
T = Time period = ?
t = time taken = 135 s
n = no. of swings in given time = 98
Therefore,
[tex]T = \frac{135\ s}{98}[/tex]
T = 1.38 s
Now, we utilize the second formula for the time period of the simple pendulum, given as follows:
[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]
where,
l = length of pendulum = 54 cm = 0.54 m
g = acceleration due to gravity on the planet = ?
Therefore,
[tex](1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}[/tex]
g = 11.2 m/s²
System A consists of a mass m attached to a spring with a force constant k; system B has a mass 2m attached to a spring with a force constant k; system C has a mass 3m attached to a spring with a force constant 6k; and system D has a mass m attached to a spring with a force constant 4k.
Rank these systems in order of increasing period of oscillation. (Use only the symbols < and =, for example A < B = C.)
Solution :
We know that the time period of oscillation of a spring mass system is given by :
[tex]$T = 2 \pi \sqrt{\frac{m}{k}}$[/tex] , where m is mass and k is the spring constant
∴ [tex]$T_A = 2 \pi \sqrt{\frac{m}{k}}$[/tex] .........(i)
[tex]$T_B = 2 \pi \sqrt{\frac{2m}{k}}$[/tex] ..........(ii)
[tex]$T_C = 2 \pi \sqrt{\frac{3m}{6k}} = 2 \pi \sqrt{\frac{m}{2k}}$[/tex] ..........(iii)
[tex]$T_D = 2 \pi \sqrt{\frac{m}{4k}}$[/tex] ...............(iv)
Comparing the equations (i), (ii), (iii) and (iv)
We get
[tex]$T_B > T_A > T_C > T_D$[/tex]
So in increasing order of time period, we get
[tex]$T_D < T_C < T_A < T_B$[/tex]
3. Two spherical objects have masses of 4.7 x 105 kg and 7.9 x 103 kg. The
gravitational attraction between them is 68 N. How far apart are their
centers?
Answer:
0.06
Explanation:
Working on the problem at hand i calculated and got my answer to around simplified at 0.06
Which of the following phenomena best demonstrates that light possesses wave characteristics?
A.The different colors of visible light
B.The transfer of electrons
C.The release of infrared light form radioactive materials
D.It shines and takes up space
The release of infrared light from radioactive materials is what demonstrates light possessing wave characteristics.
What is a Wave?This is defined as the propagation of disturbances from one place to another in an organized way.
The release of waves in a diffraction pattern is why option C was chosen as the most appropriate choice.
Read more about Wave here https://brainly.com/question/15663649
What is the speed of an object moving around a 0.75 m radius circle that completes a revolution in 0.50 seconds?
Answer:
the speed of an object is 9.42 m/s
Explanation:
The computation of the speed of an object is given below:
v = 2πr ÷ t
where
v denotes the speed
r denotes the radius
r denotes the time
So,
= 2 × 3.14 × 0.75 ÷ 0.50
= 9.42 m/s
Hence, the speed of an object is 9.42 m/s
the same would be considered and relevant too
What are three hazardous chemicals found in cigarette smoke?
Answer: I believe hydrogen cyanide, carbon monoxide, and ammonia are the three chemicals involved, but im not sure
a car initially at rest move with the constant accerates along straght line read after it's spread increase and finally related uniformly. the time interval for the three parts of the jounry are in the ratio 1:3:1 find average velocity ?
Answer:
32km per hour
Explanation:
Explanation:
In first case v = a t
==> a t = 40 km p h
Now distance covered S1 + S2 + S3
S1 = 1/2 a t^2 and S3 = 1/2 a t^2
But S2 = 3t * 40 = 120 t km
Hence total distance = at^2 + 120 t
Time taken (total) = t + 3t + t = 5 t
Hence average speed = at^2 + 120 t / 5 t
Cancelling t we have at + 120 / 5 = 40 + 120 / 5 = 160/5 = 32 km per hour
When the electrons reach the collector, they flow towards the positivly charged grid. The resulting current is measured. Note that as the electrons accelerate from the cathode toward the grid, they collide with the mercury atoms. Assume that these collisions are completely elastic. How does the collected current vary if the ΔVgrid is slowly increased? View Available Hint(s) When the electrons reach the collector, they flow towards the positivly charged grid. The resulting current is measured. Note that as the electrons accelerate from the cathode toward the grid, they collide with the mercury atoms. Assume that these collisions are completely elastic. How does the collected current vary if the is slowly increased? The current increases. The current remains constant. The current decreases.
Answer:
the current INCREASES.
Explanation:
In this experiment, the electrons are generated by a filament with very low speed, when they are subjected to a difference and potential ΔV they acquire the necessary speed to reach the regulation and the current can be measured.
Some electrons collide elastically with the atoms of the mercury gas that is much heavier and are scattered in any direction, so they do not reach the grid, by increasing the voltage this scattered electrons can acquire the necessary speed in the direction of grid to reach it and therefore are also measured, increasing the current.
Therefore, as the power difference increases, the current INCREASES.
A ball with a mass of 0.585 kg is initially at rest. It is struck by a second ball having a mass of 0.420 kg , initially moving with a velocity of 0.270 m/s toward the right along the x axis. After the collision, the 0.420 kg ball has a velocity of 0.220 m/s at an angle of 36.9 ∘ above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface.
Required:
a. What is the magnitude of the velocity of the 0.605kg ball after the collision?
b. What is the direction of the velocity of the 0.605kg ball after the collision?
c. What is the change in the total kinetic energy of the two balls as a result of the collision?
Answer:
a) [tex](v_1)=0.3989m/s[/tex]
b) [tex]\theta_1=80.5 \textdegree[/tex]
c) [tex]K.E=0.036J[/tex]
Explanation:
From the question we are told that:
Initial speed of 1st ball [tex]u_{1}=0 m/s[/tex]
Mass of 1st ball [tex]m_1=0.585kg[/tex]
Mass of 2nd ball [tex]m_2=0.420kg[/tex]
Initial speed of 2nd ball [tex]u_{2}=0.270 m/s[/tex]
Final speed of 2nd ball [tex]v_{2}=0.220 m/s[/tex]
Angle of collision [tex]\angle=36.9 \textdegree[/tex]
a)
Generally the equation for law of conservation is mathematically given by
[tex]m_1u_1+m_2u_2=m_1v_1^2+m_2v_2^2[/tex]
The final velocity [tex]v_1[/tex] is given as
[tex]0+(0.420)(0.270)=(0.585)(v_1)^2+(0.420)(0.220)^2[/tex]
[tex](v_1)^2=\frac{(0.420)(0.270)-(0.420)(0.220)^2}{0.585}[/tex]
[tex](v_1)^2=0.1591[/tex]
[tex](v_1)=0.3989m/s[/tex]
b)
Generally the equation for law of conservation is mathematically given by
[tex]m_1u_1+m_2u_2=m_1v_1cos\theta_1+m_2\theta_2[/tex]
[tex]0+(0.420)(0.270)=(0.585)(1.511)cos\theta_1+(0.420)(0.220)cos36 \textdegree[/tex]
[tex]cos\theta_1= \frac{(0.420)(0.270)-(0.420)(0.220)cos36 \textdegree}{(0.585)(0.3989)}[/tex]
[tex]cos\theta_1=0.1656[/tex]
[tex]\theta_1=80.5 \textdegree[/tex]
c)
Generally the equation for kinetic energy is mathematically given by
[tex]K.E=\frac{1}{2} mv^2[/tex]
1st Ball
[tex]K.E=\frac{1}{2} (0.585)(0.3989)^2[/tex]
[tex]K.E=0.0465J[/tex]
2nd ball
[tex]K.E=\frac{1}{2} (0.420)(0.220)^2[/tex]
[tex]K.E=0.101J[/tex]
Therefore the change in the total kinetic energy of the two balls as a result of the collision is
[tex]0.101-0.0465[/tex]
[tex]K.E=0.036J[/tex]
During the course of a demonstration the professor is called away. When he returns he finds a beaker of water that was at room temperature is now at a slightly higher temperature. There is a stirring rod on the desk and a cigarette lighter. The professor can assume that the temperature increase is due to either heat added or mechanical work done. mechanical work done on the system. heat added to the system.
Answer:
Either heat added or mechanical work done.
Explanation:
Since he found stirring rod on the desk and a cigarette lighter. This means that the beaker was probably either heated with the aid of fire from the lighter.
Also, the stirring rod could have been used to stir the water which will increase the kinetic energy which also means an increase in temperature.
Thus, it's either heat was added or mechanical work was done as a result of stirring.
On the Moon’s surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time. What percent correction is needed to account for the delay in time due to the slowing of light in Earth’s atmosphere? Assume the distance to the Moon is precisely 3.84×108 m , and Earth’s atmosphere (which varies in density with altitude) is equivalent to a layer 30.0 km thick with a constant index of refraction n = 1.000293.
Answer:
[tex]T_d=2.8*10^-^6\%[/tex]
Explanation:
From the question we are told that
Distance b/w Earth and the moon [tex]d_e=3.84×10^8 m[/tex]
Thickness of Earth's atmosphere[tex]d_t=30km[/tex]
Constant index of refraction [tex]n = 1.000293.[/tex]
Generally the equation for percentage of time delay [tex]T_d[/tex] is mathematically given as
[tex]T_d=\frac{d(n-1)}{rm}*100\%[/tex]
[tex]T_d=\frac{(35*10^3)(1.000293-1)()100)}{3.674*10^8} *\%[/tex]
[tex]T_d=2.8*10^-^6 \%[/tex]
Therefore the general percentage correction needed because of tym deley is given as
[tex]T_d=2.8*10^-^6\%[/tex]
What is the difference between 1 celcius and 1 kelvin
Answer:
One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale. The only difference between these two scales is the zero point. The zero point on the Celcius scale was defined as the freezing point of water, which means that there are higher and lower temperatures around it.
A power plant burns coal to generate electricity. Suppose that 1000 J of heat (Q) from the coal fire enters a boiler, which is kept at a constant temperature of 900 C. That 1000 J is used to boil water in a boiler. The steam from the boiler is used to drive a turbine that creates electricity (work). The excess heat is put into a cooling tower at 5 C.
Carnot efficiency If everything is perfectly efficient,
• What is the maximum efficiency that the plant could operate at?
• How much work could be done starting from that 1000 J?
Answer:
The work done will be "763 J". A further explanation is given below.
Explanation:
The given values are:
[tex]Q_1=1000 \ J[/tex]
Temperature,
[tex]T_1=900^{\circ} C[/tex]
i.e,
[tex]=1173[/tex]
[tex]T_2=5^{\circ}C[/tex]
i.e.,
[tex]=278[/tex]
As we know,
⇒ [tex]\eta =1-\frac{T_2}{T_1}[/tex]
On substituting the values, we get
⇒ [tex]=1-\frac{278}{1173}[/tex]
⇒ [tex]=\frac{1173-278}{1173}[/tex]
⇒ [tex]=\frac{895}{1173}[/tex]
⇒ [tex]=0.763[/tex]
then,
⇒ [tex]W=\eta Q_1[/tex]
⇒ [tex]=0.763\times 1000[/tex]
⇒ [tex]=763 \ J[/tex]
The gravitational force between two objects is proportional to the square of the distance between the two objects.
a. True
b. False
Answer: True!
Explanation: The force is proportional to the square of the distance between 2 point masses
Number 29 plz help physics
Answer:
a. E = 900000 J = 900 KJ
b. ΔT = 8.18 °C
c. Cost = $ 7.2
Explanation:
a.
The energy can be given by:
[tex]E = Pt[/tex]
where,
E = Energy = ?
P = Power = 500 W
t = time = (0.5 h)(3600 s/1 h) = 1800 s
Therefore,
[tex]E = (500\ W)(1800\ s)[/tex]
E = 900000 J = 900 KJ
b.
The change in temperature of room air is given by the following formula:
[tex]0.5E = mC\Delta T\\[/tex] (since 50% of energy is used to heat air)
where,
m = mass of air = 50 kg
C = specific heat of air = 1.1 KJ/kg.°C
ΔT = Change in temperature of air = ?
Therefore,
[tex](0.5)(900\ KJ) = (50\ kg)(1.1\ KJ/Kg.^oC)\Delta T[/tex]
ΔT = 8.18 °C
c.
First, we will calculate the total energy consumed:
[tex]E = (0.5\ KW)(6\ h/d)(30\ d)\\E = 90\ KWh[/tex]
Now, for the cost:
[tex]Cost = (Unit\ Cost)(Energy)\\Cost = (\$ 0.08\KWh)(90\ KWh)[/tex]
Cost = $ 7.2
Compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 10
and at a distance of 1 m from each other.
Answer:
Let's see the similarities between the two forces
* are proportional to the product of a magnitude, mass or charge
* They are inversely proportional to the square of the distance
* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.
The differences in them
* The electric force in much greater than the gravitational force
* The gravitational force is always attractive, the electrical force can be attractive or repulsive.
Explanation:
Let's start by calculating each force.
Gravitational force
F =[tex]G \frac{m_1m_2}{r^2}[/tex]
let's calculate
F = 6.67 10⁻¹¹ 1 1 / 1²
F = 6.67 10⁻¹¹ N
Electric force
F = [tex]k \frac{q_1q_2}{r^2}[/tex]
indicates that the charge is q = 10 C
F = 9 10⁹ 10 10 / 1²
F = 9 10¹¹ N
Let's see the similarities between the two forces
* are proportional to the product of a magnitude, mass or charge
* They are inversely proportional to the square of the distance
* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.
The differences in them
* The electric force in much greater than the gravitational force
* The gravitational force is always attractive, the electrical force can be attractive or repulsive.
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Explanation:
lol ....you are right.......
Carla has a stick of butter. What can Carla do to the butter to show a chemical change?
Answer:
melting butter in a saucepan, as the butter cools it turns brown
Explanation:
although melting the butter is a physical change, but the butter being cooked and turning brown Is a chemical change, because due to the heat, its particles are being broken down, which is irreversible, and shows a chemical change.
51.Shoveling snow can be extremely taxing because the arms have such a low efficiency in this activity. Suppose a person shoveling a footpath metabolizes food at the rate of 800 W. (a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process?
Complete question is;
Shoveling snow can be extremely taxing since the arms have such a low efficiency in this activity. Suppose a person shoveling a sidewalk metabolizes food at the rate of 800 W. (The efficiency of a person shoveling is 3%.)
(a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process?
Answer:
A) P_out = 24 W
B) t = 1470 s
C) Q = 1140.72 KJ
Explanation:
We are given;
Input Power; P_in = 800 W
Efficiency; η = 3% = 0.03
A) Formula for efficiency is;
η = P_out/P_in
Making P_out the subject, we have;
P_out = η•P_in
P_out = 0.03 × 800
P_out = 24 W
B) We know that;
Power = work done/time taken
Thus;
P_out = mgh/t
We are given;
m = 3000 kg
h = 1.20 m
Thus, time is;
t = (3000 × 9.8 × 1.2)/24
t = 1470 s
C) amount of heat wasted is calculated from;
Q = (P_in - P_out)t
Q = (800 - 24) × 1470
Q = 1,140,720 J
Q = 1140.72 KJ