Answer :
Considering initial velocity is 0,
It takes about 0.4 seconds.
Use the equation v-u =at
4-0 = 9.8×t
t = 4/9.8 = 0.4 seconds approximately.
Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed in m/s of a satellite in an orbit 980 km above the Earth's surface.
Answer:
564
Explanation:
In a certain region of space near earth's surface, a uniform horizontal magnetic field of magnitude B exists above a level defined to be y = 0. Below y = 0 , the field abruptly becomes zero (seethe figure). A vertical square wire loop has resistivity rho mass density rhom, diameter d, and side length l. It is initially at rest with its lower horizontal side at y = 0 and is then allowed to fall under gravity, with its plane perpendicular to the direction of the magnetic field.
a) While the loop is still partially immersed in the magnetic field (as it fallsinto the zero-field region), determine the magnetic "drag" forcethat acts on it at the moment when its speed is v.
b) Assume that the loop achieves a terminal velocity vt before its upper horizontal side exits the field. Determine a formulafor vt
c) If the loop is made of copper and B = 0.80 T find vt
Answer:
a) F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex]
b) attached below
c) 0.037 m/s
Explanation:
a) Determine the magnetic "drag" force acting at the moment
speed = v
first step: determine current in the loop
I = [tex]\frac{\pi d^2}{16pl} B lv[/tex] ----- ( 1 )
given that the current will induce force on the three sides of the loop found in the magnetic field
forces on vertical sides = + opposite
we will cancel out
hence equation 1 becomes
F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex] ( according to Lenz law we can say that the direction of force is upwards and this force will slow down the decrease in flux )
b) Determine the formula for Vt
attached below
c) Find Vt
given :
B = 0.80 T
density of copper = 8.9 * 10^3 kg/m^3
resistivity of copper = 1.68 * 10^-8 Ωm
∴ Vt = 16 ( 8.9 * 10^3 kg/m^3 ) ( 1.68 * 10^-8 Ωm ) ( 9.8 m/s^2 ) / ( 0.08 T)^2
= 0.037 m/s
reviews the general principles in this problem. A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?
Answer: θ would equal approximately 28.7°
This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.
Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°
Now if we multiply the range by 2, we get:
2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:
2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ
Thus, θ = 28.67780425
It's been awhile since I did this; though I hope it helped!
What happens in the gray zone between solid and liquid?-,-
The following 1H NMR absorptions were obtained on a spectrometer operating at 200 MHz and are given in Hz downfield from TMS. Convert the absorptions to δ units. a) 416 Hz = δ b) 1.97×103 Hz = δ c) 1.50×103 Hz = δ
Answer:
For (a): The chemical shift is [tex]2.08\delta[/tex]
For (b): The chemical shift is [tex]9.85\delta[/tex]
For (c): The chemical shift is [tex]7.5\delta[/tex]
Explanation:
To calculate the chemical shift, we use the equation:
[tex]\text{Chemical shift in ppm}=\frac{\text{Peak position (in Hz)}}{\text{Spectrometer frequency (in MHz)}}[/tex]
Given value of spectrometer frequency = 200 MHz
For (a):Given peak position = 416 Hz
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{416Hz}{200MHz}\\\\\text{Chemical shift in ppm}=2.08\delta[/tex]
For (b):Given peak position = [tex]1.97\times 10^3 Hz[/tex]
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{1.97\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=9.85\delta[/tex]
For (c):Given peak position = [tex]1.50\times 10^3 Hz[/tex]
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{1.50\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=7.5\delta[/tex]
Trong máy phát điện xoay chiều ba pha khi tổng điện áp tức thời của cuộn 1,2 là e1+e2=120V thì điện áp tức thời của cuộn 3 là
Answer:
I just noticd i dont speak this launguage
Explanation:
2. The given graph shows that the object is
(a) in non-uniform motion
(b) in uniform motion
(c) at rest
(d) in an oscillatory motion.
distance
time
Answer:
(c) at rest
Explanation:
Given
See attachment for the distance time graph
Required
What does the graph illustrate?
From the graph, we can see that the line of distance is a horizontal line.
This suggests that a time increases, the distance remains unchanged
When distance remains unchanged over time, then it means the object is at rest.
Hence, (c) is correct
What recommendations would you give to the global government to help Decrease the global effects of human impact on the environment mystery recommendations and how they will positively impact our planet
Answer:
We can help to keep it magnificent for ourselves, our children and grandchildren, and other living things besides us.
Explanation:
5 ways our governments can confront climate change
PROTECT AND RESTORE KEY ECOSYSTEMS
SUPPORT SMALL AGRICULTURAL PRODUCERS
PROMOTE GREEN ENERGY
COMBAT SHORT-LIVED CLIMATE POLLUTANTS
BET ON ADAPTATION, NOT JUST MITIGATION
Calculate the Combined resistance of the Circuit voltage across each resistor Current Passing through each resistor of 6,8,12ohms
Answer:
Sorry I don't know the answer
Let’s use these equations to compare the electrostatic force and the gravitational force in a few different situations. In each case, calculate the strength of the electrostatic force and the strength of the gravitational force.
Two electrons separated by 1 cm
q = 1.6 x 10-19 C
m = 9.1 x 10-31 kg
d = 1 cm
What is the electric force?
What is the gravitational force?
Which force will dominate the motion of the electrons?
Answer:
Electric force is the attractive force between the electrons and the nucleus. It works the same way for a negative charge, you also have an electric field around it. ... Now, like charges repel each other and opposite charges attract.
The gravitational force is a force that attracts any two objects with mass. We call the gravitational force attractive because it always tries to pull masses together, it never pushes them apart. ... This is called Newton's Universal Law of Gravitation.
electric force
This table shows clearly that the electric force dominates the motion of electrons in atoms. However, on a macroscopic scale, the gravitational force dominates. Since most macroscopic objects are neutral, they have an equal number of protons and electrons.
Explanation:
FROM THE _____ WHOLE WATER CYCLE STARTS ALL OVER AGAIN
From the water whole water cycle starts again.
Most possibly water should be the answer.A body is dropped from a height H. In how much time will it reach the ground?
Answer:
[tex]s = ut + \frac{1}{2} g {t}^{2} \\ t(u + \frac{1}{2} gt) = H \\ u + \frac{1}{2} gt = H \\ t = 2(H - u) \div g \\ t = \frac{(H - u)}{5} \\ u \: is \: speed \: or \: velocity[/tex]
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
A) 8.03 x 10^16 nuclei
B) 4.01 x 10^16 nuclei
C) 2.02 x 10^16 nuclei
D) 1.61 x 10^17 nuclei
OPTION C is the correct answer.
The radioactive decay follows the first order kinetics. The number of atoms decaying at any time is proportional to the number of atoms present at that instant. The amount of sample left is 2.02 x 10¹⁶nuclei. The correct option is C.
What is half-life?The time required for the decay of one half of the amount of the species is defined as the half-life period of a radionuclide. The half-life period is a characteristic of a radionuclide. The half lives can vary from seconds to billions of years.
The isotope decay of an atom is given by the equation:
ln [A] = -kt + ln [A]₀
The rate constant, k is:
k = ln 2 / Half life
k = ln 2 / 4.96 x 10³
k = 1.40 × 10⁻⁴ s⁻¹
t = 1.98 x 10⁴
[A]₀ = 3.21 x 10¹⁷
ln [A] = -1.40 × 10⁻⁴ × 1.98 x 10⁴ + ln [3.21 x 10¹⁷] = 37.538
[A] = 2.02 x 10¹⁶ nuclei
Thus the correct option is C.
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What is the energy equivalent of an object with a mass of 1.05g?
Answer:
The equivalent energy of an object given its mass is calculated through the equation,
E = mc²
where c is the speed of light (3 x 10^8 m/s)
Substituting the known values,
E = (1.05 g/ 1000) (3 x 10^8 m/s)²
E = 9.45x10^13 J
Explanation:
What country first colonised Ghana
Answer: Colonialism is a big topic, but it can only be understood by looking at human experiences. Formal colonialism first came to the region we today call Ghana in 1874, and British rule spread through the region into the early twentieth century. The British called the territory the “Gold Coast Colony”.
Explanation: hey, hope this hlps! oh, btw you picked the wrong subject for this question it should have been history insteat of phiscics!
A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal energy of the box, determine the magnitude of the increase.
Answer:
Explanation:
From the given information:
The initial PE [tex](PE)_i[/tex] = m×g×h
= 5 kg × 9.81 m/s² × 10 m
= 490.5 J
The change in Potential energy P.E of the box is:
ΔP.E = [tex]P.E_f -P.E_i[/tex]
ΔP.E = 0 - [tex](PE)_i[/tex]
ΔP.E = [tex]-P.E_i[/tex]
If we take a look at conservation of total energy for determining the change in the internal energy of the box;
[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]
[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]
this can be re-written as:
[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]
Here, K.E = 0
Also, 70% goes into raising the internal energy for the box;
Thus,
[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]
[tex]\Delta U =(0.70) (490.5)[/tex]
ΔU = 343.35 J
Thus, the magnitude of the increase is = 343.35 J
pls help! George pushes a wheelbarrow for a distance of 12 meters at a constant speed for 35 seconds by applying a force of 20 newtons. What is the
power applied to push this wheelbarrow?
A. 1.2 watts
B. 3.4 watts
C. 6.9 watts
D. 13 watts
Answer:
C. 6.9 watts
Explanation:
Power = work/time
if work = force×distance...
Then... power= (force×distance)/time
Power = (20×12)/35
= 6.9 watts
Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. What is the swimmers average speed and average velocity?
Answer:
The average speed is 1 m/s
The average velocity is 0
Explanation:
Given;
length of the pool, L = 50 m
time taken for the motion, t = 100 s
The total distance = 50 m + 50 m
The total distance = 100 m
The average speed = total distance / total time
= 100 / 100
= 1 m/s
The average velocity = change in displacement / change in time
change in displacement = 50 m - 50 m = 0
The average velocity = 0 / 100
The average velocity = 0
g Is a nucleus that absorbs at 4.13 δ more shielded or less shielded than a nucleus that absorbs at 11.45 δ? _________ Does the nucleus that absorbs at 4.13 δ require a stronger applied field or a weaker applied field to come into resonance than the nucleus that absorbs at 11.45 δ?
Answer: A nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded and a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field
Explanation:
While interpreting the data in NMR, the positions of signals are studied.
The nucleus/ protons having a higher value of [tex]\delta[/tex] are said to be less shielded. They are said to be upfield.
The nucleus/protons having a lower value of [tex]\delta[/tex] are said to be more shielded. They are said to be downfield.
So, a nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded by the nucleus that absorbs at [tex]4.13\delta[/tex]
Also, the less shielded nucleus/protons will require a weak applied field to come into resonance than the more shielded nucleus/protons
So, a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field to come into resonance than the nucleus that absorbs at [tex]11.45\delta[/tex]
From the top of the leaning tower of Pisa, a steel ball is thrown vertically downwards with a speed of 3.00 m/s. if the height of the tower is 200 m, how long will it take for the ball to hit the ground? Ignore air resistance.
Answer:
66,7 seconds
Explanation:
the formula for height/distance is : S=v.t
At 20 ◦C a copper wire has a resistance of 4×10−3 Ω and a temperature coefficient of resistivity of 3.9×10−3 (C◦)−1, its resistance at 100 ◦C is
A.
52.5 × 10-3 Ω
B.
5.25 × 10-3 Ω
C.
5.25 × 10-4 Ω
D.
5.25 × 10-2 Ω
E.
25.5 × 10-3 Ω
Answer:
[tex]R _{t} = R _{0}( \alpha t + 1) \\ = 4 \times {10}^{ - 3} (3.9 \times {10}^{ - 3} \times 20 + 1) \\ = 4 \times {10}^{ - 3} (1.078) \\ = 4.312 \times {10}^{ - 3} \: Ω[/tex]
Cual es l diferencia entre ruido y sonido
Answer:
E.l soni.do es un.a sensac.ión, en el órg.ano del oído, prod.ucida por el movimie/nto ondu>latorio de un m/edio elástico (normal/mente el aire), debi.do a ra.pidísimos ca/mbios de pre.sión, generado/s por el movimiento vibrat.orio d.e un cuerpo sonoro. ... /El ruido se consid/era a to/do sonid.o / o no de.seado.
Explanation:
An electric device, which heats water by immersing a resistance wire in the water, generates 20 cal of heat
per second when an electric potential difference of 6 V is placed across its leads. What is the resistance in Ω
of the heater wire? (Note: 1 cal = 4.186 J)
Select one:
a. 0.86
b. 0.17
c. 0.29
d. 0.43
Answer:
1 cal/s =4.184w
p=50 cal/s =2093w
v=12v
P = V*I
I =P/V
I = 17.43 A
P =1²*R
R = P/I²
R = 0.68Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
The question is incomplete. The complete question is :
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
Solution :
Given that :
The predicted range is 0.3503 m
Time of the fall is :
[tex]$t=\sqrt{\frac{2H}{g}}$[/tex]
[tex]v_1t= 0.35[/tex] ...........(i)
[tex]v_0t= 1.09[/tex] ...........(ii)
Dividing the equation (ii) by (i)
[tex]$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$[/tex]
∴ [tex]v_0=3.11 \ v_1[/tex]
Now loss of energy = change in the kinetic energy
[tex]$W=\frac{1}{2} m [v_0^2-v_1^2]$[/tex]
[tex]$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$[/tex]
[tex]$W=7.307\times 10^{-3} \ v_0^2$[/tex]
If f is average friction force, then
(f)(L) = W
(f) (1) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
(f) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
The Average force of friction is ( F ) = 7.307 * 10⁻³ v₀²
Given data:
Predicted range ( v₁t ) = 0.3503 m
Actual range ( v₀t ) = 1.09 m
mass = 16.3 g
First step : Determine the value of V₀
[tex]t = \sqrt{\frac{2H}{g} }[/tex] , v₁t = 0.3503 , ( v₀t ) = 1.09 m
To obtain the value of V₀
Divide ( v₀t ) by ( v₁t ) = 1.09 / 0.3503 = 3.11 v₁
∴ V₀ = 3.11 v₁
Next step : Determine the average force of friction ( f )
given that loss of energy results in a change in kinetic energy
W = [tex]\frac{1}{2} m ( vo^{2} - v1^{2} )[/tex]
= 1/2 * 16.3 * 10⁻³ * [ v₀² - [tex](\frac{v_{0} }{3.11} )^{2}[/tex] ]
∴ W = 7.307 * 10⁻³ v₀²
Average force of friction = W / Actual length
= 7.307 * 10⁻³ v₀² / 1
∴ Average force of friction ( F ) = 7.307 * 10⁻³ v₀²
Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²
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Your question has some missing data below are the missing data related to your question
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
please help very easy 5th grade work giving brainliest
Answer:
the answer is option B because opposit sides of the magnets attract each other
how do you use the coefficient to calculate the number of atoms in each molecule?
Answer:
To find out the number of atoms: MULTIPLY all the SUBSCRIPTS in the molecule by the COEFFICIENT. (This will give you the number of atoms of each element.)
Explanation:
A positively charged plastic ruler is brought close to a piece paper resting on the desk. The piece of paper was initially neutral. When the ruler was brought closer, the paper is attracted to the ruler. The surface of the paper became charged through:_________
Answer: static electricity
Explanation:
When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the positive charge in the piece of paper.
A magnetic field of 0.276 T exists in the region enclosed by a solenoid that has 517 turns and a diameter of 10.5 cm. Within what period of time must the field be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 12.6 kV
Answer:
The period the field must be reduced to zero is 9.81 x 10⁻⁵ s
Explanation:
Given;
initial value of the magnetic field, B₁ = 0.276 T
number of turns of the solenoid, N = 517 turns
diameter of the solenoid, d = 10.5 cm = 0.105 m
induced emf, = 12.6 kV = 12,600 V
when the field becomes zero, then the final magnetic field value, B₂ = 0
The induced emf is given by Faraday's law;
[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]
Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s
Lighting is the movement of?
Explanation:
Movement:refers to the changing in the lights whether it be a change in intensity, color or direction of origin.
According to Newton’s law of universal gravitation, which statements are true?
