Answer:
The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".
Explanation:
The given values are:
Potential difference,
Δv = 400 v
Radius,
r = 0.5580 cm
As we know,
⇒ [tex]W=e \Delta v[/tex]
and,
⇒ [tex]\frac{1}{2}mv^2=e \Delta v[/tex]
then,
⇒ [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]
⇒ [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]
⇒ [tex]=1.185\times 10^7 \ m/sec[/tex]
Which two chemical equations show double-replacement reactions?
A. C+02 - CO2
B. 2Li + CaCl2 - 2LiCl + Ca
I C. Ca(OH)2 + H2S04 - CaSO4 + 2H20
D. Na2CO3 + H2S - H2CO3 + Na2S
The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.
What is double replacement reaction?A double replacement reaction have two ionic compounds that are exchanging anions or cations.
From the given options, we can choose the following based on their exchange of anions or cations.
Ca(OH)2 + H2S04 - CaSO4 + 2H20Na2CO3 + H2S - H2CO3 + Na2SThus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.
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Tony ran 600 meters in 60 seconds. What was Tony's speed during the
race?
The eight plants of the Solar System orbit the Sun in a chaotic random way.
True
False
Answer:
The Solar System has plants? I assume you meant planets. If so, that is false
Explanation:
NO LINKS PLEASE
At what speed do a bicycle and its rider, with a combined mass of 90 kg
k
g
, have the same momentum as a 1500 kg
k
g
car traveling at 4.0 m/s
m
/
s
?
Answer:
2
Explanation:
. Estimate the buoyant force that air exerts on you. (To do this, you can estimate your volume by knowing your weight and by assuming that your weight density is a bit less than that of water.)
Answer:
[tex]0.886[/tex] N buoyant force is exerted by air
Explanation:
My weight is [tex]75[/tex] Kg
Weight = mass * gravity
As we know
Buoyant Force is equal to the product of density * acceleration due to gravity and volume of the body
Assuming weight density is a bit less than that of water or equal to water i.e [tex]997.77[/tex] kg/m3
Volume is equal to mass / density
[tex]= 75[/tex] Kg * g/[tex]997.777[/tex]
[tex]= 0.0751[/tex] * g
Buoyant Force
= Volume * g * density
[tex]= 0.0751 * 9.8 * 1.2041[/tex]kg/m3
[tex]= 0.886[/tex] N
A uniform magnetic field is in the positive z direction. A positively charged particle is moving in the positive x direction through the field. The net force on the particle can be made zero by applying an electric field in what direction
Answer:
We apply an electric field in the negative y direction
Explanation:
Since A uniform magnetic field is in the positive z direction and A positively charged particle is moving in the positive x direction through the field, the magnetic force acting on the positively charged particle is in the positive y direction according to Fleming's right-hand rule.
For the net force on the particle to be zero, we apply an electric field in the negative y direction to create an electric force on the positively charged particle, so as to cancel out the magnetic force.
16. Two electric bulbs marked 100W 220V and 200W 200V have tungsten
filament of same length. Which of the two bulbs will have thicker
filament?
Answer:
The second bulb will have thicker filament
Explanation:
Given;
First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V
Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V
Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m
Resistance of the first bulb:
[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]
Resistance of the second bulb:
[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]
Resistivity of the tungsten filament is given by the following equation;
[tex]\rho = \frac{RA}{L}[/tex]
where;
L is the length of the filament
R is resistance of each filament
A is area of each filament
[tex]A = \pi r^2[/tex]
where;
r is the thickness of each filament
[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]
Therefore, the second bulb will have thicker filament
The moon does not stay at the same distance from the earth.why?
Answer:
The moon does not stay at the same distance of the earth because the ortbit of the moon is slightly elliptical. If earth is not tilted at an angle of 66.5°, there will be no change in the season and the earth will have equal length of days and night.
Explanation:
mark me brainlest
The water pressure to an apartment is increased by the water company. The water enters the apartment through an entrance valve at the front of the apartment. Where will the increase in the static water pressure be greatest when no water is flowing in the system
Answer:
Option C
Explanation:
Options for the question are as follows -
A. At a faucet close to entrance valve
B. At a faucet away from the entrance valve
C. It will be the same at all faucets
D. There will be no increase in the pressure at the faucets
Solution -
The static force will be the same at all faucets and also the area of the faucets be same.
Thus, the pressure created at all faucets will be the same.
Thus, option C is correct
A spring has a spring constant of 450 N/m. How much must this spring be stretched to store 49 J of potential energy?
Answer:
W = 1/2 K x^2
x^2 = 2 * W / K = 2 * 49 J / (N/m) = .218 / m^2
x = .467 m
If a virtual image is formed 10.0 cm along the principle axis from a convex mirror of focal length-15.0 cm, how far is the object from the mirror
Answer:
U=30cm
Explanation:
All you have to do is to put
Mirror formula , 1/f=1/u + 1/v
You should be careful in sign convention .
Virtual image is negative
we take focal length of convex lens negative even if its not given and so on...
A go-cart is traveling at a rate of 25 m/sec for 20 seconds. How far will the go cart travel?
Answer:
Distance travel by go-cart = 500 meter
Explanation:
Given:
Speed of go cart = 25 m/s
Time travel = 20 seconds
Find:
Distance travel by go-cart
Computation:
Distance = Speed x time
Distance travel by go-cart = Speed of go cart x Time travel
Distance travel by go-cart = 25 x 20
Distance travel by go-cart = 500 meter
A wave has a frequency of 67 Hz and a wavelength of 7.1 meters. What is the speed of this
wave?
Answer:
475.7 m/s
Explanation:
Given,
Frequency ( f ) = 67 Hz
Wavelength ( λ ) = 7.1 m
To find : Speed ( v ) = ?
Formula : -
v = f λ
v
= 67 x 7.1
= 475.7 m/s
Therefore,
the speed of the wave is 475.7 m/s.
Help me please with both questions?
Answer:
question #1 is A
Question #2 is C
Explanation:
A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction
acting on the box is a constant 20 N. How much work is done by the gravitational force?
A. 03
OB. 10 J
C. 100
D. 50 J
Answer:
B i think
Explanation:
...
A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft/s while connected to the end of a rope. If the rope is pulled inward with a constant speed of vr = 4 ft/s, determine the speed of the box at the instant rB = 1 ft. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box
Answer:
W = 1.875 J
Explanation:
For this exercise let's use the relationship between work and kinetic energy
W = ΔK
The kinetic energy of rotational motion is
K₀ = ½ I w²
we can assume that the box is small, so it can be treated as a point object, with moment of inertia
I = m rₐ²
angular and linear velocity are related
v = w r
w = v / r
we substitute in the equation, for point A
K₀ = ½ (m rₐ²) (v / rₐ)²
K₀ = ½ m v²
For the final point B, as the system is isolated the angular momentum is conserved
initial L₀ = Io wo
final L_f = I_f w_f
L₀ = L_f
I₀ w₀ = I_f w_f
(m rₐ²) w₀ = (m [tex]r_{b} ^2[/tex]) w_f
w_f = (rₐ/r_b)² w₀
with this value we find the final kinetic energy
K_f = ½ I_f w_f²
K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)² w₀) ²
K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]
we substitute in the realcion of work
W = K_f - K₀
W = ½ m [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²
W = ½ m [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²
W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2
W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]
let's calculate
W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)
W = 0.625 (3)
W = 1.875 J
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
. Calculate
the length of this wire required to make a 10 Ω coil.
Answer:
the length of the wire is 134.62 m.
Explanation:
Given;
resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm
cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²
resistance of the wire, R = 10Ω
The length of the wire is calculated as follows;
[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]
Therefore, the length of the wire is 134.62 m.
Batteries are not perfect. They can't deliver infinite current. As the current load on a battery gets larger, the voltage output gets smaller.
a. True
b. False
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500
Answer:
8 kV
Explanation:
Here is the complete question
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?
Solution
Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V
So, Q = CV
= 500 × 10⁻⁶ F × 800 V
= 400000 × 10⁻⁶ C
= 0.4 C
Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV
V = Q/C
= 0.4 C/500 × 10⁻⁶ F
= 0.0008 × 10⁶ V
= 800 V
The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)
= 10 × 800 V
= 8000 V
= 8 kV
A 5kg cart moving to the right with a velocity of 16 m/s collides with a concrete wall and
rebounds with a velocity of 22 m/s. Is the change in momentum of the cart
Explanation:
mass, m = 5kg
initial velocity, u = 16m/s
final velocuty, v = -22m/s
change in momentum, ∆p = ?
∆p = m (v-u)
5(-22-16)
5(38)
∆p = 190kgm/s
check the calculations!
A scientist measuring the resistivity of a new metal alloy left her ammeter in another lab, but she does have a magnetic field probe. So she creates a 4.5-m-long, 2.0-mm-diameter wire of the material, connects it to a 1.5 V battery, and measures a 3.0 mT magnetic field 1.0 mm from the surface of the wire. What is the material's resistivity
Answer:
[tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex]
Explanation:
r = Radius = [tex]\dfrac{2}{2}=1\ \text{mm}[/tex]
B = Magnetic field = 3 mT
1 mm = Distance from the surface of the wire
V = Voltage
x = Distance from the probe = [tex]r+1=1+1=2\ \text{mm}[/tex]
R = Resistance
L = Length of wire = 4.5 m
Magnetic field is given by
[tex]B=\dfrac{\mu_0I}{2\pi x}\\\Rightarrow I=\dfrac{B2\pi x}{\mu_0}\\\Rightarrow I=\dfrac{3\times 10^{-3}\times 2\times \pi 2\times 10^{-3}}{4\pi 10^{-7}}\\\Rightarrow I=30\ \text{A}[/tex]
Voltage is given by
[tex]V=IR\\\Rightarrow R=\dfrac{V}{I}\\\Rightarrow R=\dfrac{1.5}{30}\\\Rightarrow R=0.05\ \Omega[/tex]
Resistivity is given by
[tex]\rho=\dfrac{RA}{L}\\\Rightarrow \rho=\dfrac{0.05\times \pi (1\times 10^{-3})^2}{4.5}\\\Rightarrow \rho=3.49\times 10^{-8}\ \Omega\text{m}[/tex]
The resistivity of the material is [tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex].
Which of the following best defines
weather?
A. the expanding or contracting of the atmosphere
B. the measurement of the amount of water vapor in the
atmosphere
C. the condition of the atmosphere at a certain time and
place
Help Resources
D. the average air temperature of a specific region
Answer:
I'd say D
Explanation:
because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)
The ear drum vibrates when struck by sound waves and directly sends a message to the brain that is then recognized as sound
True or False
Answer:
true
Explanation:
A carnival ride starts at rest and is accelerated from an initial angle of zero to a final angle of 6.3 rad by a rad counterclockwise angular acceleration of 2.0 s2 What is the angular velocity at 6.3 rad?
The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Final angular velocity of the carnival ride
The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;
ωf = ωi + 2αθ
where;
ωf is the final angular velocity of the carnival ride = ?ωi is the initial angular velocity of the carnival ride = 0α is the angular acceleration = 2.0 rad/s²θ is the angular displacement of the carnival ride = 6.3 radωf = 0 + 2(2.0) x 6.3
ωf = 25.2 rad/s
Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Learn more about angular velocity here: https://brainly.com/question/6860269
Answer: 5.0 rad/s
Explanation: Because that’s what khan said so try it out.
Tameika makes a table about sensory organs
Eye
skin
brain
tongue
Which organ should be removed from the table?
A. eye
B. skin
C. brain
D. tongue
Answer:
I think its d
Explanation:
I'm not sure I'm sorry if I'm wrong
A woman shouts at a boy who is underwater what happens to the speed of the sound wave as it moves from the air into the water
Answer:
B. it increases
Explanation:
As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).
Answer:
B is the correct answer.
Explanation:
In a nuclear fusion reaction, atoms:
split apart.
combine.
explode.
cool down.
need help ASAP!!!!!!!!!!!
Answer:
The equation says that due to variation in temperature is
delt T = .59 m/s / C = 16 C * .59 m/s = 9.44 m/s
So v = 332 m/s + 9.44 m/s = 341 m/s (to three significant figures)
Which one the answer to this question
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward and ideal spring that is attached to a wall , After the block collides with the spring, the spring is compressed a maximum distance of 0.68m . what is the speed of the block when the spring is compressed to only one-half of the maximum distance?
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall, the maximum speed of the block when the spring is compressed to one-half of the maximum distance is 4.33 m/s
From the conservation of energy; the kinetic energy of the mass is equal to the work done on the spring.
i.e.
[tex]\mathbf{\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2_{max}}[/tex]
Given that:
the mass of the block = 4.0 kg the speed at which it is moving = 5.0 m/scompression of the spring = 0.68 m∴
From the equation above, multiplying both sides with 2, we have:
[tex]\mathbf{mv^2 =kx^2_{max}}[/tex]
Making (k) the subject of the formula;
[tex]\mathbf{k = \dfrac{mv^2}{x^2_{max}}}[/tex]
[tex]\mathbf{k = \dfrac{4 \times 5^2}{0.68^2}}[/tex]
k = 216.26 N/m
However, when compressed to one-half of the maximum distance; the speed is computed as follows:
x = 0.68/2 = 0.34 m
∴
[tex]\mathbf{\dfrac{1}{2}mv_o^2 - \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{m(v_o^2 -v^2) =kx^2}[/tex]
[tex]\mathbf{(v_o^2 -v^2) =\dfrac{kx^2}{m}}[/tex]
[tex]\mathbf{(5^2 -v^2) =\dfrac{216.26 \times 0.34^2}{4.0}}[/tex]
25 - v² = 6.25
25 -6.25 = v²
v² = 18.75
[tex]\mathbf{ v= \sqrt{18.75 }}[/tex]
v = 4.33 m/s
Therefore, we can conclude that the speed of the block when the spring is compressed to only one-half of the maximum distance is 4.33 m/s
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