Answer:
1.78×10×10^10 electron
Explanation:
Electric field outside the sphere can be calculated using below expression
E= kq/ r^2..........eqn(1)
Where k= 9 × 10^9 NM^2/C^2
q= charge
E= 1440 N/C
Diameter= 30.0 cm= 0.3 m
r= radius= 0.3/2= 0.15m
If we make q subject of formula from eqn(1) we have
q= Er^2/k............eqn(2)
q= 1440 × (0.15)^2 /(9 × 10^9 )
= 2.85×10^-9C
Total charge is an integer of electron charge , then we can calculate the number of the electron using the expression below
q= Ne
Where
N = number of electron
Making N subject of the formula we have
N= q/e
Where e= electron value= 1.6× 10^-19
N=2.85×10^-9 /1.6× 10^-19
= 1.78×10×10^10 electron
Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit
Answer:
0.03 A
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12 V
Resistor (R) = 470 Ω
Current (I) =?
From ohm's law, the voltage, current and resistor are related by the following formula:
Voltage = current × resistor
V = IR
With the above formula, we can obtain the current in the circuit as follow:
Voltage (V) = 12 V
Resistor (R) = 470 Ω
Current (I) =?
V = IR
12 = I × 470
Divide both side by 470
I = 12 / 470
I = 0.03 A
Thus, the current in the circuit is 0.03 A
Answer:
0.03 A
Explanation:
Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit
0.03 A
4) Which statement about teamwork is not true?
A) Team members should not have to make personal sacrifices for the success of the team.
B) To be successful, all team members need to agree about how to achieve the goal.
C) To achieve agreement, teams must be able to communicate and negotiate.
D) Team members need to be ready to resolve conflicts in an open and honest way
Answer: A) Team should not have to make personal sacrifices for the success of the team.
Explanation:
A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to the top of the tower and hung down the center of the staircase. However, nothing is available with which to measure the length of the rope. Therefore, at the bottom of the rope a small object is attached so as to form a simple pendulum that just clears the floor. The period of the pendulum is measured to be 6.82 s. What is the
Answer:
The answer is "[tex]11.55780\ m[/tex]"
Explanation:
Using formula:
[tex]= 2 \pi f= \frac{2\pi}{T} =\sqrt{\frac{g}{L}}[/tex]
L = length of pendulum.
[tex]= T =2 \pi \sqrt{\frac{L}{g}}[/tex]
Calculate the value for L:
[tex]L= g (\frac{T}{2 \pi})^2 \\\\[/tex]
[tex]= (9.80 \ \frac{m}{s^2}) (\frac{6.82 \ s}{2 \pi})^2\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124 \ s^2}{4 \times \pi^2})\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124\ s^2}{4 \times 9.8596 })\\\\= (9.80 \ \frac{m}{s^2}) (\frac{46.5124 \ s^2}{ 39.4384 })\\\\= \frac{455.82152}{39.4384} \ m\\\\=11.55780\ m[/tex]
The height of the tower = 11.55780 m
Light of wavelength 656 nm and 410 nm emitted from a hot gas of hydrogen atoms strikes a grating with 5300 lines per centimeter. a) Determine the angular deflection of both wavelengths in the 1st and 2nd order.
Answer:
[tex]20.32^{\circ}[/tex] and [tex]44.08^{\circ}[/tex]
[tex]12.56^{\circ}[/tex] and [tex]25.77^{\circ}[/tex]
Explanation:
[tex]\lambda[/tex] = Wavelength
[tex]\theta[/tex] = Angle
m = Order
Distance between grating is given by
[tex]d=\dfrac{1}{5300}\\\Rightarrow d=0.0001886\ \text{cm}[/tex]
[tex]\lambda=656\ \text{nm}[/tex]
We have the relation
[tex]d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=20.35^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=44.08^{\circ}[/tex]
The first and second order angular deflection is [tex]20.32^{\circ}[/tex] and [tex]44.08^{\circ}[/tex]
[tex]\lambda=410\ \text{nm}[/tex]
m = 1
[tex]\theta=\sin^{-1}\dfrac{1\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=12.56^{\circ}[/tex]
m = 2
[tex]\theta=\sin^{-1}\dfrac{2\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=25.77^{\circ}[/tex]
The first and second order angular deflection is [tex]12.56^{\circ}[/tex] and [tex]25.77^{\circ}[/tex].
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 41 mm , while nonathletes' stretch only 33 mm .
Hello. Your question is incomplete. However, I managed to find it completely on the internet and I realized that you forgot to mention that the question asks you for the maximum energy difference between velovistas and non-athletes, considering that the spring constant for the tendon of the two groups is equal to 33n/mm.
To make this calculation you will need to use Hooke's law, using the formula: ¹/2*K*x², where "K" will be the value of the spring constant for the tendon and "X" will be the value of the sprinter and non-athlete terms.
So for the sprinter we will have the calculation:
¹/2*33*41² -------> 0,5*33*1681 = 27736. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
For the non-athlete we will have the calculation:
¹/2*33*33² -------> 0,5*33*1089 = 17968. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
Now, to reach the final result, you only need to subtract the two values presented by the sprinter and the non-athlete.
27736.5 - 17968.5 = 9768 Nmm
Make a poem about waves with 12 Lines and 3 Stanzas.
In a ocean full of storms
A new wave was born
Deep into that darkness flooding
Suddenly, I heard some pummeling
By the grave I saw the winds
And the sun just shined
Answer:
a friendly face that comes with waves,
the waves of all the memorial days,
and with these days we smile with pride,
as for the waves we used to ride,
given up the day has passed,
how it went away like an hour glass,
as if we knew the world was right,
just like the waves, oh so bright,
the time has come the days have passed,
the waves ashore the waves alast,
as if the friendly face was right,
the waves that rode, oh goodnight.
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:
Heeelp me faaast plllsss
Answer:
the first one
Explanation:
cartridge fuse
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!
What is the medium of the wave shown in the photograph?
A. The water
B. Kinetic energy
C. The duck
D. Gravity
Answer:
A. The water
Explanation:
i got it right on A-P-E-X
6. The rate at which velocity changes is called
O speed
O direction
O acceleration
O displacement
Two ice skaters, with masses of 50 kg and 75 kg , are at the center of a 30 m -diameter circular rink. The skaters push off against each other and glide to opposite edges of the rink. Part A If the heavier skater reaches the edge in 30 s , how long does the lighter skater take to reach the edge
Answer:
t = 20 s
Explanation:
Assuming no other forces acting on the skaters when they push off against each other, and that we can neglect friction, total momentum must be conserved.The initial momentum is just zero, because both skaters are at rest.So, when both are gliding to opposite edges of the rink, at any moment, we can write the following expression:[tex]p_{f} = m_{1} * v_{1} = m_{2} * v_{2} (1)[/tex]
where m₁ = 50 kg, m₂ = 75 kg.We know that the heavier skater reaches the edge in 30 s.Since the distance from the center to any point on the edge is just half the diameter, we can find the speed of the heavier skater as follows:[tex]v_{2} = \frac{15m}{30s} = 0.5 m/s (2)[/tex]
Replacing m₁, m₂ and v₂ in (1), we can solve for the only unknown (v₁) as follows:[tex]v_{1} = \frac{m_{2}*v_{2}}{m_{1} } = \frac{75 kg*0.5m/s}{50kg} = 0.75 m/s (3)[/tex]
Since the distance to the opposite edge from the center is the same than for the heavier skater, we can find the time needed for the lighter one to reach the edge as follows:[tex]t_{1} = \frac{15m}{0.75m/s} = 20 s (4)[/tex]A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to rotate around it. The box is 60 cmcm wide and 70 cmcm high. What minimum magnitude of the force F is needed to cause this movement
Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N
The minimum magnitude of the force needed to cause the movement is 34.3 N.
Given:
Mass of the box (m) = 5.0 kg
Width of the box (w) = 60 cm = 0.6 m
Height of the box (h) = 70 cm = 0.7 m
The torque acting is:
τ_weight = (mg) × (h÷2)
The total torque is zero. Therefore, the force is:
F × (h÷2) + (mg) × (h÷2) = 0
F = -(mg) × (h÷2) ÷ (h÷2)
F = -(5.0 × 9.8 ) × (0.7 m ÷ 2) / (0.7 m ÷ 2)
F = -34.3 N
|F| = 34.3 N
Hence, the minimum magnitude of the force F needed to cause the movement is 34.3 N.
To learn more about Force, here:
https://brainly.com/question/13191643
#SPJ6
Tony ran 600 meters in 60 seconds. What was Tony's speed during the
race?
If this is the stationary wall isn’t the ANSWER that there is no work being done? If not what is the correct answer and why? Help!!
Answer:
no work is done cause there is no movement of the wall
A bullet of mass 4.00 g is fired horizontally into a wooden block of mass 1.30 kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.170. The bullet remains embedded in the block, which is observed to slide a distance 0.240 m along the surface before stopping. Part A What was the initial speed of the bullet
Answer:
[tex]291.67\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of bullet = 4 g
[tex]m_2[/tex] = Mass of block = 1.3 kg
[tex]\mu[/tex] = Coefficient of friction = 0.17
[tex]s[/tex] = Displacement of block = 0.24 m
[tex]v_1[/tex] = Velocity of bullet
[tex]v[/tex] = Velocity of combined mass
[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
The energy balance of the system is given by
[tex]\dfrac{1}{2}(m_1+m_2)v^2=\mu(m_1+m_2)gs\\\Rightarrow v=\sqrt{2\mu gs}[/tex]
As the momentum is conserved in the system we have
[tex]m_1v_1=(m_1+m_2)v\\\Rightarrow m_1v_1=(m_1+m_2)\sqrt{2\mu gs}\\\Rightarrow v_1=\dfrac{(m_1+m_2)\sqrt{2\mu gs}}{m_1}\\\Rightarrow v_1=\dfrac{(4\times 10^{-3}+1.3)\times \sqrt{2\times 0.17\times 9.81\times 0.24}}{4\times 10^{-3}}\\\Rightarrow v_1=291.67\ \text{m/s}[/tex]
The initial speed of the bullet is [tex]291.67\ \text{m/s}[/tex].
Why is it harder to breathe on a
mountain?
A. The air pressure is so high the lungs can't expand.
B. The air is denser and oxygen can't flow easily into the
lungs.
C. The denser oxygen molecules sink below the
surrounding air.
D. The air is less dense so there are fewer oxygen
molecules.
A wave has a wavelength of 1.5 meters and period of 0.083s. What is the waves speed?
A car weighing 1,500kg possesses 20 000 units of momentum. What would be the car's new momentum if it's velocity was tripled
Answer:
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Explanation:
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:
Which particle needs to be added to this equation to show that the total numbers of neutrons and protons are not changed by the reaction? MARKLING BRAINLIEST 70 points must be correct!
Answer:
C.
Explanation:
Answer:A
Explanation:ap3x
As air pressure decreases, what happens
to the density of the atmosphere?
A. increases
B. decreases
C. stays the same
D. not enough information to tell
Answer:
I believe it is B, not 100% sure though
Explanation:
Answer:
A. increases
Explanation:
The relationship between air pressure (atmospheric pressure) and the density of the atmosphere is inversely proportional.
The Brazilian rain forest is an area with significant biodiversity. As the rain forest is replaced with agricultural land, it is reasonable to predict a reduction in -
Answers-
A: consumption of solar energy.
B: sustainability over time.
C: precipitation levels.
D: average daily temperature.
Answer:
Bb
Explanation:
true or false
The Total electric potential due to two or more charges is equal to the algebraic sum of the potentials due to the individual charges.
Answer:
i guess the answer is false
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between his clothes and Earth is 0.701. He slides so that his speed is zero just as he reaches the base. (a)How much mechanical energy is lost due to friction acting on the runner
Answer:
-589.05 J
Explanation:
Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner
So, W = ΔK
W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)
So, substituting the values of the variables into the equation, we have
W = 1/2m(v₁² - v₀²)
W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)
W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)
W = 1/2 × 72.9 kg(-16.1604 m²/s²)
W = 1/2 × (-1178.09316 kgm²/s²)
W = -589.04658 kgm²/s²
W = -589.047 J
W ≅ -589.05 J
would it be m/s or kg?
Answer:
m.s
Explanation:
I need help with this review question. I’ll give extra points.
Answer:
1.9m/s²
Explanation:
Use the equation v=u+at, where v is the final speed, u is the initial speed, a is the acceleration and t is the time.
v=u+at
15.3=0+a(8)
a=15.3/8
a= 1.9125 m/s²
Look at the diagram showing the different wavelengths in sunlight.
A diagram showing the human eye and visible light. Visible light is broken down by color with wavelength in nanometers. Red is 700, orange is 600, yellow is 580, green is 550, blue is 475, indigo is 450, violet is 400.
Which has a wavelength of 350 nanometers?
red light
violet light
infrared light
ultraviolet light
Answer:
ultraviolet light
plz mark me as brainliest.
Answer:
Ultra violet
Explanation:
Galvani wrongly believed that the frog’s leg twitched during his experiment due to _____.
Answer:
nerves
Explanation:
I think, I maybe wrong.
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