Answer:
it increases-
Explanation:
When the mass of a rocket decreases as it burns through its fuel and the force ( thrust) is constant then by newtons second law of motion
F= ma here F is constant this means that ma= constant
⇒ m= F /a this implies that mass is inversely proportional to acceleration.
its means when the mass decreases the acceleration must increase. hence the acceleration increases
Part A A conducting sphere is charged up such that the potential on its surface is 100 V (relative to infinity). If the sphere's radius were twice as large, but the charge on the sphere were the same, what would be the potential on the surface relative to infinity?
Answer:
[tex]V_{2}=\frac{V_{1}}{2}[/tex]
Explanation:
The potential of a conducting sphere is the same as a punctual charge,
First sphere:
[tex]V_{1}=k*q_{1}/r_{1}[/tex] (1)
Second sphere:
[tex]V_{2}=k*q_{2}/r_{2}[/tex] (2)
But, second sphere's radius is twice first sphere radius, and their charges the same:
[tex]q_{1}=q_{2}[/tex]
[tex]r_{2}=2*r_{1}[/tex]
If we divide the equations (1) and (2), to solve V2:
[tex]V_{2}=V_{1}*r_{1}/r_{2}=V_{1}/2[/tex]
A 77.0 kg ice hockey goalie, originally at rest, catches a 0.150 kg hockey puck slapped at him at a velocity of 22.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities (in m/s) be in this case? (Assume the original direction of the ice puck toward the goalie is in the positive direction. Indicate the direction with the sign of your answer.)
Answer:
Explanation:
For elestic collision
v₁ = [tex]\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2}[/tex]
[tex]v_2 = [tex]\frac{(m_2-m_1)u_2}{m_1+m_2} +\frac{2m_1u_1}{m_1+m_2}[/tex][/tex]
Here u₁ = 0 , u₂ = 22 m/s , m₁ = 77 kg , m₂ = .15 kg , v₁ and v₂ are velocity of goalie and puck after the collision.
v₁ = 0 + ( 2 x .15 x22 )/ 77.15
= .085 m / s
Velocity of goalie will be .085 m/s in the direction of original velocity of ball before collision.
v₂ = (.15 - 77)x 22 / 77.15 +0
= - 21.91 m /s
=Velocity of puck will be - 21.91 m /s in the direction opposite to original velocity of ball before collision.
A resistor R and another resistor 2R are connected in a series across a battery. If heat is produced at a rate of 10W in R, then in R it is produced at a rate of: A. 40 W
B. 20 W
C. 10 W
D. 5 W
Answer:
Option B
Solution:
As per the question:
Heat produced at the rate of 10 W
The resistor R and 2R are in series.
Also, in series, same current, I' passes through each element in the circuit.
Therefore, current is constant in series.
Also,
Power,[tex] P' = I'^{2}R[/tex]
When current, I' is constant, then
P' ∝ R
Thus
[tex]\frac{P'}{2R} = \frac{10}{R}[/tex]
P' = 20 W
The power dissipated in the resistor 2R is B. 20 W.
When resistors are connected in series, the current through each resistor is the same. Given two resistors, R and 2R, connected in series across a battery, we know that the power dissipated by resistor R is 10 W.
In a series circuit, the voltage across each resistor is different, but the current is the same.
The power dissipated by a resistor in a circuit is given by the formula:
P = I²R
Since the resistors are in series, the same current (I) flows through both resistors. Let's denote the current flowing through the series circuit as I. Using the given power dissipation for resistor R, we get:
10 W = I²R
To find the current:
I = √(10/R)
Next, we calculate the power dissipated by the resistor 2R:
P = I²(2R)
Substituting the value of I:
P = (10/R) × (2R) = 20 W
Thus, the power dissipated in resistor 2R is 20 W.
The correct answer is B. 20 WIf y = 0.02 sin (20x – 400t) (SI units), the wave number is
Answer:
Wave number, [tex]k=20\ m^{-1}[/tex]
Explanation:
The given equation of wave is :
[tex]y=0.02\ sin(20x-400t)[/tex]............(1)
The general equation of the wave is given by :
[tex]y=A\ sin(kx-\omega t)[/tex]..............(2)
k is the wave number of the wave
[tex]\omega[/tex] is the angular frequency
On comparing equation (1) and (2) :
[tex]k=20\ m^{-1}[/tex]
[tex]\omega=400\ rad[/tex]
So, the wave number of the wave is [tex]20\ m^{-1}[/tex]. Hence, this is the required solution.
As a hurricane passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury barometer drops by 21.4 mm from the normal height, what is the atmospheric pressure (in Pa)? Normal atmospheric pressure is 1.013 ✕ 105 Pa and the density of mercury is 13.6 g/cm3 (NOTE: This is g/cm^3, not SI units of kg/m^3).
Answer:
The atmospheric pressure is [tex]9.845 \times 10^4\ Pa[/tex]
Explanation:
There are two ways of solving this exercise:
1)
In physics you can find that mmHg is a unit of pressure.
Pressure = Force/area.
If you consider the weight of mercury as your force (mass* acceleration of gravity = density*volume* acceleration of gravity ), then
[tex]P =\frac{\rho Vg}{A} = \frac{\rho Ahg}{A} =\rho g h[/tex]
where h is the height of the mercury column and rho its density.
[tex]\Delta P= P_{atm} - P_{hur} = 21.4\ mm Hg[/tex].
if normal atmospheric pressure is [tex]1.013 \times 10^5\ Pa = 759.81\ mmHg[/tex]
then the pressure in the presence of the hurricane is
[tex]P_{hur} = 759.81 - 21.4 = 738.41\ mmHg = 9.845 \times 10^4\ Pa[/tex]
2)
Considering the definition of pressure
[tex]\Delta P = \rho g h[/tex]
where [tex]\rho = 13.6\ g/cm^3[/tex], [tex]g = 9.8\ m/s^2 =980\ cm/s^2[/tex] and [tex]h = 21.4\ mm = 2.14\ cm[/tex].
[tex]\Delta P = P_{atm} - P_{hur} = 28521,92\ g/cms^2 = 2852,192\ kg/ms^2[/tex], where [tex]kg/ms^2 = Pa[/tex].
if
[tex]P_{atm} = 1.013 \times 10^5\ Pa[/tex],
then
[tex]P_{hur} = 1.013 \times 10^5 - 2852,192 =9.845 \times 10^4\ Pa[/tex].
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 7.61 m/s. The stone subsequently falls to the ground, which is 18.1 m below the point where the stone leaves his hand. At what speed does the stone impact the ground? Ignore air resistance and use g=9.81 m/s^2 for the acceleration due to gravity.
The stone impacts the ground with a speed of approximately [tex]\(17.32 \, \text{m/s}\)[/tex].
To find the impact speed of the stone, we can use the kinematic equation that relates initial velocity [tex](\(v_0\))[/tex], final velocity [tex](\(v\))[/tex], acceleration [tex](\(g\))[/tex], and displacement [tex](\(s\))[/tex]:
[tex]\[v^2 = v_0^2 + 2gs\][/tex]
Where:
- [tex]\(v\)[/tex] is the final velocity (impact speed),
- [tex]\(v_0\)[/tex] is the initial velocity (throwing speed),
- [tex]\(g\)[/tex] is the acceleration due to gravity (9.81 m/s²),
- [tex]\(s\)[/tex] is the displacement (height the stone falls, -18.1 m, as it falls downward).
Substituting the known values:
[tex]\[v^2 = (7.61 \, \text{m/s})^2 + 2 \times (9.81 \, \text{m/s}^2) \times (-18.1 \, \text{m})\][/tex]
[tex]\[v^2 = 58.0321 - 2 \times 9.81 \times 18.1\][/tex]
[tex]\[v^2 \approx 58.0321 - 357.801\][/tex]
[tex]\[v^2 \approx -299.7689\][/tex]
Since the stone is impacting the ground, we consider the positive root:
[tex]\[v \approx \sqrt{299.7689}\][/tex]
[tex]\[v \approx 17.32 \, \text{m/s}\][/tex]
Therefore, the stone impacts the ground with a speed of approximately [tex]\(17.32 \, \text{m/s}\)[/tex].
What is the area of a circle of radius (a) 5.142 m and (b) 1.7 m?
Answer:
(a) Area = [tex]83.022m^2[/tex] (b) Area = [tex]9.07m^2[/tex]
Explanation:
We have given radius of the circle r = 5.142 m and r = 1.7 m
We have to find the area of the circle
We know that area of the circle is given by
[tex]A=\pi r^2[/tex], here r is the radius of the circle
(a) Radius = 5.142 m
So area [tex]A=\pi r^2=3.14\times 5.142^2=83.022m^2[/tex]
(b) Radius = 1.7 m
So area [tex]A=\pi r^2=3.14\times 1.7^2=9.07m^2[/tex]
A driver of a car going 75 km/h suddenly sees the lights of a barrier 35 m ahead. It takes the driver 0.60 seconds before he applies the brakes, and the average acceleration during braking is -8.5 m/s?. (A) Does the car hit the barrier? (B) What is the maximum speed at which the car could be moving and not hit the barrier 35 meters ahead?
Answer:
A) The car will hit the barrier
b) 19.82 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Distance = Speed × Time
⇒Distance = (75/3.6)×0.6
⇒Distance = 12.5 m
Distance traveled by the car during the reaction time is 12.5 m
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-\frac{75}{3.6}^2}{2\times -8.5}\\\Rightarrow s=25.53\ m[/tex]
Total distance traveled is 12.5+25.53 = 38.03 m
So, the car will hit the barrier
Distance = Speed × Time
⇒d = u0.6
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-u^2}{2\times -8.5}\\\Rightarrow s=\frac{u^2}{17}[/tex]
d + s = 35
[tex]\\\Rightarrow u0.6+\frac{u^2}{17}=35\\\Rightarrow \frac{u^2}{17}+0.6u-35=0[/tex]
[tex]10u^2+102u-5950=0\\\Rightarrow u=\frac{-51+\sqrt{62101}}{10},\:u=-\frac{51+\sqrt{62101}}{10}\\\Rightarrow u=19.82, -30.02[/tex]
Hence, the maximum velocity by which the car could be moving and not hit the barrier 35 meters ahead is 19.82 m/s.
Suppose that a steel bridge, 1000 m long, were built without any expansion joints. Suppose that only one end of the bridge was held fixed. What would the difference in the length of the bridge be between winter and summer, taking a typical winter temperature as 0°C, and a typical summer temperature as 40°C? The coefficient of thermal expansion of steel is 10.5 × 10 -6 K -1. 0.37 cm 0.11 mm 0.42 m 0.42 mm 0.11 m
Answer:
0.42 m
Explanation:
For thermal expansion, the formula used is as follows
L = L₀ ( 1 + α t )
L is length after rise of temperature by t , L₀ is length before rise of temperature , α is coefficient of thermal expansion and t is rise in temperature.
L - L₀ = L₀α t
Difference of length = L₀α t
L₀ = 1000 m , α = 10.5 x 10⁻⁶ , t = 40
Difference of length = 1000 x 10.5 x 10⁻⁶x 40
= 0.42 m .
A baseball goes from zero to 34 m/s in 0.188 s. What is its average acceleration? Answer in units of m/s^2
Answer:
[tex]a_{avg} = 180.85 m/s^{2}[/tex]
Given:
Initial velocity, u = 0 m/s
Final velocity, v = 34 m/s
Time interval, [tex]\Delta t = 0.188 s[/tex]
Solution:
Acceleration is the rate at which the velocity of an object changes.
Thus
Average acceleration, [tex]a_{avg} = \frac{\Delta v}{\Delta t}[/tex]
[tex]a_{avg} = \frac{v - u}{t - 0} = \frac{34 - 0}{0.188} = 180.85 m/s^{2}[/tex]
At cruise conditions, air flows into a jet engine at a steady rate of 60 lbm/s. Fuel enters the engine at a steady rate of 0.59 lbm/s. The average velocity of the exhaust gases is 1485 ft/s relative to the engine. If the engine exhaust effective cross-sectional area is 4.0 ft2, estimate the density of the exhaust gases in lbm/ft3.
Answer:
ρ=0.0102lbm/ft^3
Explanation:
To solve this problem we must take into account the equation of continuity, this indicates that the sum of the mass flows that enter a system is equal to the sum of all those that leave.
Therefore, to find the mass flow of exhaust gases we must add the mass flows of air and fuel.
m=0.59+60=60.59lbm/s( mass flow of exhaust gases)
The equation that defines the mass flow (amount of mass that passes through a pipe per unit of time) is as follows
m=ρVA
Where
ρ=density
V=velocity
m=mass flow
A=cross-sectional area
solving for density
ρ=m/VA
ρ=60.59/{(1485)(4)}
ρ=0.0102lbm/ft^3
A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later. How fast was the pebble going when it hit the water?
A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water [tex]2.00\ s[/tex] later. The pebble was going at a speed of [tex]19.62\ m/s[/tex] when it hit the water.
Use kinematic equations to solve this problem. The pebble is dropped from rest, so its initial velocity is 0 m/s. The time it takes for the pebble to hit the water is given as 2.00 seconds.
The speed of the pebble when it hits the water:
Given:
Initial velocity [tex](v_o) = 0 m/s[/tex]
Time [tex](t) = 2.00\ seconds[/tex]
Acceleration due to gravity [tex](a) = -9.81\ m/s^2[/tex]
Use the kinematic equation:
[tex]v = v_o + a \times t\\v = 0 + (-9.81) \times (2.00)[/tex]
Calculate the numerical value of v:
[tex]v = -19.62\ m/s[/tex]
So, the pebble was going at a speed of [tex]19.62\ m/s[/tex] when it hit the water.
To know more about the kinematic equation:
https://brainly.com/question/24458315
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The pebble was traveling at a speed of 19.6 m/s when it hit the water.
Explanation:In order to determine the speed at which the pebble hit the water, we can use the equations of motion. Assuming the acceleration due to gravity is 9.8 m/s^2 and neglecting air resistance, we can use the equation:
s = ut + (1/2)at^2
Where s is the distance traveled, u is the initial velocity, t is the time, and a is the acceleration. In this case, the distance traveled is the height of the cliff, the initial velocity is 0 (since the pebble is dropped), t is 2.00 s, and the acceleration is 9.8 m/s^2. Plugging in these values, we can solve for the initial velocity:
s = 0 * (2.00) + (1/2) * 9.8 * (2.00)^2
s = 19.6 m
Therefore, the pebble was traveling at a speed of 19.6 m/s when it hit the water.
A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree. At some instant the little creature moves with a velocity of -1.03 m/s. Then, 2.47 s later, it moves at the velocity 1.51 m/s. What is the chipmunk\'s average acceleration during the 2.47-s time interval?
Answer:
a = 1.02834008 m/s2
Explanation:
given data:
initial velocity u = -1.03 m/s
time t = 2.47 s later
final velocity v = 1.51 m/s
average acceleration is given as a
[tex]a = \frac{(v - u)}{t}[/tex]
putting all value to get required value of acceleration:
[tex]= \frac{(1.51 -(- 1.03))}{2.47}[/tex]
[tex]= \frac{1.51+1.03}{2.47}[/tex]
= 1.02834008 m/s2
A hawk flying at 38 m/s emits a cry whose frequency is 440 Hz. A wren is moving in the same direction as the hawk at 17 m/s. (Assume the speed of sound is 343 m/s.) (a) What frequency does the wren hear (in Hz) as the hawk approaches the wren? Hz (b) What frequency does the wren hear (in Hz) after the hawk passes the wren?
Answer:
frequency wren hear when approaches is 470.29 Hz
frequency wren hear after hawk pass is 415.75 Hz
Explanation:
given data
hawk velocity Vs= 38 m/s
frequency f = 440 Hz
wren velocity Vo= 17 m/s
speed of sound s = 343 m/s
to find out
What frequency wren hear and What frequency wren hear after hawk pass
solution
we apply here frequency formula that is
[tex]f1=f ( \frac{s+V_o}{s+V_s} )[/tex] .........1
here f1 is frequency hear by observer
put here all value as Vo and Vs negative because it approaches
[tex]f1=f ( \frac{s-V_o}{s-V_s} )[/tex]
[tex]f1=440 ( \frac{343-17}{343-38} )[/tex]
f1 = 470.29 Hz
so frequency wren hear when approaches is 470.29 Hz
and
after passing from equation 1 we take both Vo and Vs as positive
[tex]f1=f ( \frac{s+V_o}{s+V_s} )[/tex]
[tex]f1=440 ( \frac{343+17}{343+38} )[/tex]
f1 = 415.75 Hz
so frequency wren hear after hawk pass is 415.75 Hz
The pressure in a compressed air storage tank is 1200 kPa. What is the tank's pressure in (a) kN and m units, (b) kg, m, and s units, and (c) kg, km, and s units?
Answer:
a) 1200 kN/m²
b) 1,200,000 kg/ms²
c) 1.2 × 10⁹ kg/km.s²
Explanation:
Given:
Pressure = 1200 kPa
a) 1 Pa = 1 N/m²
thus,
1000 N = 1 kN
1200 kPa = 1200 kN/m²
b) 1 Pa = 1 N/m² = 1 kg/ms²
Thus,
1200 kPa = 1200000 Pa
or
1200000 Pa = 1200000 × 1 kg/ms²
or
= 1,200,000 kg/ms²
c) 1 km = 1000 m
or
1 m = 0.001 Km
thus,
1,200,000 kg/ms² = [tex]\frac{1,200,000}{0.001}\frac{kg}{km.s^2}[/tex]
or
= 1.2 × 10⁹ kg/km.s²
(a) In kN and m units, the pressure is [tex]$\boxed{1200 \text{ kPa}}$[/tex]. (b) In kg, m, and s units, the pressure is [tex]$\boxed{1200 \text{ kg/(m·s}^2\text{)}}.[/tex] (c) In kg, km, and s units, the pressure is [tex]\boxed{1200 \text{ kg/(km·s}^2\text{)}}[/tex].
Explanation and logic of the
To convert pressure from kPa to the required units, we need to understand the relationship between pressure, force, and area. Pressure (P) is defined as the force (F) applied per unit area (A), given by P = F/A.
(a) In the International System of Units (SI), pressure is measured in pascals (Pa), where 1 Pa = 1 N/m². Since 1 kPa = 1000 Pa, and 1 kN = 1000 N, the pressure in kN/m² is numerically the same as in kPa. Therefore, the pressure in the tank is 1200 kN/m².
(b) To express the pressure in base SI units of kg, m, and s, we need to consider that 1 N = 1 kg·m/s². Since 1 kPa = 1 kN/m², and 1 kN = 1000 N, we have:
[tex]\[ 1200 \text{ kPa} = 1200 \times \frac{1000 \text{ kg\·m/s}^2}{1 \text{ m}^2} = 1200 \times 1000 \text{ kg/(m\s}^2\text{)} \][/tex]
(c) To express the pressure in kg, km, and s, we convert the area from m² to km²:
[tex]\[ 1 \text{ m}^2 = \frac{1}{1000 \times 1000} \text{ km}^2 = \frac{1}{1000000} \text{ km}^2 \][/tex]
Thus, the pressure in kg/(km.s²) is:
[tex]\[ 1200 \text{ kPa} = 1200 \times \frac{1000 \text{ kg\·m/s}^2}{\frac{1}{1000000} \text{ km}^2} = 1200 \times 1000000 \text{ kg/(km\·s}^2\text{)} \][/tex]
However, since 1 kPa is equivalent to 1 kN/m², and 1 kN = 1000 N, we can simplify the expression by recognizing that the conversion from m² to km² results in a factor of 1/1000000, which cancels out the factor of 1000 from the conversion of kN to N, leaving us with:
[tex]\[ 1200 \text{ kPa} = 1200 \text{ kg/(km\s}^2\text{)} \][/tex]
Therefore, the pressure in the tank, when expressed in kg/(km·s²), is 1200 kg/(km·s²).
A student at the top of a building of height h throws ball A straight upward with speed v0 (3 m/s) and throws ball B straight downward with the same initial speed. A. Compare the balls’ accelerations, both direction, and magnitude, immediately after they leave her hand. Is one acceleration larger than the other? Or are the magnitudes equal? B. Compare the final speeds of the balls as they reach the ground. Is one larger than the other? Or are they equal?
Answer:same
Explanation:
Given
ball A initial velocity=3 m/s(upward)
Ball B initial velocity=3 m/s (downward)
Acceleration on both the balls will be acceleration due to gravity which will be downward in direction
Both acceleration is equal
For ball A
maximum height reached is [tex]h_1=\frac{3^2}{2g}[/tex]
After that it starts to move downwards
thus ball have to travel a distance of h_1+h(building height)
so ball A final velocity when it reaches the ground is
[tex]v_a^2=2g\left ( h_1+h\right )[/tex]
[tex]v_a^2=2g\left ( 0.458+h\right )[/tex]
[tex]v_a=\sqrt{2g\left ( 0.458+h\right )}[/tex]
For ball b
[tex]v_b^2-\left ( 3\right )^2=2g\left ( h\right )[/tex]
[tex]v_b^2=2g\left ( \frac{3^2}{2g}+h\right )[/tex]
[tex]v_b=\sqrt{2g\left ( 0.458+h\right )}[/tex]
thus [tex]v_a=v_b[/tex]
Final answer:
A. The magnitudes of the accelerations of both balls are equal, but they have opposite directions. B. The final speeds of both balls when they reach the ground will be the same.
Explanation:
A. The acceleration of both balls will be the same in magnitude but in opposite directions. Since the acceleration due to gravity acts downward, ball A will have a negative acceleration while ball B will have a positive acceleration. Therefore, the magnitudes of their accelerations will be equal.
B. When both balls reach the ground, their final speeds will also be the same. This is because the vertical motion of the balls is independent of their initial speeds when air resistance is ignored. The time it takes for them to reach the ground will be the same, and hence their final velocities will also be equal.
Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).
Answer:
a) TB = m2 * w^2 * 2*d
b) TA = m1 * w^2 * d + m2 * w^2 * 2*d
Explanation:
The tension on the strings will be equal to the centripetal force acting on the boxes.
The centripetal force is related to the centripetal acceleration:
f = m * a
The centripetal acceleration is related to the radius of rotation and the tangential speed:
a = v^2 / d
f = m * v^2 / d
The tangential speed is:
v = w * d
Then
f = m * w^2 * d
For the string connecting boxes 1 and 2:
TB = m2 * w^2 * 2*d
For the string connecting box 1 to the shaft
TA = m1 * w^2 * d + m2 * w^2 * 2*d
Two particles with positive charges q1 and q2 are separated by a distance s. Part A Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero? Express your answer in terms of some or all of the variables s, q1, q2 and k =14πϵ0. If your answer is difficult to enter, consider simplifying it, as it can be made relatively simple with some work.
The solution for the distance x from [tex]\(q_1\)[/tex] along the line connecting the charges where the total electric field is zero is given by: [tex]\[x = \frac{s \cdot (q_1 + q_1\sqrt{q_2})}{q_1 + q_2}\][/tex].
Let's go through the complete solution step by step.
Given:
- Charges: [tex]\(q_1\)[/tex] and [tex]q_2[/tex]
- Distance between charges: s
- Coulomb's constant: [tex]\(k = \frac{1}{4\pi\epsilon_0}\)[/tex]
We want to find the distance (x) from charge [tex]\(q_1\)[/tex] along the line connecting the charges where the total electric field is zero.
The electric field (E) due to a point charge (q) at a distance (r) is given by Coulomb's law:
[tex]\[E = \dfrac{k \cdot q}{r^2}\][/tex]
At a distance x from charge [tex]\(q_1\)[/tex], the electric fields due to [tex]\(q_1\) (\(E_1\))[/tex] and [tex]\(q_2\) (\(E_2\))[/tex] will have magnitudes given by:
[tex]\[E_1 = \dfrac{k \cdot q_1}{x^2}\][/tex]
[tex]\[E_2 = \dfrac{k \cdot q_2}{(s - x)^2}\][/tex]
For the total electric field to be zero, [tex]\(E_1\)[/tex] and [tex]\(E_2\)[/tex] must cancel each other out:
[tex]\[E_1 + E_2 = 0\][/tex]
Substitute the expressions for [tex]\(E_1\)[/tex] and [tex]\(E_2\)[/tex]:
[tex]\[\dfrac{k \cdot q_1}{x^2} + \dfrac{k \cdot q_2}{(s - x)^2} = 0\][/tex]
Cross-multiply and simplify:
[tex]\[q_1 \cdot (s - x)^2 = -q_2 \cdot x^2\][/tex]
Expand and rearrange:
[tex]\[q_1 \cdot (s^2 - 2sx + x^2) = -q_2 \cdot x^2\][/tex]
Solve for x:
[tex]\[q_1 \cdot s^2 - 2q_1 \cdot sx + q_1 \cdot x^2 + q_2 \cdot x^2 = 0\][/tex]
[tex]\[(q_1 + q_2) \cdot x^2 - 2q_1 \cdot sx + q_1 \cdot s^2 = 0\][/tex]
This is a quadratic equation in terms of x. Solve for x using the quadratic formula:
[tex]\[x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]
Where:
[tex]\(a = q_1 + q_2\)[/tex]
[tex]\(b = -2sq_1\)[/tex]
[tex]\(c = q_1s^2\)[/tex]
Calculate x using the quadratic formula:
[tex]\[x = \dfrac{-(-2sq_1) \pm \sqrt{(-2sq_1)^2 - 4(q_1 + q_2)(q_1s^2)}}{2(q_1 + q_2)}\][/tex]
Simplify the expression inside the square root:
[tex]\[x = \dfrac{2sq_1 \pm \sqrt{4s^2q_1^2 - 4(q_1 + q_2)(q_1s^2)}}{2(q_1 + q_2)}\][/tex]
[tex]\[x = \dfrac{2sq_1 \pm \sqrt{4s^2q_1^2 - 4q_1^2s^2 - 4q_2q_1s^2}}{2(q_1 + q_2)}\][/tex]
[tex]\[x = \dfrac{2sq_1 \pm \sqrt{-4q_2q_1s^2}}{2(q_1 + q_2)}\][/tex]
[tex]\[x = \dfrac{2sq_1 \pm 2q_1s\sqrt{-q_2}}{2(q_1 + q_2)}\][/tex]
[tex]\[x = \dfrac{s \cdot (q_1 \pm q_1\sqrt{-q_2})}{q_1 + q_2}\][/tex]
Since [tex]\(q_1\)[/tex] and [tex]\(q_2\)[/tex] are both positive charges, x will be a positive value.
Thus, the solution for the distance x from [tex]\(q_1\)[/tex] is given by [tex]\[x = \dfrac{s \cdot (q_1 + q_1\sqrt{q_2})}{q_1 + q_2}\][/tex].
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Two objects of different mass are released simtaneously
fromthe top of a 20- m tower and fall to the ground. If air
resistanceis negligible, which statement best applies?
A. the greater mass hits the ground first.
B. Both objects hit hte ground together.
C. The smaller mass hits the ground first.
D. No conclusion can be made with the information given.
Answer:b-both objects hit the ground together
Explanation:
Given
Two objects of different mass are released simultaneously from top of a tower and fall to the ground provided air resistance is negligible
then both the object hit the ground together
Because in equation of motion terms there is no unit of mass
But if the air resistance is present then the answer would have been different.
A 12 V storage battery is charged by a current of 20 A for 1 hr. A) How much power is required to charge the battery at this rate? B) How much energy has been provided during the process?
Answer:
A) 240 W
B) 864000 J
Explanation:
Hi!
We can easily calculate the required power multiplying the voltage of the process times the circulating current:
[tex]P=12V\times 20A=12(\frac{J}{coulomb})\times20(\frac{coulomb}{s})\\P = 240\frac{J}{s}=240W[/tex]
To calculate the total energy provided we need to multiply the power times the total time:
[tex]E=P\times t=240\frac{J}{s} \times 1 hr =240\frac{J}{s} \times 3600s\\ E=864 000J[/tex]
Hope this helps
Have a nice day!
Final answer:
The power required to charge the battery is 240 watts, and the energy provided during the charging process is 864,000 joules.
Explanation:
A student has asked how much power is required to charge a 12 V storage battery by a current of 20 A for 1 hour, and how much energy has been provided during the process.
A) To find the power required to charge the battery, we use the formula:
Power (P) = Voltage (V) imes Current (I)
P = 12 V imes 20 A
P = 240 W
Therefore, the power required to charge the battery at this rate is 240 watts.
B) The energy provided during the charging process can be calculated using the formula:
Energy (E) = Power (P) imes Time (t)
Since power is given in watts and time in hours, we must convert the time to seconds for consistency in units (1 hour = 3600 seconds).
E = 240 W imes 1 hour imes 3600 seconds/hour
E = 240 W imes 3600 s
E = 864,000 J
The energy provided during the charging process is 864,000 joules (or 864 kJ).
The question states: two large, parallel conducting plates are 12cm
apart and have charges of equal magnitude and opposite sign ontheir
facing surfaces. An electrostatic force of 3.9 x 10^-15 Nacts on an
electron placed anywhere between two plates (neglectingthe
fringing).
1. find the elctric field at the position of the elctron.
2. what is the potential difference between the plates?
Answer:
1. 24375 N/C
2. 2925 V
Explanation:
d = 12 cm = 0.12 m
F = 3.9 x 10^-15 N
q = 1.6 x 10^-19 C
1. The relation between the electric field and the charge is given by
F = q E
So, [tex]E=\frac{F}{q}[/tex]
[tex]E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}[/tex]
E = 24375 N/C
2. The potential difference and the electric field is related by the given relation.
V = E x d
where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.
By substituting the values, we get
V = 24375 x 0.12 = 2925 Volt
Final answer:
The electric field at the position of the electron is approximately 2.43 x 10^4 N/C, and the potential difference between the plates is about 2916 V.
Explanation:
The electric field (E) within two charged parallel plates is given by the expression E = F/q, where F is the force experienced by a charge q. Since an electron has a charge of approximately 1.602 x 10^-19 C, and it is experiencing a force 3.9 x 10^-15 N, we can calculate the electric field as E = (3.9 x 10^-15 N) / (1.602 x 10^-19 C) ≈ 2.43 x 10^4 N/C. The direction of the electric field is from the positively charged plate to the negatively charged plate.
The potential difference (V) between the plates can be found using the relationship V = E × d, where d is the distance between the plates. Given that the plates are 12 cm apart, this distance in meters is 0.12 m, so the potential difference is V = (2.43 x 10^4 N/C) × 0.12 m ≈ 2916 V.
Two 1.0 kg masses are 1.0 m apart (center to center) on a frictionless table. Each has a +10μC of charge. a) What is the magnitude of the electric force on one of the masses? b) What is the initial acceleration of one of the masses if it is released and allowed to move?
Answer:
(a). The electric force is 0.9 N.
(b). The acceleration is 0.9 m/s²
Explanation:
Given that,
Mass = 1.0 kg
Distance = 1.0 m
Charge = 10 μC
(a). We need to calculate the electric force
Using formula of electric force
[tex]F = \dfrac{kq_{1}q_{2}}{r^2}[/tex]
Put the value into the formula
[tex]F=\dfrac{9\times10^{9}\times10\times10^{-6}\times10\times10^{-6}}{1.0^2}[/tex]
[tex]F=0.9\ N[/tex]
(b). We need to calculate the acceleration
Using newton's second law
[tex]F = ma[/tex]
[tex]a =\dfrac{F}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{0.9}{1.0}[/tex]
[tex]a=0.9\ m/s^2[/tex]
Hence, (a). The electric force is 0.9 N.
(b). The acceleration is 0.9 m/s²
By applying Coulomb's law and Newton's second law, we find that the magnitude of the electric force on one of the masses is 0.0899 N and the initial acceleration of one of the masses is 0.0899 m/s^2.
Explanation:This question pertains to Coulomb's law, which deals with the force between two charges. Coulomb's law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The equation used for this law is F = k(q1*q2)/r², where F represents the force, q1 and q2 represent the charges, r represents the distance between the charges and k is Coulomb's constant (8.99 × 10⁹ N×m²/C²).
a) To find the magnitude of the electric force on one of the masses, we would use the Coulomb's law equation: F = k(q1×q2)/r², where q1 and q2 are both +10μC (which need to be converted to C by multiplying by 10⁻⁶), and r is 1.0m. This gives us F = (8.99 × 10⁹ N×m²/C²) × ((10 × 10⁻⁶ C)²) / (1.0m)² = 0.0899 N.
b) To find the initial acceleration of one of the masses, we would use Newton's second law (F = ma), where F is the force (0.0899 N from the previous calculation), m is the mass (1.0 kg), and a is acceleration. So a = F/m = 0.0899 N / 1.0 kg = 0.0899 m/s².
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Whole group time allows early childhood teachers to do all of the following except
A. Review previously covered material so that all children are getting the same information at the same time.
B. Work one on one with children in need
C. Set expectations
D. Introduce new concepts.
Answer:
B. Work one on one with children in need
Explanation:
Whole group time allows early childhood teachers to introduce new material. It ensures that each student is presented and reviewed with uniform key concepts. It also sets expectations in planning and developing the lessons because it provides baseline assessments.The only thing that seems impossible in this activity is that teacher can give individual or one-to-one attention to each student.
hence the correction will be option B Work one on one with children in need
A house is 54.0 ft long and 48 ft wide, and has 8.0-ft-high ceilings. What is the volume of the interior of the house in cubic meters and cubic centimeters? Answers needs to be in appropriate significant figures.
Explanation:
Length of house, l = 54 ft = 16.45 m
Breadth of house, b = 48 ft = 14.63 m
Height of house, h = 8 ft = 2.43 m
We need to find the volume of the interior of the house. The house is in the form of cuboid. The volume of cuboid is given by :
[tex]V=l\times b\times h[/tex]
[tex]V=16.45\ m\times 14.63\ m\times 2.43\ m[/tex]
[tex]V=584.81\ m^3[/tex]
Since, [tex]1\ m^3=1000000\ cm^3[/tex]
or the volume of the interior of the house, [tex]V=5.84\times 10^8\ cm^3[/tex]
Hence, this is the required solution.
If AB has a bearing of following 234° 51' 48" and a anti-clockwise angle from AB to C is measured as 80° Calculate the bearing AC. (Enter as numeric value of ddd.mmss e.g. 100° 20' 30" would be entered as 100.2030. Marked out of 5.00 P
Answer:
the bearing of the line AC will be 154° 51' 48"
Explanation:
given,
bearing of the line AB = 234° 51' 48"
an anticlockwise measure of an angle to the point C measured 80°
to calculate the bearing of AC.
As the bearing of line AB is calculated clockwise from North direction.
the angle is moved anticlockwise now the bearing of AC will be calculated by
= bearing of line AB - 80°
= 234° 51' 48" - 80°
= 154° 51' 48".
= 154.5148
so, the bearing of the line AC will be 154° 51' 48"
An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the positive x -direction. Another extremely long wire laying parallel to the x -axis and passing through the y -axis at ry= 1.8m carries a current of 50A running in the negative x -direction. What is the magnitude of the net magnetic field that the wires generate on the y -axis at ry= −3.510m ?
Answer:
[tex]1.232\times 10^{-5}\ T.[/tex]
Explanation:
Given:
Current through the wire, passing through the origin, [tex]I_1 = 250\ A.[/tex]Current through the wire, passing through the y axis, [tex]r_y=1.8\ m.[/tex], [tex]I_2 = 50\ A.[/tex]According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to [tex]\mu_o[/tex] times the net current threading the loop.
[tex]\oint \vec B \cdot d\vec l=\mu_o I.[/tex]
In case of a circular loop, the directions of magnetic field and the line element [tex]d\vec l[/tex], both are along the tangent of the loop at that point, therefore, [tex]\vec B\cdot d\vec l = B\ dl[/tex].
[tex]\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\[/tex]
[tex]\oint dl[/tex] is the circumference of the Amperian loop = [tex] 2\pi r[/tex]
Therefore,
[tex]B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.[/tex]
It is the magnetic field due to a current carrying wire at a distance r from it.
For the first wire, passing through the origin:
Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, [tex]r_1 = 3.510\ m.[/tex]
The magnetic field at the given point due to this wire is given by:
[tex]B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.[/tex]
For the first wire, passing through the y-axis:
Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, [tex]r_2 = 5.310\ m.[/tex]
The magnetic field at the given point due to this wire is given by:
[tex]B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.[/tex]
The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.
Thus, the net magnetic field at [tex]r_r=-3.510\ m[/tex] is given by
[tex]B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.[/tex]
Water and iron have quite different specific heats. Iron heats up more than water when the same amount of heat energy is added to it. Which one of these substances do you deduce has the higher specific heat? water
iron
both have the same specific heat
cannot be determined
Answer:
water
Explanation:
The specific heat of an object denotes the amount of energy which is required to raise the objects temperature by 1 unit of temperature and mass of the object is 1 unit.
If the same amount of heat is added to both water and iron the temperature difference in their final and initial temperatures will be different.
The difference in the initial and final temperature of iron will be higher than that of water. This means that the amount of energy which is required to raise the objects temperature by 1 unit of temperature for iron is lower than water.
So, water has higher specific heat
Also
Q = mcΔT
where
Q = Heat
m = Mass
c = Specific heat
ΔT = Change in temperature
[tex]c=\frac{Q}{m\Delta T}[/tex]
This means specific heat is inversely proportional to change in temperature.
So, for water the change in temperature will be lower than iron which means that water will have a higher specific heat.
The electric field in a particular thundercloud is 3.8 x 105 N/C. What is the acceleration (in m/s2) of an electron in this field? (Enter the magnitude.) m/s2
Answer:
Answer:
6.68 x 10^16 m/s^2
Explanation:
Electric field, E = 3.8 x 10^5 N/C
charge of electron, q = 1.6 x 10^-19 C
mass of electron, m = 9.1 x 10^-31 kg
Let a be the acceleration of the electron.
The force due to electric field on electron is
F = q E
where q be the charge of electron and E be the electric field
F = 1.6 x 10^-19 x 3.8 x 10^5
F = 6.08 x 10^-14 N
According to Newton's second law
Force = mass x acceleration
6.08 x 10^-14 = 9.1 x 10^-31 x a
a = 6.68 x 10^16 m/s^2
Explanation:
A 0.10- kg ball is thrown straight up into the air with
aninitial speed of 15m/s. Find the momentum of the ball (a) at
itsmaximum height and (b) halfway to its maximum height.
Answer:(a)0,(b)1.061 kg-m/s
Explanation:
Given
mass of ball is 0.10 kg
Initial speed is 15 m/s
Maximum height reached by ball is h
[tex]v^2-u^2=2as[/tex]
final velocity =0
[tex]-\left ( 15\right )^2=-2\times 9.81\times s[/tex]
[tex]s=\frac{15^2}{2\times 9.81}=11.467 m[/tex]
thus momentum of ball at maximum height is 0 as velocity is zero
For halfway to maximum height
[tex]v^2-u^2=2as_0[/tex]
where [tex]s_0=\frac{11.467}{2}[/tex]
[tex]s_0=5.73 m[/tex]
[tex]v^2=15^2-2\times 9.81\times 5.73[/tex]
v=10.61 m/s
Thus its momentum is
[tex]mv=0.10\times 10.61=1.061 kg-m/s[/tex]
Determine (a) how many seconds there are in 1.00 year? (b) Using your answer from part (a), how many years are there in 1.00 second?
Answer:
a) 31557600 seconds b) 3.168x10⁻⁸ years
Explanation:
1 hour is equivalent to 3600 seconds, that multiplied by 24 hours (1 day) gives 86.400 seconds
a) how many seconds there are in 1.00 year?
[tex]\frac{86400 s}{1 day} x 365,25 days[/tex] ⇒ 31557600 s
So in 1 year there are a total of 31557600 seconds.
b) how many years are there in 1.00 second?
Using the quantity from part a) it is get:
[tex]1 s x \frac{1 year}{31557600 s}[/tex] ⇒ 3.168x10⁻⁸ years
So in 1 second there are a total of 3.168x10⁻⁸ years.