Answer:
I = 0.5A
Explanation:
Hello,
Assuming the transformer is an ideal transformer, we can calculate the value of the current using the formula.
Data;
Power = 60 watt
Primary voltage = 120 volts (rms)
Power (P) = current (I) × voltage (V)
P = IV
This formula is the relationship between power, current and voltage.
Current = power / voltage
I = P / V
I = 60 / 120
I = 0.5A
The current in the primary coil is 0.5A
When using a mercury barometer , the vapor pressure of mercury is usually assumed to be zero. At room temperature mercury's vapor pressure is about 0.0015 mm-Hg. At sea level, the height hhh of mercury in a barometer is about 760 mm.Required:a. If the vapor pressure of mercury is neglected, is the true atmospheric pressure greater or less than the value read from the barometer? b. What is the percent error? c. What is the percent error if you use a water barometer and ignore water's saturated vapor pressure at STP?
Answer:
Explanation:
(a)
The true atmospheric pressure will has more value than the reading in the barometer. If Parm is the atmospheric
pressure in the tube then the resulting vapour pressure is
Patm - pgh = Prapor
The final reading ion the barometer is
pgh = Palm - Proper
Hence, the true atmospheric pressure is greater.
you can find the answer in this book
physics principles with Applications, Global Edition Problem 67P: Chapter: CH 13 Problem:67p
A 2-kg block is released from rest at the top of a 20-mlong frictionless ramp that is 4 m high. At the same time, an identical block is released next to the ramp so that it drops straight down the same 4 m. What are the values for each of the following for the blocks just before they reach ground level.
Required:
a. Gravitational potential energy Block a_____ J Block b _____ J
b. Kinetic energy Block a _____ J Block b _____
c. Speed Block a _____ J Block b _____ J
d. Momentum Block a _____ J Block b _____ J
Answer:
A.) 78.4 J for both
B.) 78.4 J for both
C.) 8.85 m/s for both
D.) 17.7 kgm/s
Explanation:
Given information:
Mass m = 2 kg
Distance d = 20 m
High h = 4 m
A.) Gravitational potential energy can be calculated by using the formula
P.E = mgh
P.E = 2 × 9.8 × 4
P.E = 78.4 J
Since the two objects are identical, the gravitational potential energy of the block for both a and b will be 78.4 J
B.) According to conservative energy,
Maximum P.E = Maximum K.E.
Therefore, the kinetic energy of the two blocks will be 78.4 J
C.) Since K.E = 1/2mv^2 = mgh
V = √(2gh)
Solve for velocity V by substituting g and h into the formula
V = √(2 × 9.8 × 4)
V = √78.4
V = 8.85 m/s
The velocities of both block will be 8.85 m/s
D.) Momentum is the product of mass and velocity. That is,
Momentum = MV
Substitute for m and V into the formula
Momentum = 2 × 8.85 = 17.7 kgm/s
Both block will have the same value since the ramp Is frictionless.
A cart of mass 350 g is placed on a frictionless horizontal air track. A spring having a spring constant of 7.5 N/m is attached between the cart and the left end of the track. The cart is displaced 3.8 cm from its equilibrium position. (a) Find the period at which it oscillates. s (b) Find its maximum speed. m/s (c) Find its speed when it is located 2.0 cm from its equilibrium position.
Answer:
(a) T = 1.35 s
(b) vmax = 0.17 m/s
(c) v = 0.056 m/s
Explanation:
(a) In order to calculate the period of oscillation you use the following formula for the period in a simple harmonic motion:
[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex] (1)
m: mass of the cart = 350 g = 0.350kg
k: spring constant = 7.5 N/m
[tex]T=2\pi \sqrt{\frac{0.350kg}{7.5N/m}}=1.35s[/tex]
The period of oscillation of the car is 1.35s
(b) The maximum speed of the car is given by the following formula:
[tex]v_{max}=\omega A[/tex] (2)
w: angular frequency
A: amplitude of the motion = 3.8 cm = 0.038m
You calculate the angular frequency:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{1.35s}=4.65\frac{rad}{s}[/tex]
Then, you use the result of w in the equation (2):
[tex]v_{max}=(4.65rad/s)(0.038m)=0.17\frac{m}{s}[/tex]
The maximum speed if 0.17m/s
(c) To find the speed when the car is at x=2.0cm you first calculate the time t by using the following formula:
[tex]x=Acos(\omega t)\\\\t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\t=\frac{1}{4.65rad/s}cos^{-1}(\frac{0.02}{0.038})=0.069s[/tex]
The speed is the value of the following function for t = 0.069s
[tex]|v|=|\omega A sin(\omega t)|\\\\|v|=(4.65rad/s)(0.038m)sin(4.65rad/s (0.069s))=0.056\frac{m}{s}[/tex]
The speed of the car is 0.056m/s
A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
If the particle is an electron [tex]E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton, [tex]E_y = 6.08 * 10^6 N/C[/tex]
Explanation:
Initial speed at the origin, [tex]u = 3 * 10^6 m/s[/tex]
[tex]\theta = 38^0[/tex] to +ve x-axis
The particle crosses the x-axis at , x = 1.5 cm = 0.015 m
The particle can either be an electron or a proton:
Mass of an electron, [tex]m_e = 9.1 * 10^{-31} kg[/tex]
Mass of a proton, [tex]m_p = 1.67 * 10^{-27} kg[/tex]
The electric field intensity along the positive y axis [tex]E_y[/tex], can be given by the formula:
[tex]E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\[/tex]
If the particle is an electron:
[tex]E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton:
[tex]E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 6.08 * 10^6 N/C[/tex]
A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 5,200 rev/min.
Required:
a. Find the kinetic energy stored in the flywheel.
b. If the flywheel is to supply energy to the car as would a 15.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.
Answer:
a
[tex]KE = 7.17 *10^{7} \ J[/tex]
b
[tex]t = 6411.09 \ s[/tex]
Explanation:
From the question we are told that
The radius of the flywheel is [tex]r = 1.50 \ m[/tex]
The mass of the flywheel is [tex]m = 430 \ kg[/tex]
The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]
The power supplied by the motor is [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]
Generally the moment of inertia of the flywheel is mathematically represented as
[tex]I = \frac{1}{2} mr^2[/tex]
substituting values
[tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]
[tex]I = 483.75 \ kgm^2[/tex]
The kinetic energy that is been stored is
[tex]KE = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]
[tex]KE = 7.17 *10^{7} \ J[/tex]
Generally power is mathematically represented as
[tex]P = \frac{KE}{t}[/tex]
=> [tex]t = \frac{KE}{P}[/tex]
substituting the value
[tex]t = \frac{7.17 *10^{7}}{11190}[/tex]
[tex]t = 6411.09 \ s[/tex]
Calculate the amount of kinetic energy the car stores if it has a mass of 1200 kg and speed of 15 m/s
Answer:
KE = 135,000 j or 135 KJ
Explanation:
KE=0.5mv^2
KE=0.5*1200*15^2
KE = 135,000 joules or 135 Kilo Joules
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands
Complete question is;
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.
Answer:
Tangential speed of foot just before it lands is; v = 5.37m/s
Explanation:
Let U (potential energy) be zero on the ground.
So, initially, U = mgh
where, h = 0.98/2 = 0.49m (midpoint of the leg)
Now just before the leg hits the floor it would have kinetic energy as;
K = ½Iω²
where ω = v/r and I = ⅓mr²
So, K = ½(⅓mr²)(v/r)²
K = (1/6) × (mr²)/(v²/r²)
K = (1/6) × mv²
From principle of conservation of energy, we have;
Potential energy = Kinetic energy
Thus;
mgh = (1/6) × mv²
m will cancel out to give;
gh = (1/6)v²
Making v the subject, we have;
v = √6gh
v = √(6 × 9.81 × 0.49)
v = √28.8414
v = 5.37m/s
A butcher grinds 5 and 3/4 lb of meat then sells it for 2 and 2/3 pounds to the customer what is the maximum amount me that the butcher can sell to the next customer
Answer:
The maximum amount of meat that the butcher can sell is [tex]3\frac{1}{12}\:lb[/tex]
Explanation:
The maximum amount can be found by taking the difference of mixed numbers.
[tex]5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\[/tex]
Best Regards!
A shell (a large bullet) is shot with an initial speed of 20 m/s, 60 degrees above the horizontal. At the top of the trajectory, the bullet explodes into two fragments of equal mass. One fragment has a speed of zero just after the explosion and simply drops straight down. How far from the gun does the other fragment land, assuming that the ground is level and that the air drag is negligible.
Answer:
17.656 m
Explanation:
Initial speed u = 20 m/s
angle of projection α = 60°
at the top of the trajectory, one fragment has a speed of zero and drops to the ground.
we should note that the top of the trajectory will coincide with halfway the horizontal range of the the projectile travel. This is because the projectile follows an upward arc up till it reaches its maximum height from the ground, before descending down by following a similar arc downwards.
To find the range of the projectile, we use the equation
R = [tex]\frac{u^{2}sin2\alpha }{g}[/tex]
where g = acceleration due to gravity = 9.81 m/s^2
Sin 2α = 2 x (sin α) x (cos α)
when α = 60°,
Sin 2α = 2 x sin 60° x cos 60° = 2 x 0.866 x 0.5
Sin 2α = 0.866
therefore,
R = [tex]\frac{20^{2}*0.866 }{9.81}[/tex] = 35.31 m
since the other fraction with zero velocity drops a top of trajectory, distance between the two fragments assuming level ground and zero air drag, will be 35.31/2 = 17.656 m
The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s) of the orbit and shaded segments are the same?
Answer: not sure
Explanation:
At least how many Calories does a mountain climber need in order to climb from sea level to the top of a 5.42 km tall peak assuming the muscles of the climber can convert chemical energy to mechanical energy with an efficiency of 16.0 percent. The total mass of the climber and the equipment is 78.4 kg. (Enter your answer as a number without units.)
Answer:
Ec = 6220.56 kcal
Explanation:
In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.
You use the following formula:
[tex]W_c=Mgh[/tex] (1)
Wc: work done by the climber
g: gravitational constant = 9.8 m/s^2
M: mass of the climber = 78.4 kg
h: height reached by the climber = 5.42km = 5420 m
You replace in the equation (1):
[tex]W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J[/tex] (2)
Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:
[tex]0.16(E_c)=4,164,294.4J[/tex]
Ec: Calories
You solve for Ec and convert the result to Cal:
[tex]E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal[/tex]
The amount of Calories needed by the climber was 6220.56 kcal
Jack and Jill went up the hill to fetch a pail of water. Jack, who’s mass is 75 kg, 1.5 times heavier than Jill’s mass, fell down and broke his crown after climbing a 15 m high hill. Jillcame tumbling after covering the same distance as Jack in 1/3rd of the time.Required:a. Who did the most work climbing up the hill? b. Who applied the most power?
Answer:
a) Jack does more work uphill
b) Numerically, we can see that Jill applied the most power downhill
Explanation:
Jack's mass = 75 kg
Jill's mass = [tex]1.5x = 75[/tex]
Jill's mass = [tex]x = \frac{75}{1.5}[/tex] = 50 kg
distance up hill = 15 m
a) work done by Jack uphill = mgh
where g = acceleration due to gravity= 9.81 m/s^2
work = 75 x 9.81 x 15 = 11036.25 J
similarly,
Jill's work uphill = 50 x 9.81 x 15 = 7357.5 J
this shows that Jack does more work climbing up the hill
b) assuming Jack's time downhill to be t,
then Jill's time = [tex]\frac{t}{3}[/tex]
we recall that power is the rate in which work id done, i.e
P = [tex]\frac{work}{time}[/tex]
For Jack, power = [tex]\frac{11036.25}{t}[/tex]
For Jill, power = [tex]\frac{3*7357.5}{t}[/tex] = [tex]\frac{22072.5}{t}[/tex]
Numerically, we can see that Jill applied the most power downhill
Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the ball, causing it to swing around the pole. What is the total initial acceleration of a tether ball on a 2.0 m rope whose angular velocity changes from 13 rad/s to 7.0 rad/s in 15 s
Answer:
a_total = 14.022 m/s²
Explanation:
The total acceleration of a uniform circular motion is given by the following formula:
[tex]a=\sqrt{a_c^2+a_T^2}[/tex] (1)
ac: centripetal acceleration
aT: tangential acceleration
Then, you first calculate the centripetal acceleration by using the following formula:
[tex]a_c=r\omega^2[/tex]
r: radius of the circular trajectory = 2.0m
w: final angular velocity of the ball = 7.0 rad/s
[tex]a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}[/tex]
Next, you calculate the tangential acceleration. aT is calculate by using:
[tex]a_T=r\alpha[/tex] (2)
α: angular acceleration
The angular acceleration is:
[tex]\alpha=\frac{\omega_o-\omega}{t}[/tex]
wo: initial angular velocity = 13 rad/s
t: time = 15 s
Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:
[tex]a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}[/tex]
Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:
[tex]a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}[/tex]
The total acceleration of the ball is 14.022 m/s²
A skydiver stepped out of an airplane at an altitude of 1000m fell freely for 5.00s opened her parachute and slowed to 7.00m/s in a negligible time what was the total elapsed time from leaving the airplane to landing on the ground
Answer:
t = 17.68s
Explanation:
In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:
[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex] (1)
y: height for a time t
yo: initial height = 1000m
vo: initial velocity = 0m/s
g: gravitational acceleration = 9.8m/s^2
t: time = 5.00 s
You replace the values of the parameters to get the values of the new height of the skydiver:
[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]
Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.
You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0
[tex]0=877.5-7.00t-4.9t^2[/tex] (2)
The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:
[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]
You use the positive value of t1 because it has physical meaning.
Finally, you sum the times of both parts of the trajectory:
total time = 5.00s + 12.68s = 17.68s
The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s
The temperature coefficient of resistivity for the metal gold is 0.0034 (C )1, and for tungsten it is 0.0045 (C )1. The resistance of a gold wire increases by 7.0% due to an increase in temperature. For the same increase in temperature, what is the percentage increase in the resistance of a tungsten wire
Answer:
% increase in resistance of tungsten = 9.27%
Explanation:
We are given:
Co-efficient of resistivity for the metal gold; α_g = 0.0034 /°C
Co-efficient of resistivity for tungsten;α_t = 0.0045 /°C
% Resistance change of gold wire due to temperature change = 7%
Now, let R1 and R2 be the resistance before and after the temperature change respectively.
Thus;
(R2 - R1)/R1) x 100 = 7
So,
(R2 - R1) = 0.07R1
R2 = R1 + 0.07R1
R2 = 1.07R1
The equation to get the change in temperature is given as;
R2 = R1(1 + αΔt)
So, for gold,
1.07R1 = R1(1 + 0.0034*Δt)
R1 will cancel out to give;
1.07 = 1 + 0.0034Δt
(1.07 - 1)/0.0034 = Δt
Δt = 20.59°C
For this same temperature for tungsten, let Rt1 and Rt2 be the resistance before and after the temperature change respectively and we have;
Rt2 = Rt1(1 + α_t*Δt)
So, Rt2/Rt1 = 1 + 0.0045*20.59
Rt2/Rt1 = 1.0927
From earlier, we saw that;
(R2 - R1)/R1) x 100 = change in resistance
Similarly,
(Rt2 - Rt1)/Rt1) x 100 = change in resistance
Simplifying it, we have;
[(Rt2/Rt1) - 1] × 100 = %change in resistance
Plugging in the value of 1.0927 for Rt2/Rt1, we have;
(1.0927 - 1) × 100 = %change in resistance
%change in resistance = 9.27%
You are trying to overhear a juicy conversation, but from your distance of 25.0 m, it sounds like only an average whisper of 25.0 dB. So you decide to move closer to give the conversation a sound level of 80.0 dB instead. How close should you come?
Answer:
r₂ = 1,586 m
Explanation:
For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m
β = 10 log (I / I₀)
where Io is the sensitivity threshold 10⁻¹² W / m²
I₁ / I₀ = [tex]e^{\beta/10}[/tex]
I₁ = I₀ e^{\beta/10}
let's calculate
I₁ = 10⁻¹² e^{25/10}
I₁ = 1.20 10⁻¹¹ W / m²
the other intensity in exercise is
I₂ = 10⁻¹² e^{80/10}
I₂ = 2.98 10⁻⁹ W / m²
now we use the definition of sound intensity
I = P / A
where P is the emitted power that is a constant and A the area of the sphere where the sound is distributed
P = I A
the area a sphere is
A = 4π r²
we can write this equation for two points of the found intensities
I₁ A₁ = I₂ A₂
where index 1 corresponds to 25m and index 2 to the other distance
I₁ 4π r₁² = I₂ 4π r₂²
I₁ r₁² = I₂ r₂²
r₂ = √ (I₁ / I₂) r₁
let's calculate
r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25
r₂ = √ (0.40268 10⁻²) 25
r₂ = 1,586 m
Strontium decays by beta decay part of the nuclear equation is shown below fill in the blank with a number? 90/38Sr -> 0/-1e 90/blankY
Answer : The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:
[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]
Explanation :
Beta decay : It is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.
The released beta particle is also known as electron.
The beta decay reaction is:
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]
The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:
[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]
Answer:
the blank is 39
Explanation: a p e x
Which three terms are needed to describe the energy a BASE jumper has as
she falls toward the ground?
O A. Potential
B. Electromagnetic
C. Gravitational
D. Kinetic
Answer:
I’m saying kinetic gravitational and electromagnetic and I will comment on this if I got it right
Explanation:.
A projectile is launched in the horizontal direction. It travels 2.050 m horizontally while it falls 0.450 m vertically, and it then strikes the floor. How long is the projectile in the air
Answer:
0.303s
Explanation:
horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m
Using equation of linear motion
Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative
-0.45 = 0 - 0.5 × 9.81×t²
0.45 / (0.5 × 9.81) = t²
t = √0.0917 = 0.303 s
A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two fragments, each 1.0 kg, move in directions that make 60o angle above and below the positive x-axis and their speeds are 60 m/s each. What is the velocity of the 4.0-kg fragment
Answer:
-15 m/s
Explanation:
The computation of the velocity of the 4.0 kg fragment is shown below:
For this question, we use the correlation of the momentum along with horizontal x axis
Given that
Weight of stationary shell = 6 kg
Other two fragments each = 1.0 kg
Angle = 60
Speed = 60 m/s
Based on the above information, the velocity = v is
[tex]1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0[/tex]
[tex]\frac{60}{2} + \frac{60}{2} - 4\ v = 0[/tex]
[tex]v = \frac{60}{4}[/tex]
= -15 m/s
Imagine you are in a small boat on a small pond that has no inflow or outflow. If you take an anchor that was sitting on the floor of the boat and lower it over the side until it sits on the ground at the bottom of the pond, will the water level rise slightly, stay the same, or lower slightly?Two students, Ian and Owen, are discussing this. Ian says that the anchor will still displace just as much water when it is sitting on the bottom of the pond as it does when it is in the boat. After all, adding the anchor to the boat causes the water level in the lake to rise, and so would immersing the anchor in the pond. So Ian reasons that both displacements would be equal, and the lake level remains unchanged.
Answer;
The pond's water level will fall.
Explanation;
Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.
When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.
Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.
When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.
Hope this Helps!!!
Buoyancy is an upward force exerted by a fluid on a body partially or completely immersed in it
The pond water level will lower slightly
According to Archimedes principle, the up thrust on the boat by the water is given by the volume of the water displaced
When a boat floats, the weight of the boat and all its contents and passengers is equal to the displaced water, so that larger boats with more wider opening can displace more water and therefore, carry more loadWith regards to lowering the anchor from the boat into the pond, the weight of the anchor is no longer carried by the boat but by the bottom of the pond, therefore, the weight of the boat reduces, and the boat rises, while the volume initially occupied by the boat is taken up by the water available, therefore, the water level lowers slightly
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A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 50.2 s. The distance between two successive crests is 30.2 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed
Explanation:
(a) The period of a wave is the time required for one complete cycle. In this case, we have the time of five cycles. So:
[tex]T=\frac{t}{n}\\\\T=\frac{50.2s}{5}\\T=10.04s[/tex]
(b) The frequency of a wave is inversely proportional to its period:
[tex]f=\frac{1}{T}\\f=\frac{1}{10.04s}\\f=0.01Hz[/tex]
(c) The wavelength is the distance between two successive crests, so:
[tex]\lambda=30.2m[/tex]
(d) The speed of a wave is defined as:
[tex]v=f\lambda\\v=(0.1Hz)(30.2m)\\v=3.02\frac{m}{s}[/tex]
what happen to the volume of liquid displaced
when the density of liquid is changed
explain ?
Answer:
Density depends on the temperature and the gap between particles of the liquid. In most of cases temperature is inversely proportional to density means if the temperature increases then the density decreases and the space between particles of that liquid is also inversely proportional to the density means if the intraparticle space increases then the density decreases.
At rest, a car's horn sounds at a frequency of 365 Hz. The horn is sounded while the car is moving down the street. A bicyclist moving in the same direction with one-third the car's speed hears a frequency of 357 Hz. What is the speed of the car?
Answer:
10.15m/s
Explanation:
The change in the frequency of sound (or any other wave) when the source of the sound and the receiver or observer of the sound move towards (or away from) each other is explained by the Doppler effect which is given by the following equation:
f₁ = [(v ± v₁) / (v ± v₂)] f ----------------------(i)
Where;
f₁ = frequency received by the observer or receiver
v = speed of sound in air
v₁ = velocity of the observer
v₂ = velocity of the source
f = original frequency of the sound
From the question, the observer is the bicyclist and the source is the car driver. Therefore;
f₁ = frequency received by the observer (bicyclist) = 357Hz
v = speed of sound in air = 330m/s
v₁ = velocity of the observer(bicyclist) = (1 / 3) v₂ = 0.33v₂
v₂ = velocity of the source (driver)
f = original frequency of the sound = 365Hz
Note: The speed of the observer is positive if he moves towards the source and negative if he moves away from the source. Also, the speed of the source is positive if it moves away from the listener and negative otherwise.
From the question, the cyclist and the driver are moving in the same direction. But then, we do not know which one is at the front. Therefore, two scenarios are possible.
i. The bicyclist is at the front. In this case, v₁ and v₂ are negative.
Substitute these values into equation (i) as follows;
357 = [(330 - 0.33v₂) / (330 - v₂)] * 365
(357 / 365) = [(330 - 0.33v₂) / (330 - v₂)]
0.98 = [(330 - 0.33v₂) / (330 - v₂)]
0.98 (330 - v₂) = (330 - 0.33v₂)
323.4 - 0.98v₂ = 330 - 0.33v₂
323.4 - 330 = (0.98 - 0.33)v₂
-6.6 = 0.65v₂
v₂ = -10.15
The value of v₂ is not supposed to be negative since we already plugged in the right value polarity into the equation.
iI. The bicyclist is behind. In this case, v₁ and v₂ are positive.
Substitute these values into equation (i) as follows;
357 = [(330 + 0.33v₂) / (330 + v₂)] * 365
(357 / 365) = [(330 + 0.33v₂) / (330 + v₂)]
0.98 = [(330 + 0.33v₂) / (330 + v₂)]
0.98 (330 + v₂) = (330 + 0.33v₂)
323.4 + 0.98v₂ = 330 + 0.33v₂
323.4 - 330 = (0.33 - 0.98)v₂
-6.6 = -0.65v₂
v₂ = 10.15
The value of v₂ is positive and that is a valid solution.
Therefore, the speed of the car is 10.15m/s
A ray in glass (n = 1.51) reaches a boundary with air at 49.2 deg. Does it reflect internally or refract into the air? Enter 0 for reflect, and 1 for refract.
Answer:
0 - Then, the ray is totally reflected
Explanation:
The ray reaches the boundary between the two mediums at 49.2°.
If the ray is totally reflected it is necessary that the crictical angle is lower that the incidet angle.
You use the following to calculate the critical angle:
[tex]\theta_c=sin^{-1}(\frac{n_2}{n_1})[/tex] (1)
n2: index of refraction of the second medium (air) = 1.00
n1: index of refraction of the first medium (glass) = 1.51
You replace the values of the parameters in the equation (1):
[tex]\theta_c=sin^{-1}(\frac{1.00}{1.51})=41.47\°[/tex]
The critical angle is 41.47°, which is lower than the incident angle 49.2°.
Then, the ray is totally reflected.
0
how much weight can a man lift in the jupiter if he can lift 100kg on the earth.calculate
Answer:
2479 NewtonSolution,
Mass=100 kg
Acceleration due to gravity(g)=24.79 m/s^2
Now,.
[tex]weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton[/tex]
hope this helps ..
Good luck on your assignment..
Immediately outside a conducting sphere(i.e. on the surface) of unknown charge Q and radius R the electric potential is 190 V, and 10.0 cm further from the sphere, the potential is 140 V. What is the magnitude of the charge Q on the sphere
Answer:
Q = 5.9 nC (Approx)
Explanation:
Given:
Further distance = 10 cm
Electric potential(V) = 190 v
Potential difference(V1) = 140 v
Find:
Magnitude of the charge Q
Computation:
V = KQ / r
190 = KQ / r.............Eq1
V1 = KQ / (r+10)
140 = KQ / (r+10) ............Eq2
From Eq2 and Eq1
r = 28 cm = 0.28 m
So,
190 = KQ / r
190 = (9×10⁹)(Q) / 0.28
53.2 = (9×10⁹)(Q)
5.9111 = (10⁹)(Q)
Q = 5.9 nC (Approx)
The figure shows an arrangement of four charged particles, with θ = 20.0° and d1 = 3.00 cm, which is the distance from the origin to a charge q1. Charge q1 is unknown, but q2= +7.00×10‒19 C and q3 = q4 = ‒2.00×10‒19 C. If there is no nett electrostatic force on q1 due to the other charges (the nett electrostatic force on q1 is zero), calculate the distance from the origin to q2, given by d2, in cm. Assume that all forces apart from the electrostatic forces in the system are negligible
Answer:
[tex]d_2=3.16cm[/tex]
Explanation:
So, in order to solve this problem, we must start by building a diagram of the problem itself. (See attached picture) And together with the diagram, we must build a free body diagram, which will include the forces that are being applied on the given charged particle together with their directions.
In this case we only care about the x-direction of the force, since the y-forces cancel each other. So if we do a sum of forces on the x-direction, we get the following:
[tex]\sum{F_{x}}=0[/tex]
so:
[Tex]-F_{12}+F_{13x}+F_{14x}=0[/tex]
Since [tex]F_{13x}=F_{14x}[/tex] we can simplify the equation as:
[tex]-F_{12}+2F_{13x}=0[/tex]
we can now solve this for [tex]F_{12}[/tex] so we get:
[tex]F_{12}=2F_{13x}[/tex]
Now we can substitute with the electrostatic force formula, so we get:
[tex]k_{e}\frac{q_{1}q_{2}}{r_{12}^{2}}=2k_{e}\frac{q_{1}q_{3}}{r_{13}^{2}}cos \theta[/tex]
We can cancel [tex]k_{e}[/tex] and [tex]q_{1}[/tex]
so the simplified equation is:
[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{r_{13}^{2}}cos \theta[/tex]
From the given diagram we know that:
[tex]cos \theta = \frac{d_{1}}{r_{13}}[/tex]
so when solving for [tex]r_{13}[/tex] we get:
[tex]r_{13}=\frac{d_{1}}{cos\theta}[/tex]
and if we square both sides of the equation, we get:
[tex]r_{13}^{2}=\frac{d_{1}^{2}}{cos^{2}\theta}[/tex]
and we can substitute this into our equation:
[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{d_{1}^{2}}cos^{3} \theta[/tex]
so we can now solve this for [tex]r_{12}[/tex] so we get:
[tex]r_{12}=\sqrt{\frac{d_{1}^{2}q_{2}}{2q_{3}cos^{3}\theta}}[/tex]
which can be rewritten as:
[tex]r_{12}=d_{1}\sqrt{\frac{q_{2}}{2q_{3}cos^{3}\theta}}[/tex]
and now we can substitute values.
[tex]r_{12}=(3cm)\sqrt{\frac{7x10^{-19}C}{2(2x10^{-19}C)cos^{3}(20^{o})}}[/tex]
which solves to:
[tex]r_{12}=6.16cm[/tex]
now, we must find [tex]d_{2}[/tex] by using the following equation:
[tex]r_{12}=d_{1}+d_{2}[/tex]
when solving for [tex]d_{2}[/tex] we get:
[tex]d_{2}=r_{12}-d_{1}[/tex]
when substituting we get:
[tex]d_{2}=6.16cm-3cm[/tex]
so:
[tex]d_{2}=3.16cm[/tex]
what is the most likely elevation of point x?
A. 150 ft
B. 200 ft
C. 125 ft
D. 250 ft
A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 60.0 cm, and the density of iron is 7.87 g∕c m cubed . Find the inner diameter in cm. Express to 3 sig figs.
Answer:
The inner diameter is 57.3 cm
Explanation:
The inner diameter of the hollow spherical iron shell can be found using the weight of the sphere ([tex]W_{s}[/tex]) and the weight of the water displaced ([tex]W_{w}[/tex]):
[tex] W_{s} = W_{w} [/tex]
[tex] m_{s}*g = m_{w}*g [/tex]
[tex] D_{s}*V_{s} = D_{w}*V_{w} [/tex]
Where D is the density and V is the volume
[tex] D_{s}*\frac{4}{3}\pi*(\frac{d_{o}^{3} - d_{i}^{3}}{2^{3}}) = \frac{4}{3}\pi*(\frac{d_{o}}{2})^{3} [/tex]
Where [tex]d_{o}[/tex] is the outer diameter and [tex]d_{i}[/tex] is the inner diameter
[tex] D_{s}*(d_{o}^{3} - d_{i}^{3}) = d_{o}^{3} [/tex]
[tex] D_{s}*d_{i}^{3} = d_{o}^{3}(D_{s} - 1) [/tex]
[tex] 7.87*d_{i}^{3} = 60.0^{3}(7.87 - 1) [/tex]
[tex] d_{i} = 57.3 cm [/tex]
Therefore, the inner diameter is 57.3 cm.
I hope it helps you!