Answer:
0.013 m/min
Explanation:
Applying,
dV/dt = (dh/dt)(dV/dh)............. Equation 1
Where
V = πr²h................ Equation 2
Where V = volume of the tank, r = radius, h = height.
dV/dh = πr²............ Equation 3
Substitute equation 3 into equation 1
dV/dt = πr²(dh/dt)
From the question,
Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14
Substitute these values into equation 3
2 = (3.14)(7²)(dh/dt)
dh/dt = 2/(3.14×7²)
dh/dt = 0.013 m/min
how much amount of heat energy is required to convert 5 kg of ice at - 5° c into 100°c steam?
Assuming no heat lost to the surrounding,
-5⁰C ice → 0⁰C ice
Specific heat capacity of ice = 2.0 x 10³ J/kg/⁰C
Q = mc∆θ
Q = 5(2.0 x 10³) x (0-(-5))
Q = 50000J
0⁰C ice → 0⁰C water
Specific latent heat of fusion of ice = 3.34 x 10⁵J/kg
Q = mLf
Q = 5(3.34 x 10⁵)
Q = 1670000J
0⁰C water → 100⁰C water
Specific heat capacity of water = 4.2 x 10³ J/kg/⁰C
Q = mc∆θ
Q = 5(4.2 x 10³) x (100-0)
Q = 2100000J
100⁰C water → 100⁰C steam
Specific latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Q = mLv
Q = 5(2.26 x 10⁶)
Q = 11300000J
Total amount of heat required
= 50000 + 1670000 + 2100000 + 11300000
= 15120000J
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
[tex]A=2.01×10^{16}\:\text{nuclei}[/tex]
Explanation:
Given:
[tex]\lambda = 4.96×10^3 s[/tex]
[tex]A_0 = 3.21x10^{17}[/tex] nuclei
t = 1.98×10^4 s
[tex]A=A_02^{-\frac{t}{\lambda}}[/tex]
[tex]A=(3.21×10^{17}\:\text{nuclei}) \left(2^{-\frac{1.98×10^4}{4.96×10^3}} \right)[/tex]
[tex]\:\:\:\:\:\:\:=2.01×10^{16}\:\text{nuclei}[/tex]
The north pole of magnet A will __?____ the south pole of magnet B
Answer:
A will attract
B will repare
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of with the horizontal
Complete Question
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of
40° 16' with the horizontal.
Answer:
[tex]d=8.01m[/tex]
Explanation:
From the question we are told that:
Length of ladder [tex]l=12.5m[/tex]
Angle [tex]\theta=40° 16'=20.26 \textdegree[/tex]
Generally the Trigonometric equation for distance d it goes up the wall is mathematically given by
[tex]d=l sin \theta[/tex]
[tex]d=12.5 sin 40.26[/tex]
[tex]d=8.01m[/tex]
A car is moving at a speed of 60 mi/hr (88 ft/sec) on a straight road when the driver steps on the brake pedal and begins decelerating at a constant rate of 10ft/s2 for 3 seconds. How far did the car go during this 3 second interval?
Answer:
219 ft
Explanation:
Here we can define the value t = 0s as the moment when the car starts decelerating.
At this point, the acceleration of the car is given by the equation:
A(t) = -10 ft/s^2
Where the negative sign is because the car is decelerating.
To get the velocity equation of the car, we integrate over time, to get:
V(t) = (-10 ft/s^2)*t + V0
Where V0 is the initial velocity of the car, we know that this is 88 ft/s
Then the velocity equation is:
V(t) = (-10 ft/s^2)*t + 88ft/s
To get the position equation we need to integrate again, this time we get:
P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0
Where P0 is the initial position of the car, we do not know this, but it does not matter for now.
We want to find the total distance that the car traveled in a 3 seconds interval.
This will be equal to the difference in the position at t = 3s and the position at t = 0s
distance = P(3s) - P(0s)
= ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)
= ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)
= (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft
The car advanced a distance of 219 ft in the 3 seconds interval.
There are two beakers of water on the table. We can compare the average kinetic energy of the water molecules in the two beakers by measuring their
A temperatures.
B volumes.
C densities.
D masses.
Answer: masses
Explanation:
Trust me
suppose the tank is open to the atmosphere instead of being closed. how does the pressure vary along
Answer:
Pressure is more in the open container than the closed one.
Explanation:
The pressure due to the fluid at a depth is given by
Pressure = depth x density of fluid x gravity
So, when the container is open, the atmospheric pressure is also add up but when the container is closed only the pressure due to the fluid is there.
So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.
hen the container is closed only the pressure due to the fluid is there.
A submarine has a "crush depth" (that is, the depth at which
water pressure will crush the submarine) of 400 m. What is
the approximate pressure (water plus atmospheric) at this
depth? (Recall that the density of seawater is 1025 kg/m3, g=
9.81 m/s2, and 1 kg/(m-s2) = 1 Pa = 9.8692 x 10-6 atm.)
Answer:
P =40.69 atm
Explanation:
We need to find the approximate pressure at a depth of 400 m.
It can be calculated as follows :
P = Patm + ρgh
Put all the values,
[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]
So, the approximate pressure is equal to 40.69 atm.
recognizing forms of energy
Answer:
hi the question isn't obvious and need a photo I guess
Question 3 of 10
Which statement describes the law of conservation of energy?
A. Air resistance has no effect on the energy of a system.
B. Energy cannot be created or destroyed.
C. The total energy in a system can only increase.
D. Energy cannot change forms.
هما
SUBMIT
Answer:
B . energy cannot be created or destroyed
Explain what a circuit breaker is and how it helps protect your house?
Explanation:
A circuit breaker is an electrical switch designed to protect an electrical circuit from damage caused by overcurrent/overload or short circuit. Its basic function is to interrupt current flow after protective relays detect a fault.
Circuit breakers have been designed to detect when there is a fault in the electricity, so it will “trip” and shut down electrical flow. ... This detection is key to preventing surges of electricity that travel to appliances or other outlets, which can cause them to break down
A student claimed that thermometers are useless because a
thermometer always registers its own temperature. How would you
respond?
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A hot air balloon is a sphere of volume 2210 m3. The density of the hot air inside is 1.13 kg/m3, while the air outside has a density of 1.29 kg/m3. The balloon itself has a mass of 240 kg. What is the TOTAL NET force acting on the balloon?
[?]N
The total net force acting on the balloon will be 24498 Newtons
Given that
Volume of the balloon = 2210 cubic meter
Density of the air inside the balloon = 1.13 kg/m3
What will be the net force exerted on the balloon ?Here force on the balloon will be equal to the weight of the air displaced by balloon
[tex]F= mass of air displaced\times gravity[/tex]
[tex]F= Density \times volume \times gravity[/tex]
[tex]F=1.13 \times 2210 \times 9.81[/tex]
[tex]F=24498 N[/tex]
The total net force acting on the balloon will be 24498 Newtons
To know more about buoyancy force follow
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A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. How far will block m drop in the first seconds after the system is released?
How long will block M move during above time?
At the time, calculate the velocity of block M
Find out the deceleration of the block M, if the connected string is
removal by cutting after the first second. Then, calculate the time
taken to contact block M and pulley.
Answer:
a) y = 0.98 t², t=1s y= 0.98 m,
b) he two blocks must move the same distance
c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m
Explanation:
For this exercise we can use Newton's second law
Big Block
Y axis
N-W = 0
N = M g
X axis
T- fr = Ma
the friction force has the expression
fr = μ N
fr = μ Mg
small block
w- T = m a
we write the system of equations
T - fr = M a
mg - T = m a
we add and resolved
mg- μ Mg = (M + m) a
a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]
a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]
a = 9.8 (6/30)
a = 1.96 m / s²
a) now we can use the kinematic relations
y = v₀ t + ½ a t²
the blocks come out of rest so their initial velocity is zero
y = ½ a t²
y = ½ 1.96 t²
y = 0.98 t²
for t = 1s y = 0.98 m
t = 2s y = 1.96 m
b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.
As the curda is in tension the two blocks must move the same distance
c) the velocity of the block M
v = vo + a t
v = 0 + 1.96 t
for t = 1 s v = 1.96 m / s
t = 2 s v = 3.92 m / s
d) the deceleration if the chain is cut
when removing the chain the tension becomes zero
-fr = M a
- μ M g = M a
a = - μ g
a = - 0.2 9.8
a = -1.96 m / s²
e) the distance to stop the block is
v² = vo² - 2 a x
0 = vo² - 2a x
x = vo² / 2a
x = 1.96² / 2 1.96
x = 0.98 m
the time to travel this distance is
v = vo - a t
t = vo / a
t = 1.96 /1.96
t = 1 s
Part AFind the x- and y-components of the vector d⃗ = (4.0 km , 29 ∘ left of +y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.d⃗ = km Part BFind the x- and y-components of the vector v⃗ = (2.0 cm/s , −x-direction).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.v⃗ = cm/s Part CFind the x- and y-components of the vector a⃗ = (13 m/s2 , 36 ∘ left of −y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.a⃗ x = m/s2
Solution :
Part A .
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, d = [tex]\text{4 km 29}[/tex] degree left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -4 x sin (29°) = -1.939 km
[tex]y[/tex] component is = 4 x cos (29°) = 3.498 km
Part B
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{v = 2 cm/s}[/tex] , [tex]\text{-x direction}[/tex]
So the [tex]x[/tex] component is = -2 cm/s
[tex]y[/tex] component is = 0
Part C
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{a = 13 m/s, 36 degree}[/tex] left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -13 x sin (36°) = -7.6412 [tex]m/S^2[/tex]
[tex]y[/tex] component is = -13 x cos (36°) = -10.517 [tex]m/S^2[/tex]
The x- and y-components of the vectors is mathematically given as as follows for each Part respectively
x= -1.939 km, y= 3.498 km
x= -2 cm/s, 0
y=, x= -7.6412m/s^2, -10.517m/s^2
What are the x- and y-components of the vectors?
Question Parameters:
Generally, we follow a basic principle where
x component= Fsin\theta
y component= Fcos\theta
Therefore
For A
x component is
x= -4 x sin (29°)
x= -1.939 km
y component is
y= 4 x cos (29°)
y= 3.498 km
For B
x component is
x= -2 cm/s
y component is
y= 0
For C
x component is
x= -13 x sin (36°)
x= -7.6412m/s^2
y component is
y= -13 x cos (36°)
y= -10.517m/s^2
Read more about Cartession co ordinate
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write down the following units in the ascending of their value A) mm nm cm um B) 1m 1cm 1km 1mm. convert the following units into SI without changing their values? A)3500g B)2.5km C)2h
Answer:
A) nm, um, mm, cm
B) 1mm, 1cm, 1m, 1km
A) 3500g, B) 2500m, C) 7200 seconds
A 64-ka base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
Required:
a. How much mechanical energy is tout due to friction acting on the runner?
b, How far does he slide?
Answer:
Explanation:
From the given information:
mass = 64 kg
speed = 3.2 m/s
coefficient of friction [tex]\mu =[/tex] 0.70
The mechanical energy touted relates to the loss of energy in the system as a result of friction and this can be computed as:
[tex]W = \Delta K.E[/tex]
[tex]\implies \dfrac{1}{2}m(v^2 -u^2)[/tex]
[tex]= \dfrac{1}{2}(64.0 \kg) (0 - (3.2 \ m/s^2))[/tex]
Thus, the mechanical energy touted = 327.68 J
According to the formula used in calculating the frictional force
[tex]F_r = \mu mg[/tex]
= 0.70 × 64 kg× 9.8 m/s²
= 439.04 N
The distance covered now can be determined as follows:
d = W/F
d = 327.68 J/ 439.04 N
d = 0.746 m
The velocity-time graph of a body is given. What quantities are represented by (a) slope of the graph and (b) area under the graph?
Answer:
a) acceleration
b) displacement
Explanation:
The velocity-time graph is a graph of velocity versus time. The velocity (m/s) would be on the Y-axis while time (s) would be on the X-axis.
a) The slope of a graph is given by: change in Y-axis/change in X-axis = ΔY/ΔX
In a velocity-time graph, ΔY = change in velocity and ΔX = change in time.
Hence, the slope of a velocity-time graph becomes: change in velocity/change in time.
Also, acceleration = change in velocity/change in time.
Hence, the slope of a velocity-time graph = acceleration.
b) Assuming that the area under a velocity-time graph is a rectangle, the area is given as:
Area of a rectangle = length x breadth
= velocity x time (m/s x s)
Also, displacement = velocity x time (m)
Hence, the area under a velocity-time graph of a body would give the displacement of the body.
The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?
Answer:
7.9 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
Take the fact that mass is inversely proportional to accelertation:
m ∝ a
Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:
[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]
Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:
[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]
Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:
0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]
Plug everything in:
7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]
(1590kg the initial weight plus the weight of the added passenger)
What word chemical equation describes this chemical reaction?
Answer : sodium + chlorine → sodium chloride
A typical incandescent light bulb consumes 75 W of power and has a mass of 20 g. You want to save electrical energy by dropping the bulb from a height great enough so that the kinetic energy of the bulb when it reaches the floor will be the same as the energy it took to keep the bulb on for 2.0 hours. From what height should you drop the bulb, assuming no air resistance and constant g?
Answer:
h = 2755102 m = 2755.102 km
Explanation:
According to the given condition:
Potential Energy = Energy Consumed by Bulb
[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]
where,
h = height = ?
P = Power of bulb = 75 W
t = time = (2 h)(3600 s/1 h) = 7200 s
m = mass of bulb = 20 g = 0.02 kg
g = acceleration due to gravity = 9.8 m/s²
Therefore,
[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]
h = 2755102 m = 2755.102 km
As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.
Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is
[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]
where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
What is (a) the x component and (b) the y component of the net electric field at the square's center
Answer:
What is (a) the x component and (b) the y component of the net electric field at the square's center
When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.
Answer:
angular acceleration.
Explanation:
Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.
Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.
A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)
Answer:
K = 0 J
Explanation:
Given that,
The mass of the particle, m = 1.2 mg
The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]
We need to find the kinetic energy of the particle at time t = 0 s.
At t = 0 s, the particle is at rest, v = 0
So,
[tex]K=\dfrac{1}{2}mv^2[/tex]
If v = 0,
[tex]K=0\ J[/tex]
So, the kinetic energy of the particle at time t = 0 s is 0 J.
What is significant about the primary colors of pigments?
They can be mixed together to make almost any other color.
Any two primary colors of pigments combine to make white pigment.
Each primary color of pigment absorbs all other colors.
Any two primary colors of pigments combine to make black pigment.
Answer:
They can be mixed together to make almost any other color.
Explanation:
All the three primary colors can mix to form white color.
Blue and red mix to form a black color.
12) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen, what things could you do to help yourself (before calling me over to assist you?)
Answer:
See the answer below
Explanation:
After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem
1. Drop a few drops of immersion oil on the slide and view again under high the power objective.
2. If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.
An inductor of inductance 0.02H and capacitor of capatance 2uF are connected in series to an a.c. source of frequency 200 Hz- Calculate the Impedance in the circuit . TC
Explanation:
Given:
L = 0.02 H
C = [tex]2\:\mu \text{F}[/tex]
f = 200 Hz
The general form of the impedance Z is given by
[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]
Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to
[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]
[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]
An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration
Answer:
a= -80.357 m/s
Explanation:
use the formula
vf^2=vi^2+2a(xf-xi)
Plug in givens
0=(7.50)^2+2a(0.350m)
solve for acceleration
a= -80.357 m/s
A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?
Answer:
7.8% of the original volume.
Explanation:
From the given information:
Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C
Pressure [tex]P_1[/tex] = 240 kPa
Temperature [tex]T_2[/tex] = 45° C
At initial temperature and pressure:
Using the ideal gas equation:
[tex]P_1V_1 =nRT_1[/tex]
making V_1 (initial volume) the subject:
[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]
[tex]V_1 = \dfrac{nR*295}{240}[/tex]
Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:
the final volume [tex]V_2[/tex] can be computed as:
[tex]V_2 = \dfrac{nR*318}{240}[/tex]
Now, the change in the volume ΔV = V₂ - V₁
[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]
[tex]\Delta V = \dfrac{23nR}{240}[/tex]
∴
The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:
[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]
[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]
[tex]= 0.078[/tex]
= 7.8% of the original volume.
