Answer:
130.22 g
Explanation:
Parameters given:
Mass of water Mw = 225 g
Mass of stirrer Ms = 40 g
Mass of silver M(S) = 410 g
By applying the law of conservation of energy:
(McCc + MsCs + MwCw)ΔTw = M(S)C(S)ΔT(S)
where Mc = Mass of cup
Cc = Specific heat capacity of aluminium cup = 900 J/gC
Cs = Specific heat capacity of copper stirrer = 387 J/gC
Cw = Specific heat capacity of water = 4186 J/gC
ΔTw = change in temperature of water = 32 - 27 = 5 °C
C(S) = Specific heat capacity of silver = 234 J/gC
ΔT(S) = change in temperature of silver = 88 - 32 = 56 °C
Therefore:
[(Mc * 900) + (40 * 387) + (225 * 4186)] * 5 = 410 * 234 * 56
(900Mc + 957330) * 5 = 5276700
900Mc + 957330 = 5276700 / 5 = 1074528
900Mc = 1074528 - 957330
900Mc = 117198
Mc = 117198/ 900
Mc = 130.22 g
The mass of the cup is 130.22 g.
An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 60 W incandescent bulb can be replaced by a 12 W LED bulb. Both produce 800 lumens of light. Assuming the cost of electricity is $0.29 per kilowatt-hour, how much does it cost (in dollars) to run the LED bulb for one year if it runs for four hours a day?
Answer:
C = $5.08
it costs $5.08 to run the LED bulb for one year if it runs for four hours a day
Explanation:
Given;
Power of Led bulb P = 12 W
Rate r = $0.29 per kilowatt-hour
Time = 4 hours per day
The number of hours used in a year is;
time t = 4 hours per day × 365 days per year
t = 1460 hours
The energy consumption of Led bulb in a year is;
E = Pt
E = 12 W × 1460 hours
E = 17520 watts hour
E = 17.52 kilowatt-hour
The cost of the energy consumption is;
C = E × rate = Er
C = 17.52 × $0.29
C = $5.08
it costs $5.08 to run the LED bulb for one year if it runs for four hours a day
An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination of the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 10.0 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination? Consider: east to the right, west to the left, north upwards and south downwards
Answer:
θ = 4.78º
with respect to the vertical or 4.78 to the east - north
Explanation:
This is a velocity compound exercise since it is a vector quantity.
The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed
v_fly² = v_nort² + v_air²
v_nort² = v_fly² + - v_air²
Let's use trigonometry to find the direction of the plane
sin θ = v_air / v_fly
θ = sin⁻¹ (v_air / v_fly)
let's calculate
θ = sin⁻¹ (10/120)
θ = 4.78º
with respect to the vertical or 4.78 to the north-east
A block of mass M rests on a block of mass M1 which is on a tabletop. A light string passes over a frictionless peg and connects the blocks. The coefficient of kinetic friction between the blocks and between M1 and the tabletop is the same. A force F pulls the upper block to the left and the lower block to the right. The blocks are moving at a constant speed. Required:Determine the mass of the upper block. (Express your answer to three significant figures.)
Answer:
M = 3.66 kg
Explanation:
Here is the complete question
A block of mass M rests on a block of mass M1 = 5.00 kg which is on a tabletop. A light string passes over a frictionless peg and connects the blocks. The coefficient of kinetic friction μ
k at both surfaces equals 0.330. A force of F = 56.0 N pulls the upper block to the left and the lower block to the right. The blocks are moving at a constant speed. Determine the mass of the upper block. (Express your answer to three significant figures.)
Solution
The forces on mass M are
F - μMg = Ma (1)
The forces on mass M₁ are
F - μ(M + M₁)g = M₁a (2) (since both weights act downwards on M)
From (1) a = (F - μMg)/M
Substituting a into (2), we have
F - μ(M + M₁)g = M₁((F - μMg)/M)
Cross-multiplying M we have
MF - μ(M + M₁)Mg = M₁F - μMM₁g
Expanding the bracket, we have
MF - μM²g + μM₁Mg = M₁F - μMM₁g
We now collect like terms
MF - μM²g + μM₁Mg + μMM₁g = M₁F
MF - μM²g + 2μM₁Mg - M₁F = 0
- μM²g + 2μM₁Mg + MF - M₁F = 0
Dividing through by -1, we have
- μM²g + (2μM₁g + F)M - M₁F = 0
μM²g - (2μM₁g + F)M + M₁F = 0
M² - (2M₁ + F/μg)M + M₁F/μg = 0
We now have a quadratic equation in M. We now substitute the values of the variables int o the quadratic equation to get
M² - (2(5 kg) + 56 N/(0.33 × 9.8 m/s²))M + (5 kg × 56 N)/(0.33 × 9.8 m/s²) = 0
M² - (10 kg) + 56 N/3.234 m/s²)M + (5 kg × 56 N)/(3.234 m/s²) = 0
M² - (10 kg + 17.32 kg) M + 86.58 kg = 0
M² - 27.32 kg M + 86.58 kg = 0
Using the quadratic formula
with a = 1, b = -27.32 and c = 86.58,
[tex]M = \frac{-(-27.32) +/- \sqrt{(-27.32)^{2} - 4 X 1 X 86.58} }{2 X 1} \\= \frac{27.32 +/- \sqrt{746.38 - 346.32} }{2}\\= \frac{27.32 +/- \sqrt{400.06} }{2}\\= \frac{27.32 +/- 20.001 }{2}\\= \frac{27.32 + 20.001 }{2} or \frac{27.32 - 20.001 }{2}\\= \frac{47.32 }{2} or \frac{7.32 }{2}\\\\=23.66 or 3.66[/tex]
Since M cannot be greater than M₁ for M to move over M₁, we take the smaller number.
So, M = 3.66 kg
How much electrical energy is used by a 75 W laptop that is operating for 12
minutes?
"1 watt" means 1 joule of energy per second.
75 W means 75 joules/sec .
Energy = (75 Joule/sec) x (12 min) x (60 sec/min)
Energy = (75 x 12 x 60) (Joule-min-sec / sec-min)
Energy = 54,000 Joules
A 0.140-kg baseball is thrown with a velocity of 27.1 m/s. It is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original velocity. How long were the bat and ball in contact?
Answer:
About [tex]1.795 \times 10^{-3}[/tex] seconds
Explanation:
[tex]\Delta p=F \Delta t[/tex], where delta p represents the change in momentum, F represents the average force, and t represents the change in time.
The change of velocity is:
[tex]37-(-27.1)=64.1m/s[/tex]
Meanwhile, the mass stays the same, meaning that the change in momentum is:
[tex]64.1\cdot 0.14kg=8.974[/tex]
Plugging this into the equation for impulse, you get:
[tex]8.974=5000\cdot \Delta t \\\\\\\Delta t= \dfrac{8.974}{5000}\approx 1.795 \times 10^{-3}s[/tex]
Hope this helps!
charged particles from the solar winds ultimately cause ___. a. the earth to maintain it's magnetic field b. the earth to change shape c. the auroras d. strong winds on earth
Answer:
The auroras C.
Explanation:
New evidence increasingly emphasizes that __________.
How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for a slit width of 0.110 mm, using light of wavelength 582 nm?
Answer:
6
Explanation:
We are given that
[tex]\theta=2.12^{\circ}[/tex]
Slid width,a=0.110 mm=[tex]0.11\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
Wavelength,[tex]\lambda=582 nm=582\times 10^{-9}[/tex] m
[tex]1nm=10^{-9} m[/tex]
We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.
[tex]asin\theta=\frac{2N+1}{2}\lambda[/tex]
Using the formula
[tex]0.11\times 10^{-3}sin(2.12)=\frac{2N+1}{2}(582\times 10^{-9})[/tex]
[tex]2N+1=\frac{0.11\times 10^{-3}sin(2.12)\times 2}{582\times 10^{-9}}[/tex]
[tex]2N+1=13.98[/tex]
[tex]2N=13.98-1=12.98[/tex]
[tex]N=\frac{12.98}{2}\approx 6[/tex]
Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern
A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the acceleration due to gravity is 9.81 m/s 2. The puck experiences no significant air resistance as it falls. The height of the platform above the ground is 2.00 m. What is the angle below the horizontal of the velocity of the puck just before it hits the ground
Answer:
θ = 12.60°
Explanation:
In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})[/tex] (1)
θ: angle below he horizontal
vy: y component of the velocity just after the puck hits the ground
vx: x component of the velocity
The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:
[tex]v_x=v_o[/tex]
vo: initial velocity = 28.0 m/s
The y component is calculated with the following equation:
[tex]v_y^2=v_{oy}^2+2gy[/tex] (2)
voy: vertical component of the initial velocity = 0m/s
g: gravitational acceleration = 9.8 m/s^2
y: height
You solve the equation (2) for vy and replace the values of the parameters:
[tex]v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}[/tex]
Finally, you use the equation (1) to find the angle:
[tex]\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°[/tex]
The angle below the horizontal is 12.60°
The angle below the horizontal of the velocity of the puck just before it hits the ground is 12.60°.
Given the following data:
Initial velocity = 28.0 m/s Acceleration due to gravity = 9.81 [tex]m/s^2[/tex]Displacement (height) = 2.00 meters.To find the angle below the horizontal of the velocity of the puck just before it hits the ground:
First of all, we would determine the horizontal and vertical components of the hockey puck.
For horizontal component:
[tex]V_y^2 = U_y^2 + 2aS\\\\V_y^2 = 0^2 + 2(9.81)(2)\\\\V_y^2 = 39.24\\\\V_y = \sqrt{39.24} \\\\V_y = 6.26 \; m/s[/tex]
For vertical component:
[tex]V_x = U_x\\\\V_x = 28.0 \;m/s[/tex]
Now, we can find the angle by using the formula:
[tex]\Theta = tan^{-1} (\frac{V_y}{V_x} )[/tex]
Substituting the values, we have:
[tex]\Theta = tan^{-1} (\frac{6.26}{28.0} )\\\\\Theta = tan^{-1} (0.2236)\\\\\Theta = 12.60[/tex]
Angle = 12.60 degrees.
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A cyclist rides a bicycle with a wheel radius of 0.500 m across campus. A piece of plastic on the front rim makes a clicking sound every time it passes through the fork. If the cyclist counts 320 clicks between her apartment and the cafeteria, how far has she traveled
Answer:
1500 clicks
Explanation:
Well if .5 times 320 is done then that would leave 150 then g(1) = 1² + 1 = 2 and lead to 150 being multipled to 1500 clicks
A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor
Answer:
9.6 Ns
Explanation:
Note: From newton's second law of motion,
Impulse = change in momentum
I = m(v-u).................. Equation 1
Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.
Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)
Substitute into equation 1
I = 2.4[2.5-(-1.5)]
I = 2.4(2.5+1.5)
I = 2.4(4)
I = 9.6 Ns
The magnitude of impulse will be "9.6 Ns".
According to the question,
Mass,
m = 2.4 kgFinal velocity,
v = 2.5 m/sInitial velocity,
u = -1.5 m/sBy using Newton's 2nd law of motion, we get
→ Impulse, [tex]I = m(v-u)[/tex]
By substituting the values, we get
[tex]= 2.4[2.5-(1.5)][/tex]
[tex]= 2.4(2.5+1.5)[/tex]
[tex]= 2.4\times 4[/tex]
[tex]= 9.6 \ Ns[/tex]
Thus the above answer is right.
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If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how fast must the fish spit the water to hit its target? The insect is at the highest point of the trajectory of the spit water. Use g = 10 m/s2.
Answer:
The speed is [tex]v = 9.8 \ m/s[/tex]
Explanation:
From the question w are told that
The angle made is [tex]\theta = 30^o[/tex]
The distance above the surface of the water is [tex]h_{max} = 1.2 \ m[/tex]
The value of [tex]g = 10 \ m/s^2[/tex]
The maximum height attained by the fish is mathematically evaluate as
[tex]h_{max} = \frac{v^2 sin ^2 \theta }{2g }[/tex]
Making v which is the speed of the fish the subject of the formula
[tex]v = \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }[/tex]
substituting values
[tex]v = \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2 } }[/tex]
[tex]v = 9.8 \ m/s[/tex]
What happens when both sides of the tug-of-war rope are equal in force?What is the net force in this scenario?
The net force is zero
What will happen to an astronaut when the jets produce these four forces
A car travels 2500 m in 8 minutes. Calculate the speed at which the car travelled
Answer:
5.95m/s to 2 decimal places
Explanation:
In physics speed is measured in metres per second so convert 8mins to seconds
8x60=420 seconds
The formula needed:
Speed (m/s)= Distance (m)/Time (s)
2500/420=5.95m/s
An object with a mass m slides down a rough 37° inclined plane where the coefficient of kinetic friction is 0.20. If the plane is 10 m long and the mass starts from rest, what will be its speed at the bottom of the plane?
Answer: 9.312 m/s
Explanation:
The friction force (opposite to the motion) is Fa = μ*m*g*cos(α) with μ = kinetic friction. The force that makes the motion is
F = m*g*sin(α).
The Newton's law gives:
F - Fa = m*a
m*g*sin(α) - μ*m*g*cos(α) = m*a
g*sin(α) - μ*g*cos(α) = a so a = 4.335 m/s²
It's a uniformly accelerated motion:
Space
S = 0.5*a*t²
10 = 0.5*a*t²
=> t = 2.148 s
Velocity
V = a*t = 9.312 m/s.
We have that the speed at the bottom of the plane is
[tex]v-9.3m/s[/tex]
From the question we are told that:
Angle of slide [tex]\theta =3.7 \textdegree[/tex]
Coefficient of kinetic friction [tex]\mu= 0.20[/tex]
Length [tex]L=10m[/tex]
Generally, the equation for acceleration along the slide is mathematically given by
[tex]a=gsin \theta-\mu cos\theta[/tex]
[tex]a=(9.8sin37-0.20*9.8*cos37[/tex]
[tex]a=4.33m/s^2[/tex]
Therefore
Velocity v is is mathematically given by
[tex]v=\sqrt{2as}[/tex]
[tex]v=\sqrt{2*4.33*10}[/tex]
[tex]v-9.3m/s[/tex]
In conclusion
The speed at the bottom of the plane is
[tex]v-9.3m/s[/tex]
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g 12. Spring A has a spring constant of 100 N/m and spring B has a spring constant of 200 N/m, and both have the same displacement of 0.20 m. The spring potential energy of spring A is _____ that of spring B. twice half the same as None of the above
Answer:
A. twiceExplanation:
Spring potential Energy is expressed as E = 1/2 ke²
k = spring constant
e = displacement of the string
If Spring A has a spring constant of 100 N/m and displacement of 0.2m, the spring potential energy will be;
= 1/2 * 100 * 0.2²
= 50 * 0.04
= 2Joules
If spring B has a spring constant of 200 N/m and displacement of 20m, the spring potential energy will be;
= 1/2 * 200 * 0.2²
= 100 * 0.04
= 4Joules
From the values gotten, it can be seen that the spring potential energy of spring B is twice that of A
Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box of mass M. There is friction between the surfaces of blocks 2M and 3M so the three blocks accelerate together to the right.
Which block has the smallest net force acting on it?
A) M
B) 2M
C) 3M
D) The net force is the same for all three blocks Submit
Answer:
A) M
Explanation:
The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:
Box with mass M
[tex]\Sigma F = F - F' = M\cdot a[/tex]
Box with mass 2M
[tex]\Sigma F = F' - F'' = 2\cdot M \cdot a[/tex]
Box with mass 3M
[tex]\Sigma F = F'' = 3\cdot M \cdot a[/tex]
On the third equation, acceleration can be modelled in terms of F'':
[tex]a = \frac{F''}{3\cdot M}[/tex]
An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.
[tex]F' = 2\cdot M \cdot a + F''[/tex]
[tex]F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''[/tex]
[tex]F' = \frac{5}{3}\cdot F''[/tex]
Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:
[tex]F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)[/tex]
[tex]F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''[/tex]
[tex]F = 2\cdot F''[/tex]
[tex]F'' = \frac{1}{2}\cdot F[/tex]
Afterwards, F' as function of the external force can be obtained by direct substitution:
[tex]F' = \frac{5}{6}\cdot F[/tex]
The net forces of each block are now calculated:
Box with mass M
[tex]M\cdot a = F - \frac{5}{6}\cdot F[/tex]
[tex]M\cdot a = \frac{1}{6}\cdot F[/tex]
Box with mass 2M
[tex]2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F[/tex]
[tex]2\cdot M \cdot a = \frac{1}{3}\cdot F[/tex]
Box with mass 3M
[tex]3\cdot M \cdot a = \frac{1}{2}\cdot F[/tex]
As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.
Select the correct answer. When waves travel through water, why do the moving water particles continue to return to their starting position? A. Waves transfer matter, but not energy. B. Waves transfer both matter and energy. C. Waves transfer energy, but not matter. D. Waves don’t transfer matter or energy.
Waves transfer energy, but not matter. (C)
Answer:
C. Waves transfer energy not matter
Explanation:
wave is a disturbance
The index of refraction for a certain type of glass is 1.645 for blue light and 1.609 for red light. A beam of white light (one that contains all colors) enters a plate of glass from the air, nair≈1, at an incidence angle of 38.55∘. What is the absolute value of ????, the angle in the glass between blue and red parts of the refracted beams?
Answer:
blue θ₂ = 22.26º
red θ₂ = 22.79º
Explanation:
When a light beam passes from one material medium to another, it undergoes a deviation from the path, described by the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the incident and transmitted media refractive indices and θ are the angles in the media
let's apply this equation to each wavelength
λ = blue
in this case n₁ = 1, n₂ = 1,645
sin θ₂ = n₁/ n₂ sin₂ θ₁
let's calculate
sin θ₂ = 1 / 1,645 sint 38.55
sin θ₂ = 0.37884
θ₂ = sin⁻¹ 0.37884
θ₂ = 22.26º
λ = red
n₂ = 1,609
sin θ₂ = 1 / 1,609 sin 38.55
sin θ₂ = 0.3873
θ₂ = sim⁻¹ 0.3873
θ₂ = 22.79º
the refracted rays are between these two angles
A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right, and friction. The weight of the crate is 400 N. If the 600-N force is removed, the resultant force acting on the crate is:
Answer:
200 N.
It seems very difficult but it really is not, as the wording just makes it seem difficult.
We have a force of 600 N, a force of 200N, and friction. We're leaving the direction of the forces aside. The weight of the crate is 400N, but this is totally irrelevant, as it is the crate itself, and not a force acting upon it. Now we have 600N and 200N. Taking away the 600N simply leaves us with 200N.
A total charge of 62 nC is uniformly distributed throughout a non-conducting sphere with a radius of 5.00 cm. The electric potential at r = 15.0 cm , relative to the potential far away, is:________
Answer:
2790 J/C
Explanation:
charge on sphere Q = 62 nC = [tex]62*10^{-9} C[/tex]
radius of the sphere r = 5.0 cm = 0.05 m
distance away from reference point d = 15.0 cm = 0.15 m
total distance of charge relative reference point R = r + d = 0.05 + 0.15 = 0.2 m
electric potential V is given as
[tex]V = \frac{kQ}{R}[/tex]
where k = Coulumb's constant = [tex]9*10^{9}[/tex] kg⋅m³⋅s⁻⁴⋅A⁻²
[tex]V = \frac{9*10^{9} * 62*10^{-9} }{0.2}[/tex] = [tex]\frac{9*62}{0.2}[/tex]
V = 2790 J/C
Suppose that when you move the north pole of a bar magnetic toward a coil it induces a positive current in the coil. The strength of the field produced by an electromagnetic can be controlled electronically. Suppose you place a coil near the north pole of an electromagnet and increase the field while keeping everything stationary. Which one of the following will happen? a) A positive current will be induced in the coilb) A negative current will be induced in the coil c) No current will be induced in the coil since there is no relative motion.
Answer:
a) A positive current will be induced in the coil
Explanation:
Electromagnetic induction is the induction of an electric field on a conductor due to a changing magnetic field flux. The change in the flux can be by moving the magnet relative to the conductor, or by changing the intensity of the magnetic field of the magnet. In the case of this electromagnets, the gradual increase in the the electromagnet's field strength will cause a flux change, which will in turn induce an electric current on the coil.
According to Lenz law, the induced current acts in such a way as to negate the motion or action that is producing it. A positive current will be induced on the coil so as to repel any form of attraction between the north pole of the electromagnet and the coil. This law obeys the law of conservation of energy, since work has to be done to move the move them closer to themselves.
Identify the following as combination, decomposition, replacement, or ion exchange reactions: Al(s) + 3 Cl2(g) → 2 AlCl3(s) Ca(OH)2(aq) + H2SO4(aq) → CaSO4(aq) + 2 H2O(l
Answer:
2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)
This is a combination reaction.
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)
This is a replacement reaction.
Explanation:
A combination reaction is a reaction in which two reagents are combined into one product. The reaction has the following general form:
A + B → AB
where A and B represent any two chemical substances.
2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)
This is a combination reaction because a single compound forms from two or more reacting species.
Double Substitution, Double Displacement or Metastasis Reactions are those in which two elements found in different compounds exchange their positions forming two new compounds. These chemical reactions do not present changes in the number of oxidation or relative load of the elements. So they are not considered redox reactions.
The solvent of the double displacement reactions usually is water and the reagents and products are usually ionic compounds (cations or anions are exchanged), although they can also be acids or bases.
In general, this type of reaction can be expressed as:
AB + CD ⇒ AD + CD
In the reaction:
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)
This is a replacement reaction because it is a double replacement reaction in which the ions are exchanged to form new compounds.
A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a surface area of 9.50 m2. The total power it absorbs from the star is 4400 W. Assuming the surface is an ideal absorber and radiator, calculate the equilibrium temperature of the asteroid (in K).
Answer:
T = 300.6K
Explanation:
In this problem, since the asteroid is an ideal absorber, we can approximate it to a black body, and use Stefan's law
P = σ A e T⁴
where P is the absorbed power, A the area of the asteroid, and the emissivity that for a black body is worth 1 and sigma the Stefan_boltzmann constant 5,670 10⁻⁸ W / m² K⁴
they ask us for the temperature of the asteroid
T = [tex]\sqrt[4]{(P / \sigma A e)}[/tex]
let's calculate
T = (4400 / (5,670 10⁻⁸ 9.50 1)
T =(81.6857 108)
T = 3,006 102 K
T = 300.6K
According to Oxford Dictionaries, a spit take is an act of suddenly spitting out liquid one is drinking in response to something funny or surprising. In spit takes, a gauge pressure is applied in the mouth, p1, so that liquid flows through pursed lips forming a column of liquid with radius r2 = 4 mm. If the liquid travels at v2 = 3.1 m/s outside the body, and if the column's area is 10x larger inside the mouth, what is p1 in Pa?
Answer:
The pressure is [tex]p_1 = 4051.4 \ Pa[/tex]
Explanation:
From the question we are told that
The gauge pressure at the mouth is [tex]p_1[/tex]
The radius of the column is [tex]r_2 = 4 \ mm = 0.004 \ m[/tex]
The speed of the liquid outside the body is [tex]v_2 = 3.1 \ m/s[/tex]
The area of the column is [tex]A_2[/tex]
The area inside the mouth [tex]A_1 = 10 A_2[/tex]
Generally according to continuity equation
[tex]v_1 A_1 = v_2 A_2[/tex]
=> [tex]v_ 1 = v_2 * \frac{A_2}{A_1}[/tex]
=> [tex]v_ 1 = 3.1 * \frac{1}{10}[/tex]
=> [tex]v_ 1 = 0.31 \ m/s[/tex]
So
[tex]A_1 = 10A_2[/tex]
=> [tex]\pi * r_1^2 = 10(\pi * r_2^2)[/tex]
=> [tex]r_1 = 10 * r_2[/tex]
substituting values
[tex]r_1 = 10 * 0.004[/tex]
[tex]r_1 =0.04 \ m[/tex]
Now the height of inside the mouth is [tex]h_1 = d = 2r_1 = 2* 0.04 = 0.08\ m[/tex]
Now the height of the column is [tex]h_2 = d = 2r_2 = 2* 0.004 = 0.008\ m[/tex]
Generally according to Bernoulli's equation
[tex]p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ][/tex]
Now [tex]\rho = 1000 \ kg m^{-3}[/tex] which is the density of water
[tex]p_2[/tex] is the gauge pressure of the atmosphere which is zero
So
[tex]p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-[/tex]
[tex][(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)][/tex]
[tex]p_1 = 4051.4 \ Pa[/tex]
The initial pressure of the gauge pressure at the mouth is 4757 pascals.
The principle of continuity equation asserts that in a steady flow state, the quantity of fluid flowing at the inlet is equivalent to the quantity of fluid at the outlet given that there is a constant mass flow rate.
It can be expressed by using the formula:
[tex]\mathbf{A_1v_1=A_2v_2}[/tex]
where;
The speed of the liquid inside the body is [tex]\mathbf{v_1}[/tex] = ???The speed of the liquid outside the body is [tex]\mathbf{v_2}[/tex] = 3.1 m/sArea outside the column is = A₂Area inside the column is [tex]\mathbf{A_1}[/tex] = 10A₂∴
Making v₁ the subject of the formula:
[tex]\mathbf{v_1 = \dfrac{A_2}{A_1}\times v_2}[/tex]
[tex]\mathbf{v_1 = \dfrac{1}{10}\times3.1}[/tex]
[tex]\mathbf{v_1 =0.31 \ m/s}[/tex]
Since the area are equivalent to each other
[tex]\mathbf{A_1 = 10A_2}[/tex]
[tex]\mathbf{\pi r^2_1 = 10\times \pi r^2_2}[/tex]
[tex]\mathbf{ r^2_1 = 10\times r^2_2}[/tex]
where;
r₂ = radius of the column. = 4mm = 0.004 m∴
[tex]\mathbf{ r_1 = 10\times 0.004}[/tex]
[tex]\mathbf{ r^2_1 = 0.04 \ m}[/tex]
However, in fluid dynamics, Bernoulli's equation can be applied for the estimation of the initial gauge pressure by using the expression:
[tex]\mathbf{\dfrac{1}{2} \rho v_1^2 + h_1 \rho g + p_1 = \dfrac{1}{2}\rhov_2^2 + h_2 \rho g +p_2}[/tex]
Since the gauge pressure of the atmosphere [tex]\mathbf{p_2}[/tex] = 0
∴
The initial gauge pressure applied at the mouth can be determined as:
[tex]\mathbf{p_1 = \dfrac{1}{2} \rho ( v_2^2-v_1^2)}[/tex]
here;
[tex]\mathbf{\rho = 1000 \ kg/m^3}[/tex]
[tex]\mathbf{p_1 = \dfrac{1}{2} \times 1000 \ kg/m^3 ( 3.1^2-0.31^2)}[/tex]
[tex]\mathbf{p_1 =500 \ kg/m^3 (9.5139)}[/tex]
[tex]\mathbf{p_1 =4756.95 \ Pa}[/tex]
[tex]\mathbf{p_1 \simeq4757 \ Pa}[/tex]
Learn more about the principle of continuity equation here:
https://brainly.com/question/14619396
A car travels 13 km in a southeast direction and then 16 km 40 degrees north of east. What is the car's resultant direction?
Answer:
21.48 km 2.92° north of east
Explanation:
To find the resultant direction, we need to calculate a sum of vectors.
The first vector has module = 13 and angle = 315° (south = 270° and east = 360°, so southeast = (360+270)/2 = 315°)
The second vector has module 16 and angle = 40°
Now we need to decompose both vectors in their horizontal and vertical component:
horizontal component of first vector: 13 * cos(315) = 9.1924
vertical component of first vector: 13 * sin(315) = -9.1924
horizontal component of second vector: 16 * cos(40) = 12.2567
vertical component of second vector: 16 * sin(40) = 10.2846
Now we need to sum the horizontal components and the vertical components:
horizontal component of resultant vector: 9.1924 + 12.2567 = 21.4491
vertical component of resultant vector: -9.1924 + 10.2846 = 1.0922
Going back to the polar form, we have:
[tex]module = \sqrt{horizontal^2 + vertical^2}[/tex]
[tex]module = \sqrt{460.0639 + 1.1929}[/tex]
[tex]module = 21.4769[/tex]
[tex]angle = arc\ tangent(vertical/horizontal)[/tex]
[tex]angle = arc\ tangent(1.0922/21.4491)[/tex]
[tex]angle = 2.915\°[/tex]
So the resultant direction is 21.48 km 2.92° north of east.
Sophie throws a tennis ball down from a height of 1.5 m at an angle of 450 with respect to vertical. She drops another tennis ball from the same height. Use the Energy Interaction Model to predict which ball will hit the ground with greater speed.
Given that,
Height =1.5 m
Angle = 45°
We need to find the greater speed of the ball
Using conservation of energy
[tex]P.E_{i}+K.E_{f}=P.E_{f}+K.E_{f}[/tex]
[tex]mgh+\dfrac{1}{2}mv_{i}^2=mgh+\dfrac{1}{2}mv_{f}^2[/tex]
Here, initial velocity and final potential energy is zero.
[tex]mgh=\dfrac{1}{2}mv_{f}^2[/tex]
Put the value into the formula
[tex]9.8\times1.5=\dfrac{1}{2}v_{f}^2[/tex]
[tex]v_{f}^2=2\times9.8\times1.5[/tex]
[tex]v_{f}=\sqrt{2\times9.8\times1.5}[/tex]
[tex]v_{f}=5.42\ m/s[/tex]
Hence, the greater speed of the ball is 5.42 m/s.
The force a spring exerts on a body is a conservative force because:
a. a spring always exerts a force parallel to the displacement of the body.
b. the work a spring does on a body is equal for compressions and extensions of equal magnitude.
c. the net work a spring does on a body is zero when the body returns to its initial position.
d. the work a spring does on a body is equal and opposite for compressions and extensions of equal magnitude.
e. a spring always exerts a force opposite to the displacement of the body.
Answer:
c. the net work a spring does on a body is zero when the body returns to its initial position
Explanation:
A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.
An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.
A disk between vertebrae in the spine is subjected to a shearing force of 640 N. Find its shear deformation taking it to have the shear modulus of 1.00 109 N/m2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.30 cm in diameter.
Answer:
3.08*10^-6 m
Explanation:
Given that
Total shearing force, F = 640 N
Shear modulus, S = 1*10^9 N/m²
Height of the cylinder, L = 0.7 cm
Diameter of the cylinder, d = 4.3 cm
The solution is attached below.
We have our shear deformation to be 3.08*10^-6 m