A spherical brass shell has an interior volume of 1.26 x 10-3 m3. Within this interior volume is a solid steel ball with volume of 0.63 x 10-3 m3. The space between the steel ball and the inner surface of the brass shell's filled completely with mercury. A hole's drilled through the brass, and the temperature of the arrangement in increased by 18 C°. How much mercury spills out?
Answer:
ΔV = 2.04 x 10⁻⁶ m³ = 2.04 cm³
Explanation:
First we will find the initial volume of mercury by subtracting the volume of the steel ball from the interior volume of the brass shell:
V = Vs - Vb
where,
V = Initial Volume of Mercury = ?
Vb = Volume of brass shell = 1.26 x 10⁻³ m³
Vs = Volume of Steel Ball = 0.63 x 10⁻³ m³
Therefore,
V = (1.26 x 10⁻³ m³) - (0.63 x 10⁻³ m³)
V = 0.63 x 10⁻³ m³
Now, we find the change in volume of mercury that will spill out due to increase in temperature:
ΔV = βVΔT
where,
ΔV = Amount of Mercury Spilled out = ?
β = Coefficient of Volumetric Expansion of mercury = 1.8 x 10⁻⁴ °C⁻¹
ΔT = Increase in temperature = 18 °C
Therefore,
ΔV = (1.8 x 10⁻⁴ °C⁻¹)(0.63 x 10⁻³ m³)(18 °C)
ΔV = 2.04 x 10⁻⁶ m³ = 2.04 cm³
The volume of mercury that spills out due to increase in temperature is [tex]2.04 \times 10^{-6} \ m^3[/tex].
The given parameters;
outer volume, V₁ = 1.26 x 10⁻³ m³inner volume, V₂ = 0.63 x 10⁻³ m³change in temperature, Δt = 18 ⁰CThe volume of mercury in the given shell is calculated as;
[tex]V= V_1 - V_2\\\\V= 1.26 \times 10^{-3} \ - \ 0.63 \times 10^{-3} \\\\V = 6.3 \times 10^{-4} \ m^3[/tex]
The change in the volume of the mercury is calculated as follows;
[tex]\Delta V = \beta V \Delta t\\\\[/tex]
where;
β is the coefficient of volumetric expansion of mercury[tex]\Delta V = (1.8 \times 10^{-4} ) \times (6.3\times 10^{-4}) \times 18\\\\\Delta V = 2.04 \times 10^{-6} \ m^3[/tex]
Thus, the volume of mercury that spills out due to increase in temperature is [tex]2.04 \times 10^{-6} \ m^3[/tex].
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Which of the following is an example of projectile motion
A student sitting at their desk staring at their test
An eraser resting on the desk
A pencil mid air falling toward the ground
O A back pack placed underneath the desk
Giving brainlist ASAP!
Check all that apply!!
Answer:
metal seat belt and door handle
Explanation:
bc its all medal it _____ heat (forgot the word mybad)
The stage in cell division in which chromosomes replicate themselves.*
Interphase
Metaphase
Telophase
Anaphase
Answer:
interphase
Explanation:
If you measure the properties of an unknown material and find that it has a density of 3.13 g/cm3 (+/- 0.1 g/cm3) and boils at 1415 °C (+/- 5 ºC), which material would it most likely be?
A) Potassium iodide
B) Sulfur
C) Iodine
D) Lead nitrate
E) Copper chloride
F) Sodium chloride
G) Copper nitrate
Answer:
The correct option is A
Explanation:
For a substance to boil at 1415 °C shows the substance is an electrovalent/ionic substance (hence the elements are not the answer). Electrovalent/ionic substances are characterized by high melting and boiling points. Also, with a density of 3.13 g/cm³ (with a +/- 0.1 g/cm³), the correct answer is potassium iodide. This is because potassium iodide has a density of 3.12 g/cm³ (which is - 0.1 g/cm³). It should also be noted that pure potassium iodide boils at 1,330 °C, however the presence of impurities could increase the boiling point of the salt, thus an impure potassium iodide could have a density of 3.12 g/cm³ and a temperature around 1415 °C
"Potassium iodide" would it most likely be. A further explanation is provided below.
Given values are:
Density,
3.13 g/cm³Boiling point,
1415 CDensity can be change by,
+/-0.1 g/cm³Uncertainty in temperature,
+/-5°CWe can take the value of density from:
3.03 (3.13-0.1) to,3.23 (3.13+0.1)We can take the value of temperature from:
1410 C (1415-5) to,1420 C (1415+5)After checking the ranges, we find "Potassium iodide" have density as well as temperature in our range.
Thus the above answer is right.
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The nose is relation to the foot
RLLY?
Explanation:
ALSO
MY DAD ALWAYS TELLS ME THAT IF YOUR FEET ARE COLD THAT YOU NEED TO PUT A HAT ON?!?!?!?!?!
Select the correct answer from each drop-down Box menu Debbie places to shopping carts in a car corral she pushes the first car which then pushes a second car what force is being exerted Debbie exerts blank force on the first shopping cart the first card exerts blank force on the second card
Answer: Debbie exerts applied force on the first cart.
Explanation:
The first cart exert applied force on the second cart.
How do you deal with unanswered questions after death?
Answer: you have loved ones who help you get through it and help answer the questions they know.
Answer:
you have loved ones who help you get through it and help answer the questions they know.
Explanation:
What holds a moon in its orbit?
OA. its own internal magnetism
OB. the heat generated by the planet it orbits
OC. the gravitational pull of the star it orbits
OD. the gravitational pull of the planet it orbits
Answer:
D
Explanation:
The gravitational pull of the planet it orbits holds a moon in its orbit.
6. A painter weighing 800 newtons stands on a single board scaffold that is supported at each end
by a vertical rope from above. If one rope has a tension of 550 newtons and the other has a
tension of 450 newtons, what is the weight of the board?
Answer:
The weight of the board is 200 N
Explanation:
Translational Equilibrium
An object is in translational equilibrium if the velocity of its translational motion is constant or if it's at rest.
The painter is standing on the single board scaffold, obviously at rest. This means the sum of all translational forces must be zero.
If the tensions of both ropes are T1 and T2, the weight of the painter is Wp, and the weight of the board is Wb, then:
T1 + T2 - Wp - Wb = 0
Both tensions go vertically up and both weights go vertically down.
Solving for Wb:
Wb = T1 + T2 - Wp
Wb = 550 N + 450 N - 800 N
Wb = 200 N
The weight of the board is 200 N
Unpolarized light passes through two Polaroid sheets. The transmission axis of the second polarizing filter makes an angle of 340 with the axis of the first polarizer. What fraction of the original unpolarized light is transmitted through the second polarizing filter?
This question is incomplete, the complete question is;
Unpolarized light passes through two Polaroid sheets. The transmission axis of the second polarizing filter makes an angle of 34° with the axis of the first polarizer. What fraction of the original unpolarized light is transmitted through the second polarizing filter?
Answer: the fraction of the original unpolarized light transmitted through the second polarizing filter is 0.343
Explanation:
We know that when Unpolarized light passes through first polarizer, it's intensity will be reduced to half i.e
I1 = I0/2
Now after that when light passes through 2nd polarizer, Intensity will be;
I2 = I1 × (cos A1)²
but I1 = I0/2
so
I2 = (I0/2) × (cos A1)²
I2/I0 = 0.5 × (cos A1)²
In the question we were given that; The transmission axis of the second polarizing filter makes an angle of 34° with the axis of the first polarizer
so A = 34°
so we substitute
I2/I0 = 0.5 × (cos 34)²
I2/I0 = 0.343
Therefore the fraction of the original unpolarized light transmitted through the second polarizing filter is 0.343
Explain how you would make a cast of a shoeprint in left in the mud
Answer:
Put some melted plastic over the print and let it harden
Explanation:
I think that's how the police do it
Which change of state is shown in the model?
-condensation
-deposition
-boiling
-freezing
Answer: deposition
Explanation:
Try process of elimination. We know it's not boiling because that is basically water turning into gas. It's not condensation because it's not water which collects as droplets on a cold surface when humid air is in contact with it.
Answer:
deposition
Explanation:
A golfer attempts a birdie putt, sending the 0.050-kg ball toward the hole with a speed of 1.4 m/s. Unfortunately, the ball does a "horseshoe" and goes halfway around the rim of the hole. The ball ends up heading back toward the golfer with a speed of 1.3 m/s.If we take the positive direction to be from the golfer to the hole, what is the impulse delivered to the ball by the hole?
Answer:
I = - 0.135 kg m/s
Explanation:
The formula to calculate the impulse of delivered to the ball is:
I = Ft
where,
I = Impulse delivered to the ball by hole = ?
F = Force = ma (By Newton's Second Law)
t = time
Therefore,
I = mat
where,
m = mass of ball = 0.05 kg
a = acceleration of ball = (Vf - Vi)/t
Therefore,
I = m[(Vf - Vi)/t]t
I = m(Vf - Vi)
where,
Vf = Final velocity = - 1.3 m/s
Vi = Initial Velocity = 1.4 m/s
Therefore,
I = (0.05 kg)(- 1.3 m/s - 1.4 m/s)
I = - 0.135 kg m/s
What happens when a charged balloon is brought close to a wall?
The balloon becomes negatively charged.
The charges in the wall rearrange themselves.
The wall and the balloon repel each other.
The wall becomes negatively charged.
Answer:
a the balloon becomes negatively charged
When a charge balloon brought closer to the wall, they will repel each other. Option C is correct.
Electric Charge:
It shows the degree at which the atom or object has less or more electron than proton.
An atom with extra electrons than proton has negative charge while atoms will less electron than proton has positive charge.
Opposite charge always attracts while similar charges repel each other. Since both wall and balloon are negatively charged.
Therefore, when a charge balloon brought closer to the wall, they will repel each other.
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A wire carrying a current is shaped in the form of a circular loop of radius 4.0 mm. If the magnetic field strength at its center is 1.1 mT with no external magnetic fields contributing to it, what is the magnitude of the current that flows through the wire? A) 17 A B) 28 A C) 14 A D) 7.0 A
Answer:D) 7.0 A
Explanation:
The magnetic field strength at the center of a circular loop is calculated using the formulae
B = μ₀I / 2R
where B = ,magnetic field strength= 1.1mT = 1.1 X 10^-3T
μ₀= magnetic constant or the permeability of free space=4π X 10^-7 H/m
R = radius = 4.0mm =4.0 x 10^-3m
I= Current = ?
B=μ₀I / 2R
I = B2R /μ₀
I = 1.1 X 10^-3T X 2 X 4.0 x 10^-3m / 4π X 10^-7 H/m
I =7.00 A
Therefore magnitude of the current that flows through the wire is 7.0 A
need help with these 4 ASAP. if u can answer even 1 of them that would be amazing! formula sheet provided
1. a) a mass of 5.0kg is at rest at the top of an inclined plane, 5.0m above the ground. if the inclined plane is frictionless, calculate the final velocity of the mass as it reaches the bottom of the inclined plane?
b) if the mass slides down the plane and encounters a spring with a force constant of 8000.00 N/m by how much will the spring be compressed when the mass comes to a stop?
c) if the mass is found to move down the inclined plane at a constant velocity what is the force of friction between the ramp and the mass? (the angle of incline is 30°)
2. a 12 m long ramp has its high end 4.0m above the low end. a crate of mass 10kg is given a sharp push so that it begins to slide down the ramp from the top of the ramp with an initial speed of 4.0m/s. the force of friction along the ramp is 20N. determine the speed of the crate at the bottom of the ramp.
3. a 7.0 kg crate is dropped from a window 8.0m above the ground. if the average air resistance on the crate is 20.0N, what is the crates kinetic energy and velocity upon reaching the ground?
4. a 0.25 kg puck moving with a constant speed across a 2.0m stretch of horizontal frictionless ice. how much work is done?
you may want to google the answers cuz that is a lot
Cliff divers at Acapulco jump into the sea from a cliff 36 meters high. At the level of the sea, there is an outcropping of rock that sticks out a horizontal distance of 6 meters. With what minimum horizontal velocity must the cliff divers leave the top of the cliff if they are to miss this rocky outcropping?
Answer:
The cliff drivers must leave the top of the cliff at 2.21 m/s
Explanation:
Horizontal Motion
When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.
The range or maximum horizontal distance traveled by the object can be calculated as follows:
[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]
If the range is known and we are required to find the initial speed of launch, then we can solve the above equation for v as follows:
[tex]\displaystyle v=d\cdot\sqrt{\frac {g}{2h}}[/tex]
Cliff drivers jump into the sea from a height of h=36 m and must pass over an obstacle d=6 meters away from the cliff.
Let's calculate the speed needed:
[tex]\displaystyle v=6~m\cdot\sqrt{\frac {9.8~m/s^2}{2\cdot 36~m}}[/tex]
[tex]\displaystyle v=6~m\cdot\sqrt{\frac {9.8~m/s^2}{72~m}}[/tex]
[tex]v=2.21~m/s[/tex]
The cliff drivers must leave the top of the cliff at 2.21 m/s
what force is needed to accelerate 300kg rock as a rste of 4m/s
Answer:
60 maybe
Explanation:
Answer:
1200 newton
and that's 4m/s² not 4 m/s
A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.
What minimum spacing must there be between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope?
Assume the telescope's resolution is limited only by diffraction and that it is recording light with a wavelength of 500 nm.
Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
Explanation:
Given that;
diameter of the mirror d = 1.7 m
height h = 180 km = 180 × 10³ m
wavelength λ = 500 nm = 5 × 10⁻⁹ m
Now Angular separation from the peak of the central maximum is expressed as;
sin∅= 1.22 λ / d
sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7
sin∅ = 3.588 × 10⁻⁷
we know that;
sin∅ = object separation / distance from telescope
object separation = sin∅ × distance from telescope
object separation = 3.588 × 10⁻⁷ × 180 × 10³
object separation =6.45 × 10⁻² m
then we convert to centimeter
object separation = 6.45 cm
Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
a ball is thrown downward from the top of roof with a speed of 25 m/s. after 2 s, its velocity will be:
Answer:
12.5
Explanation:
first, v=d/t
v=25/2
v=12.5
The final velocity of the ball is 44.6 m/s.
The given parameters;
initial velocity of the ball, u = 25 m/stime of motion of the ball, t = 2 sThe velocity of the ball increases as the ball moves downward and it will become maximum before the ball hits the ground.
This velocity is known as the final velocity of the ball.
The final velocity of the ball is calculated as follows;
v = u + gt
where;
v is the final velocity of the ballg is the acceleration due t gravity = 9.8 m/s²v = 25 + 9.8(2)
v = 44.6 m/s
Thus, the final velocity of the ball is 44.6 m/s.
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A car is accelerating at 30 m/s2, if the car is 400 kg how much force
must be supplied from the engine?
3.1
power
Every day a certain household uses a 4.4 KW electric stove for 70 minutes, seven 150 W light bulbs for 7
hours each and miscellaneous appliances amounting to 1.8 KW/h. The 4 KW hot water cylinder switches
on for a total of 120 minutes during the day. If electricity cost 50c a KW/h. what is the monthly electricity
bill. (5 marks)
Answer:
The value is [tex]C = 30729\ c[/tex]
Explanation:
From the question we are told that
The power rating of the stove is [tex]P = 4.4 KW = 4.4 *10^{3} \ W[/tex]
The duration of its use everyday is [tex]t_1 = 70 \ minutes = 1.167 \ hours[/tex]
The rating of the light bulbs is [tex]P_2' = 150 W[/tex]
The number is [tex]n = 7[/tex]
The power rating of the total bulb is [tex]P_2 = 7 * 150 = 1050 \ W[/tex]
The duration of its use everyday is [tex]t_2 = 7 hours[/tex]
The power rating of miscellaneous appliance [tex]P_3 = 1.8 \ KW = 1.8 *10^{3} \ W[/tex]
The duration of its use everyday is [tex]t_3 = 1 hour[/tex]
The power rating of hot water [tex]P = 4 KW = 4 *10^{3} \ W[/tex]
The duration of its use everyday is [tex]t_4 = 120 \ minutes = 2 hours[/tex]
Generally the total electrical energy used in 1 month is mathematically represented as
[tex]E = P_1 * t_1 * 30 + P_2 * t_2 * 30 + P_3 * t_3 * 30+ P_4 * t_4 * 30[/tex]
=> [tex]E = 4.4*10^{3} * 1.167 * 30 + 1050 * 7 * 30 + 1.8 *1 * 30+ 4*10^{3} * 2 * 30[/tex]
=> [tex]E = 614598 \ W \cdot h[/tex]
=> [tex]E = 614.6 \ K W \cdot h[/tex]
Generally the monthly electricity bill is mathematically represented as
[tex]C = 614.6 * 50[/tex]
=> [tex]C = 30729\ c[/tex]
a rocket leaving the lanch pad is ana exmaple of which law
Answer:
Newton's Third Law
The boat goes forward, you go backward! A rocket can lift off from a launch pad only when it expels gas out of its engine. The rocket pushes on the gas, and the gas in turn pushes on the rocket.
Explanation:
Newton's Third Law
The boat goes forward, you go backward! A rocket can lift off from a launch pad only when it expels gas out of its engine
Explanation:
A car is initially moving at 100 m/s and decides to gun it. They accelerate at a rate of 20 m/s/s for 10 seconds. How far did they travel during this time period?
Answer:
dx = 2000 [m]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]dx =v_{o}*t +\frac{1}{2} *a*t^{2}[/tex]
where:
dx = distance [m]
Vo = initial velocity = 100 [m/s]
a = acceleration = 20 [m/s²]
t = time = 10 [s]
dx = 100*10 + 0.5*20*10²
dx = 2000 [m]
Note: the positive sign in the above equation means that the car is increasing its velocity.
which transition metals have more in common than other
a) alkali metals
b) lanthanides
c) actinides
d) lanthanides and actinides
Answer:
A. alkali metals
Explanation:
im not sure about that
Answer:
c
Explanation:
haha signs d d Sd Didi d do f idi
This diagram shows two different forces acting on a skateboarder. The
combined mass of the skateboard and the person is 65.2 kg. Based on this
information, what is the acceleration of the skateboarder?
Answer:
Option C. 0.36 m/s² to the left.
Explanation:
The following data were obtained from the question:
Mass of skateboarder = 65.2 Kg
Force in the right direction (Fᵣ) = 31.8 N
Force in the left direction (Fₗ) = 55.2 N
Acceleration of the skateboarder =.?
Next, we shall determine the net force acting on the skateboarder. This can be obtained as illustrated below:
Force in the right direction (Fᵣ) = 31.8 N
Force in the left direction (Fₗ) = 55.2 N
Net force (Fₙ) =?
Fₙ = Fₗ – Fᵣ
Fₙ = 55.2 – 31.8
Fₙ = 23.4 N to the left
Thus, the net force acting on the skateboarder is 23.4 N to the left.
Finally, we shall determine the acceleration of the skateboarder. This can be obtained as follow:
Mass of skateboarder = 65.2 Kg
Net force (Fₙ) = 23.4 N to the left
Acceleration (a) of the skateboarder =.?
Net force = mass × acceleration
23.4 N = 65.2 × a
Divide both side by 65.2
a = 23.4 / 65.2
a = 0.36 m/s² to the left.
Therefore, the acceleration of the skateboarder is 0.36 m/s² to the left.
Answer:
C. 0.36 [tex]m/s^{2}[/tex] to the left
Explanation:
A driver of a car going 40 m/s (approx. 90 mph) suddenly sees the lights of a barrier 40.0 m ahead. It takes the driver 0.75 s before he applies the brakes (this is known as reaction time). Once he does begin to break the decelerates at a rate of 10.0 m/s^2 (Be careful of signs). Does he hit the barrier?
Answer:
Explanation:
Distance of barrier = 40 m
Distance covered during reaction time = .75 x 40 = 30 m .
distance covered during deceleration can be calculated as follows .
v² = u² - 2 a s
0 = 40² - 2 x 10 x s
s = 80 m
Total distance covered before coming to rest = 30 + 80 = 110 m
Distance of barrier = 40 m , so it will hit the barrier .
A student travels 200 m north and then 500 m east to get to school. What is the shortest distance that he could travel between home and school?
A.700
B.300
C.539
D.458
Answer:
B???
Explanation:
guess
The shortest distance that the student could travel between home and school is approximately 539 meters.
Hence Option C) 539m is the correct answer.
Given the data in the question;
As illustrated in the image below,
Distance traveled by student towards north; [tex]b = 200m[/tex]Distance traveled by student towards east; [tex]a = 500m[/tex]We get a right angled triangle and the hypotonus "c" is the shortest distance from the student's house to her school.
Now, using Pythagorean theorem:
[tex]a^2 + b^2 = c2[/tex]
We substitute our values into the expression and solve for "c"
[tex](500m)^2 + (200m)^2 = c^2\\\\250000m^2 + 40000m^2 = c^2\\\\c^2 = 290000m^2\\\\c = \sqrt{290000m^2}\\\\c = 539 m[/tex]
The shortest distance that the student could travel between home and school is approximately 539 meters.
Hence Option C) 539m is the correct answer.
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A 50 kg pitcher throws a baseball with a mass of 0.15 kg, if the ball is thrown with a positive velocity of 35 m/s, and there is no net force on the system, what is the velocity of the pitcher?
Answer: A. -0.1 m/s
Explanation:
Answer:
The Answer is the first one, A. Or -0.1 m/s.
Explanation:
