Answer: the average density (in kg/m3) of the log is 400 kg/m³
Explanation:
Given that;
a log with uniform-diameter floats with 60.0% of its length above water.
from the Archimedes principle formula;
friction submerged = Average density of object / density of fluid. --- equ 1
We know that; density of water = 1000 kg/m³
now the log floats with 60.0% of its length above water
so submerged fraction is 100% - 60% = 40% = 0.40
from our equation;
friction submerged = Average density of object / density of fluid
we substitute
0.40 = Average density of object / 1000 kg/m³
Average density of object = 0.40 × 1000
Average density of object = 400 kg/m³
Therefore; the average density (in kg/m3) of the log is 400 kg/m³
A car accelerates uniformly from rest and reaches a speed of 9.9 m/s in 11.4 s. The diameter of a tire is 86.9 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.
Answer:
Number of revolutions = 20.71 rev.
Explanation:
Given the following data;
Initial speed, u = 0m/s
Final speed, v = 9.9m/s
Time, t = 11.4secs
Diameter = 86.9cm to meters = 86.9/100 = 0.869m
To find the acceleration;
Acceleration, a = (v - u)/t
Acceleration, a = (9.9 - 0)/11.4
Acceleration, a = 9.9/11.4
Acceleration, a = 0.87m/s²
Now we would find the distance covered by the tire using the second equation of motion.
S = ut + ½at²
S = 0(11.4) + ½*0.87*11.4²
S = 0 + 0.435*129.96
S = 56.53m
The circumference of the tire is calculated using the formula;
Circumference = 3.142 * diameter
Circumference = 3.142 * 0.869
Circumference = 2.73m
Number of revolutions = distance/circumference
Number of revolutions = 56.53/2.73
Number of revolutions = 20.71 rev.
Therefore, the number of revolutions the tire makes during this motion is 20.71 revolutions.
1. Amy uses 30N of force to push a lawn mower 10 meters. How much work does she do?
100 POINTS. PLEASE EXPLAIN
Answer:
Explanation:
Note the charge balls on the top and bottom row are identical. So those charges cancel each other out. The only charges in the net electric field are the two in the middle row.
Electric field strength = k*Q/r^2
= (8.99 *10^9) * (3-(-3)) * 5*10^(-6) / (2*0.5)^2
= 269700 N/C
Answer:
Explanation:
(b) cuz the 1st n 3nd row cancel out, net electric field will go from +3Q to -3Q. the direction is right.
An athlete runs on track at a constant speed of 75 meters/min for 15 s. What is the total distance he covered during that time?
Answer:
45
Explanation:
Determine the gravitational field of Earth at a height 2.88 x 10^8 m above its surface (the height of the moon above Earth). Earth's mass is 6.0 x 10^24 kg and its radius is 6.4 x 10^6 m.
Answer:
12
Explanation:
Two charges q1 and q2 are separated by a distance r and apply a force F to each other. If both charges are doubled, and the distance between them is halved, the new force between them is
Answer:
The new force between them is increased by a factor of 16.
Explanation:
According to Coulombs law, the force of attraction between two (2) charges is given by the formula;
F = Kq1q2/r²
Given the following data;
q1 = 2q1
q2 = 2q2
r = r/2
Substituting into the equation, we have;
F = 2q1*2q2/(r/2)²
F = 4q1q2/r²/4
F = 4q1q2 * 4/r²
F = 16q1q2/r²
Therefore, the new force between them is increased by a factor of 16.
Two force vectors are oriented such that the angle between their directions is 46 degrees and they have the same magnitude. If their magnitudes are 2.81 newtons, then what is the magnitude of their sum
Answer:
F = 5.17 N
Explanation:
If we know the magnitudes of both vectors, and the angle between them, we can find the magnitude of their sum, applying the cosine theorem, as follows:[tex]F =\sqrt{F_{1} ^{2} +F_{2} ^{2} + 2*F_{1} * F_{2} * cos \theta} (1)[/tex]
Replacing by the givens, F₁ = F₂ = 2.81 N, θ = 46º, we get:[tex]F =\sqrt{2.81 N ^{2} +2.81 N ^{2} + 2*2.81N* 2.81N* cos 46} = 5.17 N (2)[/tex]
A jumbo egg (80 grams) is dropped from a height of 15 meters onto a 1 inch of foam. Using kinematics, determine the velocity of the egg the instant before impact.
Answer:
the velocity of the egg the instant before impact is 17.15 m/s.
Explanation:
Given;
mass of the egg, m = 80 g = 0.08 kg
height through which the egg was dropped, h = 15 m
The velocity of the egg before impact will be maximum, and the final velocity is given by the following kinematic equation;
v² = u² + 2gh
where;
u is the initial velocity of the egg = 0
v is the final velocity of the egg before impact
v² = 0 + 2 x 9.8 x 15
v² = 294
v = √294
v = 17.15 m/s
Therefore, the velocity of the egg the instant before impact is 17.15 m/s.
Suppose the posted designated speed for a highway ramp is to be 30 mph and the radius of the curve is 700 ft. At what angle must the curve be banked? No Friction
Answer:
4.92°
Explanation:
The banking angle θ = tan⁻¹(v²/rg) where v = designated speed of ramp = 30 mph = 30 × 1609 m/3600 s = 13.41 m/s, r = radius of curve = 700 ft = 700 × 0.3048 m = 213.36 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the variables into the equation, we have
θ = tan⁻¹(v²/rg)
= tan⁻¹((13.41 m/s)²/[213.36 m × 9.8 m/s²])
= tan⁻¹((179.8281 m²/s)²/[2090.928 m²/s²])
= tan⁻¹(0.086)
= 4.92°
3.
A person drives north 6 blocks, then drives west 6 blocks.
The displacement is a straight line from the starting point to the finish in a
direction
O A northwesterly
O B southwesterly
O c northeasterly
OD southeasterly
While driving his sports car at 20.00 m/s down a four lane highway, a
man comes up behind a very slow school bus. The man is in a hurry so he
decides to pass the bus. If the man's car can accelerate at 6.00 m/s^2,
how long will it take for him to reach a speed of 35.00 m/s in order to pass
the bus? Round to the nearest hundredth. (2 decimal places) Type the
number only.
Answer:
2.50 s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 20 m/s
Acceleration (a) = 6 m/s²
Final velocity (v) = 35 m/s
Time (t) =?
Thus, we can obtain the time taken for the man to reach a speed of 35 m/s as follow:
v = u + at
35 = 20 + 6t
Collect like terms
35 – 20 = 6t
15 = 6t
Divide both side by 6
t = 15/6
t = 2.50 s
Thus, it will take the man 2.50 s to reach a speed of 35 m/s.
A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fast!!!)
Answer:let initial velocity u=14m/s
Final velocity v=20m/s
Time taken t=30
Acceleration =a
V=u +at
a= (20-14)/30
a=0.2m/s^2
Explanation:
Acceleration is the change in velocity with respect to time.
Which statements correctly describe the formula or name of a compound? Select all that apply.
OA. The formula of nitrogen trifluoride is NF 3
B. The formula of ammonia is NH3.
C. The name of AlF, is trialuminum fluoride.
D. The formula of calcium chloride is CaCl2
E. The name of Li, Se is lithium selenate.
OF. The formula of dinitrogen monoxide is NO
2
Results
G. The formula of sulfur trioxide is 30.
Cho
Аа
H. The formula of magnesium hydroxide is Mg(OH)2
G.
Answer:
A, B, D, and H
Explanation:
Statements A, B, D and H are all correct except the following:
Statement C is incorrect. The name of [tex] AlF_3 [/tex] is aluminium fluoride NOT "trialuminium fluoride".
Statement E is incorrect. The name of [tex] Li_2Se [/tex] is lithium selenide NOT "lithium selenate".
Statement F is incorrect. Dinitrigen monoxide, also known as nitrous oxide has a formula of [tex] N_2O [/tex] NOT [tex] NO_2 [/tex].
Statement G is incorrect. Sulfur trioxide formula is [tex] SO_3 [/tex].
PLEASE ANSWER-Why are loose electrons needed for heat conduction?
Answer:
Why do metals conduct heat so well? The electrons in metal are delocalised electrons and are free moving electrons so when they gain energy (heat) they vibrate more quickly and can move around, this means that they can pass on the energy more quickly.
A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a total contact area of 4 cm 2 with the ground. Calculate the pressure he exerts on the ground (include units), and write your answer to two significant figures.
Answer:
1500000 Pa
Explanation:
The formula for pressure is force per unit area.
P=F/A where F is force and A is area
Given that ;
F= mass * acceleration due to gravity
F= 60 * 9.81 = 588.6 = 589 N
A= area = 4cm² = 0.0004 m²
P= F/A = 589 / 0.0004
P= 1471500
P=1500000 Pa
A 56-kg woman contestant on a reality television show is at rest at the south end of a horizontal 149-kg raft that is floating in crocodile-infested waters. She and the raft are initially at rest. She needs to jump from the raft to a platform that is several meters off the north end of the raft. She takes a running start. When she reaches the north end of the raft she is running at 4.2 m/s relative to the raft. At that instant, what is her velocity relative to the water
Answer:
3.04 m/s
Explanation:
As the woman moves north, the raft moves south.
We then make the following assumptions for easier calculations
Let the motion of the woman = x
Let the motion of the raft = 4.2 - x
Using law of momentum, we see that
m*v = m1*v1, where
m = 56 kg
v = x
m1 = 149 kg
v1 = (4.2 -x)
Substituting the values into the equation, we have
56 * x = 149 (4.2 - x), opening the bracket
56x = 623 - 149x collecting like terms
205x = 623 Divide by 205
x = 623/205
x = 3.04 m/s
This then means that our answer is 3.04 m/s
A physics grad student has a machine that measures how
many electrons are whizzing through a ring. If she measures 2
Coulombs of charge flowing through the ring every second,
what is the current passing through the ring?
1 = _A
Answer:
2A
Explanation:
Given parameter:
Quantity of charge = 2C
Time taken = 1s
Unknown:
The current passing through the ring = ?
Solution:
Current can also be defined as the quantity of charge that passes through a conductor per unit of time:
I = [tex]\frac{q}{t}[/tex]
So;
I = [tex]\frac{2}{1}[/tex] = 2A
Help can’t find the answer no where
i'm stuck on that question also
A toy car, mass of 0.025 kg, is traveling on a horizontal track with a velocity of 5 m/s. If
the track then starts to climb upwards, how high up the track can the car reach?
Answer:
1.25 m
Explanation:
This is the vertical height not the distance along the slope.
[tex]K=U\\\frac{1}{2}mv^{2} = mgh\\h = \frac{v^{2}}{2g}=\frac{25}{20}=1.25 m[/tex]
The height the car can reach if the the track starts to climb upwards is 1.2742 meters up.
What is kinetic and potential energy?Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or its relation with its surrounding systems.
P.E. = mass × g × height
K.E. = 0.5 × mass × (velocity)²
Given that the toy car has a mass of 0.025 kg and is traveling on a horizontal track with a velocity of 5 m/s. Now, the car starts to climb up vertically, therefore, the kinetic energy will be converted to potential energy.
Kinetic Energy = Potential Energy
0.5 × mass × (velocity)² = mass × g × height
Cancel mass from both the sides,
0.5 × (velocity)² = g × height
0.5 × (5 m/s)² = 9.81 m/sec² × height
height = 1.2742 meters
Hence, the car will travel 1.2742 meters up.
Learn more about Kinetic and Potential Energy here:
https://brainly.com/question/15764612
#SPJ5
A car with mass 1141 kg, moving at 16 m/s, strikes a(n) 2916 kg car at rest. If the two cars stick together, with what speed do they move
Answer:
V = 4.49 m/s
Explanation:
Given that,
Mass of the car 1, m₁ = 1141 kg
Initial speed of car 1, u₁ = 16 m/s
Mass of car 2, m₂ = 2916 kg
Initial speed of car 2, u₂ = 0
We need to find the speed of the cars if they stick together. Let the speed be V. The momentum will remain conserved in the process. Using the conservation of momentum to find it.
m₁u₁ + m₂u₂ = (m₁+m₂)V
[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\=\dfrac{1141\times 16+2916 \times 0}{(1141 +2916 )}\\\\=4.49\ m/s[/tex]
So, the required speed of the two cars is 4.49 m/s.
If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0o. (b) What is the minimum coefficient of friction needed for a frightened dri
Answer:
a) The ideal speed = 16.21 m/s
b) Minimum co-efficient of friction = 0.216
Explanation:
From the given information:
The ideal speed can be determined by considering the centrifugal force component and the gravity component.
[tex]\dfrac{mv^2}{r}cos \theta = mg sin \theta[/tex]
[tex]v = \sqrt {gr \ tan \theta}[/tex]
[tex]= \sqrt{(9.8 \ m/s^2) (100) \ tan 15^0}[/tex]
= 16.21 m/s
(b)
Let assume that it requires 25 km/h to take the same curve.
Then, using the equilibrium conditions;
[tex]mg \ sin \theta = \dfrac{mv^2}{r} cos \theta + \mu ((\dfrac{mv^2}{r}) sin \theta + mg cos \theta)[/tex]
[tex]\mu = \dfrac{mg sin \theta - \dfrac{mv^2}{r} cos \theta }{((\dfrac{mv^2}{r}) sin \theta + mg cos \theta) }[/tex]
[tex]\mu = \dfrac{g sin \theta - \dfrac{ v^2}{r} cos \theta }{((\dfrac{v^2}{r}) sin \theta + g cos \theta) }[/tex]
[tex]\mu = \dfrac{(9.8 \ m/s^2 ) sin (15^0) - \dfrac{ \dfrac{(25 \times 10^3}{3600} \ m/s)^2 }{100 \ m } cos (15^0) }{((\dfrac{(\dfrac{25 \times 10^3}{3600} )^2}{100}) sin 15^0 + (9.8 \ m/s^2) cos 15^0 ) }[/tex]
[tex]\mathbf{\mu = 0.216}[/tex]
What is the wavelength of a wave that has a speed of 350 meters second and a frequency of 140 hert??
Wavelength is the distance between identical points (adjacent crests) in the adjacent cycles of a waveform signal propagated in space or along a wire. In wireless systems, this length is usually specified in meters (m), centimeters (cm) or millimeters (mm).
Hope this helped
A 5kg box is sliding down a ramp with a rough surface as seen below. The height of the ramp is 20m and the distance the box travels down the ramp (from A to B) is 15m. At point A the velocity of the box is 8 m/s. If the velocity at point B is 3m/s, what was the impulse caused by friction? If the force of friction is 5N, how long did it take the box to slide the 15 m?
Answer:
the answer is b luv .
Explanation:
A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a total contact area of 4 cm 2 with the ground. Calculate the pressure he exerts on the ground (include units), and write your answer to two significant figures.
Answer:
P = 1471500 [Pa]
Explanation:
We must remember that pressure is defined as the relationship between Force over the area.
[tex]P=F/A[/tex]
where:
P = pressure [Pa] (units of pascals)
F = force [N] (units of Newtons)
A = area of contact = 4 [cm²]
But first we must convert from cm² to m²
[tex]A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }[/tex]
A = 0.0004 [m²]
Also, the weight should be calculated as follows:
[tex]w = m*g[/tex]
where:
m = mass = 60 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now replacing:
[tex]w = 60*9.81\\w = 588.6[N][/tex]
And the pressure:
[tex]P=588.6/0.0004\\P=1471500 [Pa][/tex]
Because 1 [Pa] = 1 [N/m²]
The Earth (geosphere) comprises several major layers, differing in chemical and mineral composition, material strength, and other physical properties. Which of the layers listed below constitutes the largest proportion of Earth's volume?
a) Lithosphere
b) Crust
c) Outer core ·
d) Inner core
e) Mantle
Answer:
Option E:
The mantle
Explanation:
The earth's mantle is the mushy, semi-solid portion of the earth that makes up most of the earth's volume. The mantle extends for a depth of about 2800km downwards into the earth, making it the largest internal portion of the earth. It makes up about 84 percent of the earth's structure, leaving the core and the crust with 15 percent and 1 percent respectively.
Due its nature, convection currents are set up predominantly in the mantle of the earth, which leads to movements in the upper layers of the earth (the crust).
The mantle is large enough for the lighter crust to float on its surface.
which hand has a negatively charged?
Answer:
The dryer sheet is negatively charged and your hand is positively charged
Explanation:
For which medical procedure would Doppler ultrasound be most useful?
A.
Finding a lung tumor
B.
Fixing a pulled muscle
C.
Locating a broken bone in a finger
D.
Detecting a blockage in a heart artery
Doppler ultrasound would be most useful in detecting a blockage in a heart artery.
What are the clinical uses of Doppler ultrasound?By monitoring the rate of change in pitch, a Doppler ultrasound may calculate how quickly blood flows (frequency). A sonographer with training in ultrasound imaging applies pressure to your skin with a tiny, hand-held instrument (transducer) roughly the size of a bar of soap across the area of your body being scanned, moving from one place to another as required.
As an alternative to more invasive treatments like angiography, which involves injecting dye into the blood arteries to make them visible on X-ray images, this test may be performed.
Your doctor may use a Doppler ultrasound to assess for artery damage or to keep track of specific vein and artery therapies.
Learn more about Doppler technology here:
https://brainly.com/question/6109735
#SPJ2
A 2.0 kg block of ice with a speed of 8.0 m/s makes an elastic collision with another block of ice that is at rest. The first block of ice proceeds in the same direction as it did initially, but with a speed of 2.0 m/s. What is the mass of the second block? (Hint: Use the conservation of kinetic energy to solve for the second unknown variable.)
Answer:
6kg
Explanation:
According to conservation of kinetic energy
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses of the bodies
u1 and u2 are the initial velocities
v is the final velocity
Given
m1 =2kg
u1 = 8.0m/s
m2 = ?
u2 = 0m/s (second ice at rest)
v = 2.0m/s
Substitute into the formula
2(8)+m2(0) = (2+m2)(2)
16+0 = 4+2m2
16-4= 2m2
12 = 2m2
m2 =12/2
m2 = 6kg
Hence the mass of the second block is 6kg
A Discuss the possibility of fracture of two leg bones that have a length of about 70cm and an average area of
about 4cm
2 when a 80kg person jump from a height of 300cm.
Noting: The breaking stress of the bone ϬB =1.5×108 N/m2 , and
Young’s modulus for the bone is Y=1.5×1010 N/m2
Answer: The bones won't fracture.
Explanation: Stress, in Physics, is a quantity describing forces that can cause deformation. Strain is the measure of how muc an object can be stretched or deformed. The ratio between stress and strain is called Young's modulus or elastic modulus
Breaking Stress of Bone is the maximum stress a bone can take before a rupture occur.
To determine if a person will break his/her bones by jumping from a height, we determine the energy necessary for that jump and compare it with the energy necessary to break a bone.
The energy for breaking a bone is calculated as
[tex]E=\frac{Al_{0}\sigma_{B}^{2}}{2Y}[/tex]
A is the area in m²
l₀ is length in m
[tex]\sigma_{B}[/tex] is breaking stress in N/m²
Y is Young's modulus in N/m²
Calculating energy to break a bone:
[tex]E=\frac{4.10^{-4}.7.10^{-1}.(1.5.10^{8})^{2}}{2.(1.5.10^{10})}[/tex]
[tex]E=210[/tex] J
This is the energy necessary to break one leg bone, so as there are 2, energy will be 420 Joules.
Potential energy gained by jumping is calculated as
E = m.g.h
m is mass in kg
g is acceleration due to gravity in m/s²
h is height in m
Calculating
E = 80.(9.8)(0.3)
E = 235.2 J
Comparing the two energies, potential energy for jumping is less than maximum energy a bone can absorve without breaking, so the leg bones won't suffer a fracture.
g A box of mass 10 kg attached to a spring is pulled to a maximum of 44 cm. The box is released. What is the speed when the box reaches a point 10 cm from the equilibrium position
Answer:
The speed when the box reaches a point 10 cm from the equilibrium position is 2.02 m/s.
Explanation:
Mass attached to the spring, m = 10 kg
maximum displacement of the spring, A = 0.44 m
The spring constant is calculated from Hook's law;
F = kx
mg = kx
k = (mg) / x
k = (10 x 9.8) / 0.44
k = 222.73 N/m
The angular speed of the spring is calculated as;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega =\sqrt{\frac{222.73}{10} } \\\\\omega = 4.72 \ rad/s[/tex]
The speed when the box reaches a point 10 cm from the equilibrium position is calculated as;
[tex]v = \omega \sqrt{A^2-x^2} \\\\v = 4.72\sqrt{0.44^2-0.1^2}\\\\v = 2.02 \ m/s[/tex]
Therefore, the speed when the box reaches a point 10 cm from the equilibrium position is 2.02 m/s.