Answer:
.5m per second to the west
Explanation:
distance divided by time
The velocity of a Froghopper flea, which is represented by a vector, has a direction and a magnitude. If a coordinate system is oriented where the x-axis is horizontal, and the y-axis is vertical, is it possible to write expressions for the x and y components of the velocity vector in terms of the vector's magnitude and direction.
Answer:
Explanation:
The solution of the question cab e found in attachment below:
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.
A child is twirling a 0.0113-kg ball on a string in a horizontal circle whose radius is 0.147 m. The ball travels once around the circle in 0.388 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, by what factor does the centripetal force increase
Answer:
0.435 N
The centripetal force increases by a factor of 4
Explanation:
Let the centripetal force be F
F =mv^2/r
m= mass of the object
v = linear velocity of the object
r = radius of the circular path
but v = 2πr/T = 2 * 3.142 * 0.147 / 0.388
v = 2.38 m/s
F = 0.0113 * (2.38)^2/0.147
F= 0.435 N
if v= 2v
Then;
F = m(2v)^2/r
F = m4v^2/r
F= 4mv^2/r
The centripetal force increases by a factor of 4
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.
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.
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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.
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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:
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:
A Impulse = – 25 Ns
B. Time = 5 s
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 5 Kg
Initial velocity (u) = 8 m/s
Final velocity (v) = 3 m/s
Impulse (I) =?
Time (t) =?
A. Determination of the Impulse.
Mass (M) = 5 Kg
Initial velocity (u) = 8 m/s
Final velocity (v) = 3 m/s
Impulse (I) =?
I = Ft = M(v – u)
I = M(v – u)
I = 5 (3 – 8)
I = 5 × – 5
I = – 25 Ns
NOTE: the negative sign indicates that the net force is acting in the negative direction.
B. Determination of the time.
Impulse (I) = 25 Ns
Force (F) = 5 N
Time (t) =?
I = Ft
25 = 5 × t
Divide both side by 5
t = 25 / 5
t = 5 s
Thus, it will take 5 s for the box to slide through the 15 m long ramp.
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:
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.
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.
Exercise 3 In an equation y mx+c; y and x have dimensions of length and c is constant. What are the dimensions of m'
A. Mass
B. Length
C. Time
D. m is Dimensionless
Answer:
D. m is Dimensionless
Explanation:
The equation of a straight line is given as:
y = mx + c
Dimension of y = l
x = l
c has no dimension
So;
if we do a dimensional analysis:
L = m L + 0
m = 1
So, m has no dimension
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
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.
You have to deliver some 5.0-kg packages from your home to two locations. You drive for 2.0 h at 30 mi/h due east (call this segment 1 of your trip), then turn around and drive due west for 30 min at 28 mi/h (segment 2). Use a coordinate system with the positive x axis aimed toward the east and the origin at your home.
A. What is your position vector at the instant you reach the end of segment 1?
B. What is your position vector at the instant you reach the end of segment 2?
C. Calculate your displacement during segment 2.
Answer:
A. 60 mi.
B. 46 mi.
C. -14 mi.
Explanation:
A)
Assuming that we use a coordinate system with the positive x-axis aimed toward the east, and the origin at the start point, we can find the position vector at the end of the segment 1, applying the definition of average velocity, as follows:[tex]v_{avg1} = \frac{x_{1f} -x_{1o}}{t} (1)[/tex]
where x₁f = final position at the end of segment 1, x₁₀ = initial position at the start of the segment 1 = 0, t = time traveled during segment 1 = 2.0 h, and v₁, average velocity during segment 1 = 30 mi/h due east.Replacing by the givens, and solving for x₁f, we get:[tex]x_{1f} = v_{1avg} * t =30 mi/h * 2.0 h = 60 mi due east. (2)[/tex]
B)
In order to find the position at the end of the segment 2, we can use the same equation (1), but taking into account that the initial position will not be zero, but the final position at the end of the segment 1, i.e., 60 mi due east.Replacing by the givens, and solving for x₂f, we get:[tex]x_{2f} = x_{1f} + v_{2avg} * t = (-28 mi/h) * 0.5 h = 60 mi - 14 mi = 46 mi (3)[/tex]
C)
The displacement during the segment two, is simply the difference between the final and initial positions for this segment.Since x₂₀= x₁f = 60 mi, and x₂f = 46 mi, we find that the displacement is as follows:Δx = x₂f - x₂₀ = 46 mi - 60 mi = - 14 mi.a car traveled at 100 km h for 3 hours how far did it travel
100 km
300 km/h
300 km
3 km/h
Answer:
300km
Explanation:
Given parameters:
Speed of the car = 100km/h
Time taken for the travel = 3hrs
Unknown:
How far did it travel = ?
Solution:
To solve this problem, we must understand that;
Speed = [tex]\frac{distance}{time}[/tex]
Distance = speed x time
Distance = 100km/hr x 3hr = 300km
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 plane drops a package for delivery. The plane is flying horizontally at a speed of 120\,\dfrac{\text m}{\text s}120 s m 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, and the package travels 255\,\text m255m255, start text, m, end text horizontally during the drop. We can ignore air resistance.
Answer:
-22.1
Explanation:
1 / 4
Step 1. List horizontal (xxx) and vertical (yyy) variables
xxx-direction yyy-direction
t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text
a_x=0a
x
=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a
y
=−9.8
s
2
m
a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction
\Delta x=255\,\text mΔx=255mdelta, x, equals, 255, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text
v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=?v
y
=?v, start subscript, y, end subscript, equals, question mark
v_{0x}=120\,\dfrac{\text m}{\text s}v
0x
=120
s
m
v, start subscript, 0, x, end subscript, equals, 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v
0y
=0v, start subscript, 0, y, end subscript, equals, 0
Note that there is no horizontal acceleration, so v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript. The time is the same for the xxx and yyy directions.
Also, the package has no initial vertical velocity.
Our yyy-direction variable list has too many unknowns to solve for \Delta yΔydelta, y directly. Since both the yyy- and xxx-directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:
\Delta x=v_xtΔx=v
x
tdelta, x, equals, v, start subscript, x, end subscript, t
Once we know ttt, we can solve for \Delta yΔydelta, y using the kinematic equation that does not include the unknown variable v_yv
y
v, start subscript, y, end subscript:
\Delta y=v_{0y}t+\dfrac {1}{2}a_yt^2Δy=v
0y
t+
2
1
a
y
t
2
delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared
Hint #22 / 4
Step 2. Find ttt from horizontal variables
\begin{aligned}\Delta x&=v_xt \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ t&=\dfrac{255\,\text m}{120\dfrac{\text m}{\text s}} \\\\ &=2.125\,\text s \end{aligned}
Δx
t
t
=v
x
t
=
v
0x
Δx
=
120
s
m
255m
=2.125s
Hint #33 / 4
Step 3. Find \Delta yΔydelta, y using ttt
Using ttt to solve for \Delta yΔydelta, y gives:
\begin{aligned}\Delta y&=v_{0y}t+\dfrac{1}{2}a_yt^2 \\\\ &=\cancel{ (0 )t}+\dfrac{1}{2}\left (-9.8\dfrac{\text m}{\text s^2}\right )\left(2.125\,\text s\right)^2 \\\\ &=-22.1\,\text m \end{aligned}
Δy
=v
0y
t+
2
1
a
y
t
2
=
(0)t
+
2
1
(−9.8
s
2
m
)(2.125s)
2
=−22.1m
Hint #44 / 4
The correct answer is -22.1\,\text m−22.1mminus, 22, point, 1, start text, m, end text.
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]
You want to build a snowman, so you accelerate a 2kg snowball across your yard at a rate of 0.5m/s2. Calculate the amount of force you applied to your friend.
Answer:
4
Units:
Newtons
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.
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
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°
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.
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]
A projectile is launched with an initial velocity of
200 meters per second at an angle of 30° above the
horizontal. What is the magnitude of the vertical
component of the projectile's initial velocity by?
(1) 200 m/s x cos 30°
(2) 200 m/s X sin 30°
(3) (200 m/s)/(cos 30 °)
(4) (200 m/s)/(sin 30 °)
The magnitude of the vertical component of the projectile's initial velocity is 200 m/s × sin 30°.
The diagrammatic representation of the velocity of the projectile can be seen in the attached image below.
From the diagram, let consider the ΔOAP where Vector OP makes an ∠θ = 30° to the horizontal x-axis.
where;
|OP| = magnitude of projectile velocity|OA| = magnitude of the horizontal component|OB|/|AP| = vertical component of the projectile∴
Using trigonometric approach for ΔOAP;
[tex]\mathbf{sin\theta = \dfrac{AP}{OP}}[/tex]
[tex]\mathbf{AP =OP\times sin \theta}}[/tex]
AP = 200 × sin 30°
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Padded dashboards in cars are safer in an accident than nonpadded ones because an occupant hitting the dash has:________.
(a) increased time of impact.
(b) decreased impulse.
(c) decreased impact force.
(d) a and b above.
(e) a and c above.
Answer:
(e) a and c above.
Explanation:
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = Mass * Velocity [/tex]
The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.
Padded dashboards in cars are safer in an accident than non-padded ones because an occupant hitting the dashboard of an automobile car has an increased time of impact and a decreased impact force because the force or shock experienced is high and happens rapidly over a short period of time, thus, the occupant has less time and velocity while absorbing the momentum of the car in the course of the collision.
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.
A dog sees a flowerpot sail up and then back past a window H high. If the total time the pot is in sight is t seconds, find the height above the window that the pot rises. (Let H = 2 m and t = 1.0 s, find the height above the window-- after you have found an algebraic solution.)
Answer:
maximum height = 0.1225 m
Explanation:
given data
H = 2m
t = 1 sec
solution
we consider here velocity of pot at lower side is u
and final velocity is v with acceleration a
time take is t/2
so
height h = u × [tex]\frac{t}{2}[/tex] - 0.5 × g × [tex](\frac{t}{2})^2[/tex] .................1
here
u = [tex]\frac{2}{t} \times ( h + \frac{gt^2}{8} )[/tex]
and
v = u +a t/2 .........................2
v = u + g t/2
v = [tex]\frac{2h}{t} + \frac{gt}{4} - \frac{gt}{2}[/tex]
v = [tex]\frac{2h}{t} - \frac{gt}{4}[/tex]
so that
maximum height is = [tex]\frac{v^2}{2g}[/tex]
maximum height = [tex]\frac{(\frac{2h}{t} - \frac{gt}{4})^2}{2g}[/tex]
put here value of h and t
maximum height = [tex]\frac{(\frac{2(2)}{1} - \frac{g(1)}{4})^2}{2g}[/tex]
maximum height = 0.1225 m
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].