Answers
Answer:
Explanation:
you can say the law of superpoition can tell us that each rock layer is older than the one above it. So, the relative age of the rock or fossil in the rock or fossil in the rock is older if it is farther down in the rock layers. hope it helps
Related Questions
Mechanical energy is the most concentrated form of energy.
a. true
b. false
Answers
On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the cart begins to experience an average frictional force of 15N throughout the ride.
Find:
a) The total energy at (A)
b) The velocity at (B)
c) The velocity at (C)
d) Can the cart make it to Point (D)? Why or why not?
Answers
When grip strength increases:
a. action potential voltage increases.
b. action potential frequency decreases.
c. action potential frequency increases.
d. action potential frequency increases.
e. the number of active motor units increases.
Answers
Answer:
e. the number of active motor units increases.
Explanation:
There is a direct relationship between the number of active motor units and the grip strength in a given scenario. For example, increase in the grip strength leads to increase in the number of active motor units. In the other-hand, the decrease in grip strength leads to the decrease in the number of active motor units.
In this experiment, you will use a track and a toy car to explore the concept of movement. You will measure the time it takes the car to travel certain distances, and then complete some calculations. In the space below, write a scientific question that you will answer by doing this experiment.
Answers
Answer: if weight affects how fast they go?
Explanation:
Answer:
How can we change the speed of a toy car on a racetrack to describe the car’s motion?
Explanation:
thats the sample respond
calculate the resistance if a potential difference of 3v exists when a current of 2A flows through the conductor
Answers
Answer:
Resistance = 1.5 ohms
Explanation:
Given:
Potential difference = 3v
Current flow = 2 A
Find:
Resistance
Computation:
Resistance = Potential difference / Current flow
Resistance = 3 / 2
Resistance = 1.5 ohms
In high air pressure the molecules are
A-Warm and moving fast
b-Close together and moving slowly
c-far apart and moving slowly
d-hot and moving rapidly
Answers
A car's initial speed of 15m/s is running with the acceleration of 32m/
s2 in 8 seconds. What is the car's final velocity?
Answers
Explanation:
15m/s
acceleration= (+)
so, 15m/s +32m/s=47m/s
42m/s. X 8 = 336
Select the correct answer.
Why are the requirements to become a congressional representative and a senator different?
A.
The framers didn’t want senators to worry about their constituents as often as representatives do.
B.
The framers gave senators more lawmaking powers and responsibilities than representatives.
C.
The framers wanted senators to be more mature and more experienced than representatives.
D.
The framers thought senators and representatives should have reasons to argue with each other.
Answers
The framers wanted senators to be more mature and more experienced than representatives. (C)
Answer:
C -> The framers wanted senators to be more mature and more experienced than representatives.
which one is odd copper,plastic,rubber
Answers
Answer:
It's plastic.
trust me it's plastic, i've rad it somewhere.
All of them have something that's not like the others.
-- Rubber is the only one on the list that has two repeated letters.
-- Plastic is the only one on the list thagt has no repeated letters.
-- Plastic is the only one on the list that has no 'r' in its name.
-- Copper is the only one on the list that is an element, not a compound.
-- Copper is the only good electrical conductor on the list.
-- Plastic is the only one on the list with more than six letters in its name.
-- Rubber is the only one on the list with no 'p' in its name.
-- Plastic is the only one on the list that doesn't end in "-er".
How fast were both runners traveling after 4 seconds?
40
Distance (in yards)
30
20
10
1
2.
3
0
Time in seconds
Answers
Answer:
they were fast ⛷⛷
A student's backpack has a mass of 9.6 kg. The student applies a force of 94.08 N [up] while walking through 1.4 km [E] to get to school. Calculate the work done by the student on the backpack
Answers
The student does zero work on the backpack because the upward force applied by the student is acting perpendicular to the backpack's displacement parallel to the ground.
A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. (a) Calculate the rate of heat loss per unit length for a calm day. (b) Calculate the rate of heat loss on a breezy day when the wind speed is 8
Answers
Answer:
Heat loss per unit length = 642.358 W/m
The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m
Explanation:
From the information given:
Diameter D [tex]= 100 mm = 0.1 m[/tex]
Surface emissivity ε = 0.8
Temperature of steam [tex]T_s[/tex] = 150° C = 423K
Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]
Velocity of wind V = 8 m/s
To calculate average film temperature:
[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]
[tex]T_f = \dfrac{423+293}{2}[/tex]
[tex]T_f = \dfrac{716}{2}[/tex]
[tex]T_f = 358 \ K[/tex]
To calculate volume expansion coefficient
[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]
From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;
Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s
Thermal conductivity k = 30.608 × 10⁻³ W/m.K
Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s
Prandtl no. Pr = 0.698
Rayleigh No. for the steam line is determined as follows:
[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]
[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]
[tex]Ra_{D} = 5.224 \times 10^6[/tex]
The average Nusselt number is:
[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]
[tex]Nu_D = 23.29[/tex]
However, for the heat transfer coefficient; we have:
[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]
[tex]h_D = 7.129 \ Wm^2 .K[/tex]
Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]
Now;
To determine the heat loss using the formula:
[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]
[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]
Now; here we need to determine the Reynold no and the average Nusselt number:
[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]
However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;
[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]
[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]
[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]
SO, the heat transfer coefficient for forced convection is determined as follows afterward:
[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]
Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:
[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]
A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of 0.64 m and a mass of 0.48 kg. The rod has a length of 1.78 m and a mass of 0.50 kg. The rod is placed on a fulcrum (pivot) at X = 0.34 m from the left end of the rod.
(a) Calculate the moment of inertia (click for graphical table) of the contraption around the fulcrum. kg m2
(b) Calculate the torque about the fulcrum, using CCW as positive. N.m
(c) Calculate the angular acceleration of the contraption, using CCW as positive. rad/s2
(d) Calculate the linear acceleration of the right end of the rod, using up as positive. m/s2
Answers
The image of this hollow sphere and uniform rod is missing, so i have attached it.
Answer:
A) J = 0.7443 kg•m²
B) T = 1.9169 N•m CCW
C) α = 2.5754 rad/s²
D) a = 3.966 m/s²
Explanation:
A) The moment of inertia J of the contraption around the fulcrum is given by the formula;
J = Jℓ + Jr
Let's calculate Jℓ
Jℓ = [((0.34²/3) × 0.50 × 0.34)/1.78] + (0.48 × (0.34 + 0.64)²)
Jℓ = 0.4647 kg•m²
Now, let's Calculate Jr
Jr = ((1.78 - 0.34)²/3) × ((1.78 - 0.34)/1.78) × 0.50
Jr = 0.2796 kg•m²
Thus;
J = 0.4647 + 0.2796
J = 0.7443 kg•m²
(b) Using CCW as positive, Torque in Nm is calculated as;
T = Tℓ - Tr
Let's calculate Tℓ
Tℓ = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81
Tℓ = 4.7739 N•m CCW
Now, let's Calculate Tr;
Tr = [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81
Tr = 2.857 N•m CW
Thus;
T = 4.7739 - 2.857
T = 1.9169 N•m CCW
(c) The angular acceleration α of the contraption, using CCW is gotten from the formula;
α = T/J
α = 1.9169/0.7443
α = 2.5754 rad/s²
(d) The linear acceleration a of the right end of the rod, using up as positive is given by;
a = α*(1.78 - 0.34)
a = 2.5754 × 1.54
a = 3.966 m/s²
A) the moment of inertia of the contraption is 0.7443 kgm²
B) The torque about the fulcrum is 1.9169 Nm
C) Angular acceleration of the contraption is 2.5754 rad/s²
D) The linear acceleration of the contraption is 3.966 m/s²
Moment of inertia:(A) The moment of inertia I of the contraption around the fulcrum is given by :
[tex]I = [(0.34^2/3) \times 0.50 \times 0.34)/1.78 + (0.48 \times (0.34 + 0.64)^2)] + [(1.78 - 0.34)^2/3) \times (1.78 - 0.34)/1.78) \times 0.50][/tex]
I = 0.4647 + 0.2796
I = 0.7443 kgm²
(B) Using CCW as positive, Torque in Nm is given by;
T = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81 - [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81
T = 4.7739 - 2.857
T = 1.9169 Nm
(C) The angular acceleration (α) of the contraption is given by:
α = T/I
since, torque is defined as T = Iα
α = 1.9169/0.7443
α = 2.5754 rad/s²
(D) The linear acceleration (a) of the right end of the rod
a = αr
where r is the distance from the pivot
a = α × (1.78 - 0.34)
a = 2.5754 × 1.54
a = 3.966 m/s²
Learn more about moment of inertia:
https://brainly.com/question/6953943?referrer=searchResults
What is it called when the moon passes through the penumbra of Earth’s shadow?(1 point)
total lunar eclipse
total solar eclipse
partial lunar eclipse
partial solar eclipse
Answers
Answer: I'm not sure, but I think it would be a total lunar eclipse
When the moon passes through the penumbra of Earth’s shadow it is referred to as partial lunar eclipse. The correct option is C.
What is partial lunar eclipse?A partial lunar eclipse occurs when the moon is not completely immersed in the umbra of the earth's shadow.
During a partial solar eclipse, the Moon, Sun, and Earth do not align perfectly straight, and the Moon casts only the penumbra of its shadow on Earth. From our vantage point, it appears that the Moon has eaten the Sun.
A shadow's penumbra is the lighter outer edge. Partial solar eclipses are caused by the Moon's penumbra, while penumbral lunar eclipses are caused by the Earth's penumbra. The penumbra is a type of lighter shadow.
The Penumbra is a half-shadow region that occurs when an object only partially covers a light source.
Thus, the correct option is C.
For more details regarding lunar eclipse, visit:
https://brainly.com/question/28679608
#SPJ2
A 50 N force causes a spring to compress 0.09 m. What is the spring constant? What is the potential energy of the spring?
Answers
Resolution
using hooke's relation
F = K . d
50N = k . 0.09m
k = 50N / 0.09m
k = 5555.56 N/m
Calculating the potential energy of the spring
Ep = 1/2 k . x²
Ep = 1/2 (5555.56 N/m) (0.09m)²
Ep = 22.5 Joules
Answer
the spring constant? =
k = 5555.56 N/m
potential energy of the spring?
Ep = 22.5 Joules
The Potential energy of the spring is 2.25 J
What is the Potential energy of spring?This is the energy stored in spring due to its elastic properties.
To calculate the potential energy of the spring, we use the formula below.
Formula:
E = Fe/2................ Equation 1Where:
E = Potential energy of the springF = Force applied to the springe = compression.From the question,
Given:
F = 50 Ne = 0.09 mSubstitute these values into equation 1
E = 50(0.09)/2E = 2.25 J.Hence, The Potential energy of the spring is 2.25 J
Learn more about the potential energy of spring: https://brainly.com/question/3510901
Define Mechanical advantage
fe effort of 2125N is used to lift a Lead of 500N
through a Verticle high of 2.N using a buly System
if the distance Moved by the effort is 45m
Calculate 1. Work done on the load
2. work done by the effort
3. Efficiency of the System
Answers
Answer:
1) 1000Nm
2) 95,625Nm
3) 1.05%
Explanation:
Mechanical Advantage is the ratio of the load to the effort applied to an object.
MA = Load/Effort
1) Workdone on the load = Force(Load) * distance covered by the load
Workdone on the load = 500N * 2m
Workdone on the load = 1000Nm
2) work done by the effort = Effort * distance moves d by effort
work done by the effort = 2125 * 45
work done by the effort = 95,625Nm
3) Efficiency = Workdone on the load/ work done by the effort * 100
Efficiency = 1000/95625 * 100
Efficiency = 1.05%
Hence the efficiency of the system is 1.05%
The gravitational potential energy of an object is defined as the energy it has due to its position in a gravitational field. A ball with a weight of 50 N is lifted to a height of 1 meter. Which graph correctly represents the change in gravitational potential energy (shaded in blue) as it is lifted to this height?
Answers
Answer:
athletic
Explanation:
because internet system has been down since we were in few days
What is the weight of a 48kg rock?
Answers
Answer:
48kg
Explanation:
why doesn't a radio operating with two batteries function when one of the batteries is reversed?
Answers
Answer:
If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.
Explanation:
Sam moves an 800 N wheelbarrow 5 meters in 15 seconds. How much work did he do?
Answers
Answer:
work done= force × displacement
=800×5
=4000J
Explanation:
The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.
How much work is done when 100 N of force is applied to a rock to move it 20 m
Answers
Answer: 2000 J
Explanation: work W = F s
(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroid, travelling at a speed of 3 m/s. The payload has a mass of 29 kg. If the spring is to be compressed initially an amount of 1.4 m, what stiffness ks must the spring be designed to have
Answers
Answer:
ks= 133.2 N/m
Explanation:
Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.So, we can write the following equation:[tex]\Delta U + \Delta K = 0 (1)[/tex]
where ΔU = -1/2*k*(Δx)² (2)and ΔK = 1/2*m*v² (3)Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:[tex]k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)[/tex]
1. What types of natural phenomena could serve as time standards?
Answers
Answer:
The movement of Sun and moon
Explanation:
When the sun rise.it is am and when it sets .it is pm.
Calculate the kinetic energy of an 80,000 kg airplane that is flying with a velocity of 167 m/s.
Answers
Answer:
1115560000 J
Explanation:
1/2 * 80,000 * 167^2 m/s = 1115560000 J
३.रात में घूमने वाला write one word substitute
Answers
Explanation:
रात में घूमने वाला arthaarat निशाचर
An object is accelerated by a net force in which direction?
A. at an angle to the force
B. in the direction of the force
C. in the direction opposite to the force
D. Any of these is possible.
Answers
Answer:
B. in the direction of the force
Explanation:
Sana nakatulong
At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?
Answers
Answer:
8.45 m
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 90 Kg
Initial velocity (u) = 13 m/s
Final velocity (v) = 0 m/s
Height (h) =?
NOTE: Acceleration due to gravity (g) = 10 m/s²
The height of the hill can be obtained as follow:
v² = u² – 2gh (since the cart is going against gravity)
0² = 13² – (2 × 10 × h)
0 = 169 – 20h
Rearrange
20h = 169
Divide both side by 20
h = 169/20
h = 8.45 m
Therefore, the height of the hill is 8.45 m
How much energy would be required to move the earth into a circular orbit with a radius 2.0 kmkm larger than its current radius
Answers
Answer:
[tex]3.52\times 10^{25}\ \text{J}[/tex]
Explanation:
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]
m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]
[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]
[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]
Energy required is given by
[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]
The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].
Is there a way to see moon and the sun at once?
Answers
a sharp image is formed when light reflects from a
Answers
Answer:
Regular reflection
Explanation:
Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.
i hope this helps a bit.
According to the context, a sharp image is formed when light reflects from a regular reflection.
What is regular reflection?It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.
This reflection of light happens when the angles that the two rays determine with the surface are equal.
Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.
Learn more about regular reflection here: https://brainly.com/question/3778324
#SPJ2
