Answer:
The magnification of an image is equal to the ratio of the image height to the object height.
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
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].
if the water measures -5 feet at low tide and 3ft at high tide what is the tidal range
Answer:
8 feet
................
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?
३.रात में घूमने वाला write one word substitute
Explanation:
रात में घूमने वाला arthaarat निशाचर
How much work is done when 100 N of force is applied to a rock to move it 20 m
Answer: 2000 J
Explanation: work W = F s
How fast were both runners traveling after 4 seconds?
40
Distance (in yards)
30
20
10
1
2.
3
0
Time in seconds
Answer:
they were fast ⛷⛷
Calculate the kinetic energy of an 80,000 kg airplane that is flying with a velocity of 167 m/s.
Answer:
1115560000 J
Explanation:
1/2 * 80,000 * 167^2 m/s = 1115560000 J
1. What types of natural phenomena could serve as time standards?
Answer:
The movement of Sun and moon
Explanation:
When the sun rise.it is am and when it sets .it is pm.
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
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
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
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%
3
How does the electrical conductivity of metals
and metalloids change with an increase in
temperature?
Answer:
In metals there are free electrons at normal temperature so when we increase temperature it resistivity gets increases,so conductivity decreases,while in semiconductor the electrons are not free so when we increase the temperature the covalent bonds begin to break and the electron becomes free so conductivity get.
Explanation:
Calculate the momentum of a 10 kg bowling ball rolling at 2m/s towards north.
Answer:
momentum=mass x velocity= 10 x 2 = 20kgm/s
A ball is sitting at the top of a ramp. As the ball rolls down the ramp, the potential energy of the ball decreases, what happens to the potential energy as the ball moves
Answer:
the potential energy decreases as it is converted to kinetic energy.
Explanation:
As things move, their potential energy converts to kinetic energy to power them along. When a ball rolls down the top of a ramp, all the potential energy it accumulated at the top of the ramp converts to kinetic energy to help it roll down. In other words, its potential energy decreases as its kinetic energy increases.
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?
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
(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
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]
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
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.
which causes magnets to stick to metal
Answer:
Steel
Explanation:
Steel is a metal that magnets stick to because iron can be found inside steel
Answer:Magnets stick to any metal that contains iron, cobalt or nickel.
Explanation:Iron is found in steel, so steel attracts a magnet and sticks to it. Stainless steel, however, does not attract a magnet.
Mechanical energy is the most concentrated form of energy.
a. true
b. false
The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. The water pressure in the line at the street is 130 psig. If I flow 10 gpm through the pipe, what pressure would I expect when I get to my house. My house is 10 ft higher in elevation than the water line at the street
Answer:
The right solution is "126 Psi".
Explanation:
The given values are:
P₁ = 130 psig
i.e.,
= [tex]130\times 6.894[/tex]
= [tex]896.22 \ Kpa[/tex]
or,
= [tex]896.22\times 10^3 \ Pa[/tex]
Z₂ = 10ft
= 3.05 m
[tex]\delta[/tex] = 1000 kg/m³
According to the question,
Z₁ = 0
V₁ = V₂
As we know,
⇒ [tex]\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2[/tex]
On substituting the values, we get
⇒ [tex]\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2[/tex]
⇒ [tex]\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05[/tex]
⇒ [tex]P_2=866330 \ P_a[/tex]
i.e.,
⇒ [tex]=866330\times 0.000145[/tex]
⇒ [tex]=126 \ Psi[/tex]
(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60m/s and is moving in a circle of radius 50m?
Answer:
5.0/s
Explanation:
Answer:
b and a it is this that abewsr
According to Newton's first law, an object at rest will _____.
never move
stay at rest forever
start moving
stay at rest unless moved by force
Sam moves an 800 N wheelbarrow 5 meters in 15 seconds. How much work did he do?
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.
Is there a way to see moon and the sun at once?
a sharp image is formed when light reflects from a
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
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5. How much heat is needed to warm .052 kg of gold from 30°C to 120°C? Note: Gold has a specific heat of 136
J/kg °C
Answer:
Q = 636.48 J
Explanation:
Given that,
The mass of gold, m = 0.052 kg
The temperature increase from 30°C to 120°C.
The specific heat of gold is 136 J/kg °C.
We need to find the heat needed to warm the gold. The formula for heat needed is given by :
[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]
So, 636.48 J of heat is needed to warm gold.
why doesn't a radio operating with two batteries function when one of the batteries is reversed?
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:
A reaction occurs when a compound breaks down. This reaction has one reactant and two or more products. Energy, as from a battery, is usually needed to break the compound apart.
Answer:
decomposition
Explanation:
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 cm from the axis of rotation at the knee. How much force must the knee extensors exert to produce an angular acceleration at the knee of 1 rad/s2 , given a mass of the lower leg and foot of 4.5 kg, and a radius of gyration of 23 cm
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
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
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
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]
