Answers
Related Questions
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
Answers
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.
When measuring espresso for a drink, which instrument would give the
greatest precision?
Answers
How many mL is an espresso?
One shot of espresso is generally about 30–50 ml (1–1.75 oz), and contains about 63 mg of caffeine (3). Important point: The “golden ratio” for espresso is this: a single shot is 30 to 44 mL (1 to 1.5 ounces) of water and 7 grams of coffee
A uniform-density 7 kg disk of radius 0.20 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 42 N through a distance of 0.9 m. Now what is the angular speed
Answers
Answer:
The angular speed is 23.24 rad/s.
Explanation:
Given;
mass of the disk, m = 7 kg
radius of the disk, r = 0.2 m
applied force, F = 42 N
distance moved by disk, d = 0.9 m
The torque experienced by the disk is calculated as follows;
τ = F x d = I x α
where;
I is the moment of inertia of the disk = ¹/₂mr²
α is the angular acceleration
F x r = ¹/₂mr² x α
The angular acceleration is calculated as;
[tex]\alpha = \frac{2Fr}{mr^2} \\\\\ \alpha = \frac{2F}{mr}\\\\\alpha = \frac{2 \times 42 }{7 \times 0.2} \\\\\alpha = 60 \ rad/s^2[/tex]
The angular speed is determined by applying the following kinematic equation;
[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta[/tex]
initial angular speed, ωi = 0
angular distance, θ = d/r = 0.9/0.2 = 4.5 rad
[tex]\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta} \\\\\omega _f = \sqrt{2 \times 60 \times 4.5} \\\\\omega _f = 23.24 \ rad/s[/tex]
Therefore, the angular speed is 23.24 rad/s.
The mass of the Moon is 7.3x1022 kg and its radius is 1738 km. What is the strength of the gravitational field on the
surface of the Moon? (Do all required steps)
Answers
Answer:
1.61 N/kg
Explanation:
Take the universal gravitational constant G as 6.67 × 10^(-11) Nm²/kg²
The required gravitational field strength
= 6.67 × 10^(-11) × 7.3 × 10^(22) / (1738000)²
= 1.61194103 N/kg
= 1.61 N/kg (corr. to 3 sig. fig.)
Four rods that obey Hooke's law are each put under tension.
a. A rod 50.0 cm long with cross-sectional area 1.00 mm2 and with a 200 N force applied on each end.
b. A rod 25.0 cm long with cross-sectional area 1.00 mm2 and with a 200 N force applied on each end.
c. A rod 20.0 cm long with cross-sectional area 2.00 mm2 and with a 100 N force applied on each end. Order the rods according to the tensile stress on each rod, from smallest to largest.
Order the rods according to the tensile stress on each rod, from smallest to largest.
1. c < b < a
2. a=b
3. a< b < c
4. c
Answers
Answer:
c < a<=b
Explanation:
Tensile stress = (force) /Area
for A:
Tensile stress = 200/1 =200N/mm²
For B:
Tensile stress = 200/1 =200N/mm2
For C:
Tensile stress = 100/2 =50N/mm²
Ranking from smallest to largest we have;
C<A<=B which is option 4
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:
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.
A snail traveled 3.12 meters in 27.13 minutes. What is the snails speed in meters per second
Answers
Answer:
0.002 m/s
Explanation:
27.13(60) = 1,627.8 seconds
3.12/1,627.8 = 0.00191 ≈ 0.002 = s
Are you sure you're not looking for cm/s?
TRUE OR FALSE
2 QUESTIONS
please HELP ASAP
Answers
1. false
2. true
I hope this helps ^-^
True
please mark brainliest i hope this helped!
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
What would happen if the molecules in a sample moving entirely ?
Answers
Answer:
Molecular scale. The story begins a long time ago
when the idea that molecules are in constant motion
was first discovered. Part of the evidence that you can
see in everyday life was discovered by Robert Brown
about 150 years ago when he used a microscope to
watch how tiny dust particles move.
So how fast do molecules move? It all depends upon
the molecule and its state: molecules in a solid state
move slower than in a liquid state, and much slower
than gas molecules. One estimate puts gas molecules
in the range of 1,100 mph at room temperature. Cool
them down to almost absolute zero and they slow
down to less than 0.1 mph (slower than the average
couch potato). The fact that they are always moving
makes it a challenge to see molecules and make stuff
out of them, but it’s a challenge that scientists
work hard to figure out.
Explanation:
Is there a way to see moon and the sun at once?
Answers
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
Answers
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]
Would sound travel faster in an oven or a freezer?
Answers
Answer:
An Oven
Explanation:
The heat is higher, so it moves faster. Shile in a freezer the particles are extremely slow!
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%
an inventor makes a clock using a brass rod and a heavy mass as a pendulum.WHAT Happens when the clock get colder?
Answers
An inventor makes a clock using a brass rod and a heavy mass as a pendulum. when the clock gets colder then the time clock would gain time
What is thermal expansion?The expansion of any material due to the variation of the temperature is known as thermal expansion. It varied differently for different materials according to their corresponding values of the coefficient of the thermal expansion.
As given in the problem statement that an inventor makes a clock using a brass rod and a heavy mass as a pendulum, when the clock gets colder then the length of the brass decreases due to thermal expansion.
The length of the pendulum gets reduced which further results in the reduction in the time period, as per the formula of the time period for the pendulum
T = 2π√(L/g)
As the length of the brass gets reduced. This means the pendulum of the clock moves faster and the clock would gain time
Thus, if a pendulum made of a heavy mass and a brass rod is used to create a clock by an inventor. The time clock would advance in time as the clock get colder
Learn more about Thermal expansion here
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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.
When6-2 He He-6 undergoes beta decay, the daughter is?
Answers
Answer: The daughter is named Susie.
Explanation: LIL SUSIE!!!
HUH? DIDN'T UNDERSTAND THE QUESTION!
HAVE A GREAT DAY!!!!!
Answer:6/3 Li
Explanation:
I’m not sure what the person under me is talking about but yeah
(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]
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.
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]
Water enter the horizontal, circular cross-sectional, sudden-contraction nozzle sketched below at section (1) with a uniformly distributed velocity of 30 ft/s and a pressure of 80 psi. The water exits from the nozzle into the atmosphere at section (2) where the uniformly distributed velocity is 100 ft/s. Determinethe axial component of the anchoring force required to hold the contraction in place.
Answers
This question is incomplete, the missing image is uploaded along this answer.
Answer:
the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
Explanation:
Given the data in the question and as illustrated in the image below;
first we calculate the area at section 1
A₁ = (πD²)/4
we substitute
A₁ = (π(3 in)²)/4
A₁ = 7.06858 in²
we know that; 1 ft = 12 in
A₁ = ( 7.06858 / (12²) ) ft²
A₁ = ( 7.06858 / 144 ) ft²
A₁ = 0.0491 ft²
now, we write the elation for area at section 2
A₂ = πd²/4
here, d is the diameter at section 2
next, we use the conservation of mass equation between the two section;
m" = pV₁A₁ = pV₂A₂
we calculate the mass flow rate;
m" = pV₁A₁
= (1.94[tex]\frac{slug}{ft^2}[/tex]) × 30[tex]\frac{ft}{s}[/tex] × 0.0491 ft²
= 2.8576 slug/s
Now, Apply the linear momentum along the horizontal direction for the control volume between 1 - 2
-pV₁A₁V₁ = pV₂A₂V₂ = P₁A₁ - F[tex]_A[/tex] - P₂A₂
m"( V₂ - V₁ ) = P₁A₁ - F[tex]_A[/tex] - P₂A₂
F[tex]_A[/tex] = P₁A₁ - P₂A₂ - m"( V₂ - V₁ )
we substitute
F[tex]_A[/tex] = ((80×[tex]\frac{144 in^2}{1 ft^2}[/tex])×0.0491 ft²) - (0×(πd²/4)) - 2.8576( 100 - 30 )ft/s
F[tex]_A[/tex] = 565.632 - 0 - 200.032
F[tex]_A[/tex] = 565.632 - 200.032
F[tex]_A[/tex] = 365.6 lb
Therefore, the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
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:
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A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red light of wavelength 635 nm appear green to him, with a wavelength of 550 nm. The police officer writes out a traffic citation for speeding. How fast was the physicist traveling, according to his own testimony
Answers
Answer:
Explanation:
[tex]\lambda[/tex] = Observed wavelength = 550 nm
[tex]\lambda'[/tex] = Actual wavelength = 635 nm
c = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
v = Velocity of the physicist
Doppler shift is given by
[tex]f=\sqrt{\dfrac{c+v}{c-v}}f'\\\Rightarrow \dfrac{c}{\lambda}=\sqrt{\dfrac{c+v}{c-v}}\dfrac{c}{\lambda'}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}=\dfrac{c+v}{c-v}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}=\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}(1-\dfrac{v}{c})=1+\dfrac{v}{c}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}(1-\dfrac{v}{c})=1+\dfrac{v}{c}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}-1=\dfrac{v}{c}(1+\dfrac{\lambda'^2}{\lambda^2})\\\Rightarrow v=\dfrac{c(\dfrac{\lambda'^2}{\lambda^2}-1)}{1+\dfrac{\lambda'^2}{\lambda^2}}[/tex]
[tex]\Rightarrow v=\dfrac{3\times 10^8\times (\dfrac{635^2}{550^2}-1)}{1+\dfrac{635^2}{550^2}}\\\Rightarrow v=42817669.77\ \text{m/s}[/tex]
The physicist was traveling at a velocity of [tex]42817669.77\ \text{m/s}[/tex].
3
How does the electrical conductivity of metals
and metalloids change with an increase in
temperature?
Answers
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:
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].
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 3.0-kilogram object initially at rest explodes and splits into three fragments. One fragment has a mass of 0.50 kg and flies off along the negative x axis at a speed of 2.8 m/s, and another has a mass of 1.3 kg and flies off along the negative y axis at a speed of 1.5 m/s.
Required:
What are the speed and direction of the third fragment?
Answers
Answer:
The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.
Explanation:
The object is not under the influence of any external force, meaning that Principle of Momentum Conservation to calculation of the velocity of the third fragment:
[tex](m_{1}+m_{2}+m_{3})\cdot \vec {v}_{o} = m_{1}\cdot \vec v_{1} + m_{2}\cdot \vec v_{2} + m_{3}\cdot \vec v_{3}[/tex] (1)
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex] - Masses of the first, second and third fragments, in kilograms.
[tex]\vec v_{o}[/tex] - Initial velocity of the object, in meters per second.
[tex]\vec v_{1}[/tex], [tex]\vec v_{2}[/tex], [tex]\vec v_{3}[/tex] - Velocities of the first, second and third fragments, in meters per second.
If we know that [tex]m_{1} = 0.5\,kg[/tex], [tex]m_{2} = 1.3\,kg[/tex], [tex]m_{3} = 1.2\,kg[/tex], [tex]\vec v_{o} = (0,0)\,\left[\frac{m}{s} \right][/tex], [tex]\vec v_{1} = \left(-2.8, 0\right)\,\left[\frac{m}{s} \right][/tex] and [tex]\vec v_{2} = \left(0,-1.5\right)\,\left[\frac{m}{s} \right][/tex], the velocity of the third fragment is:
[tex](-1.4,0) + (0,-1.95) + 1.2\cdot \vec v_{3} = (0,0)[/tex]
[tex]1.2\cdot \vec v_{3} = (1.4,1.95)[/tex]
[tex]\vec v_{3} = (1.167, 1.625)\,\left[\frac{m}{s} \right][/tex]
The speed of the third fragment is the magnitude of the result found above:
[tex]v_{3} = 2\,\frac{m}{s}[/tex]
And the direction of the third fragment is:
[tex]\theta_{3} = \tan^{-1} \left(\frac{1.625}{1.167}\right)[/tex]
[tex]\theta_{3} \approx 54.316^{\circ}[/tex]
The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.
३.रात में घूमने वाला write one word substitute
Answers
Explanation:
रात में घूमने वाला arthaarat निशाचर
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
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