Answer:
[tex]E_0=0.173N/C[/tex]
Explanation:
From the question we are told that:
Power [tex]P=50kw=>50*10^3w[/tex]
Distance [tex]d=10km=10000m[/tex]
Generally the equation for Intensity is mathematically given by
[tex]I=\frac{P}{4\pi d^2} w/m^2[/tex]
[tex]I=\frac{50*10^3}{4 \pi 10000^2} w/m^2[/tex]
[tex]I=3.98*10^{-5}w/m^2[/tex]
Generally Intensity is also
[tex]I=\frac{1}{2}cE_0^2e[/tex]
Where
[tex]e=8.854*10^{-12}Nm^2/c^2[/tex]
Therefore
[tex]E_0=\sqrt{\frac{2I}{c *e}}[/tex]
[tex]E_0=\sqrt{\frac{2*3.98*10^{-5}}{3*10^8 *8.854*10^{-12}}}[/tex]
[tex]E_0=0.173N/C[/tex]
A cylindrical wire has a length of 2.80 m and a radius of 1.03 mm. It carries a current of current of 1.35 A, when a a voltage of 0.0320 V is applied across the ends of the wire. From this information calculate the resistance of the wire.
Answer:
0.023 Ohms
Explanation:
Given data
Length= 2.8m
radius= 1.03mm
current I= 1.35 A
voltage V= 0.032V
We know that from Ohm's law
V= IR
Now R= V/I
Substitute
R= 0.032/1.35
R= 0.023 Ohms
Hence the resistance is 0.023 Ohms
Definition of distance in physics
Answer:
Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
Explanation:
The standard unit of brightness is called the candela.
True
False
Answer:
TRUE
Explanation:
Cho hệ thống thùng lắc có mô hình tại vị trí đang xét như hình vẽ
Answer:
I can't understand this language!!!Answer:
vdhdbdnnsnsbdhhshzbhshsbbsbd is not ask you to be able and r in the exam qq and
A 3.0 kg block is pushed by a 14 N force. If µ = 0.6, will the block move?
Answer:
The block will not move.
Explanation:
We'll begin by calculating the frictional force. This can be obtained as follow:
Coefficient of friction (µ) = 0.6
Mass of block (m) = 3 Kg
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (R) = mg = 3 × 10 = 30 N
Frictional force (Fբ) =?
Fբ = µR
Fբ = 0.6 × 30
Fբ = 18 N
From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.
An object moving with a constant
acceleration changes its velocity from
10ms' to 20 ms' in five seconds. What is the
distance travelled in five seconds
Answer:
Acceleration:
[tex]{ \tt{a = \frac{v - u}{t} }} \\ { \tt{a = \frac{20 - 10}{5} }} \\ { \tt{a = 2 \: m {s}^{ - 2} }}[/tex]
From third equation:
[tex]{ \bf{ {v}^{2} = {u}^{2} + 2as}} \\ { \tt{s = \frac{ {20}^{2} - {10}^{2} }{2 \times 2} }} \\ = { \tt{s = 75 \: m}}[/tex]
Answer:
Formula = m/s
Explanation:
The answer is 10 m / 5 seconds = 2 meters distance
The answer is 20 m / 5 seconds = 4 meters distance
You are a detective investigating why someone was hit on the head by a falling flowerpot. One piece of evidence is a smartphone video taken in a 4th-floor apartment, which happened to capture the flowerpot as it fell past a window. In a span of 8 frames (captured at 30 frames per second), the flowerpot falls 0.84 of the height of the window. You visit the apartment and measure the window to be 1.27 m tall.
Required:
Assume the flowerpot was dropped from rest. How high above the window was the flowerpot when it was dropped?
Answer:
0.37 m
Explanation:
Given :
Window height, [tex]h_1[/tex] = 1.27 m
The flowerpot falls 0.84 m off the window height, i.e.
[tex]h_2[/tex] = (1.27 x 0.84 ) m in a time span of [tex]$t=\frac{8}{30}$[/tex] seconds.
Assuming that the speed of the pot just above the window is v then,
[tex]h_2=ut+\frac{1}{2}gt^2[/tex]
[tex]$(1.27 \times 0.84) = v \times \left( \frac{8}{30} \right) + \frac{1}{2} \times 9.81 \times \left( \frac{8}{30} \right)^2$[/tex]
[tex]$v=\left(\frac{30}{8}\right) \left[ (1.27 \times 0.84) - \left( \frac{1}{2} \times 9.81 \times \left( \frac{8}{30 \right)^2 \right) \right]}$[/tex]
[tex]$v= 2.69$[/tex] m/s
Initially the pot was dropped from rest. So, u = 0.
If it has fallen from a height of h above the window then,
[tex]$h = \frac{v^2}{2g}$[/tex]
[tex]$h = \frac{(2.69)^2}{2 \times 9.81}$[/tex]
h = 0.37 m
De que esta hecho el sol? plisss ayuda.no necesito un texto de 100 reglones, puede ser resumido en solo 2 renglones
Describing Uses ñ Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with her? Why or why not?
If a car drives 10 mph South, this is an example of a:
A. Displacement
B. Velocity
C. Speed
D. Distance
Answer:
杰杰伊杜杜杜伊格富尔杰迪耶赫分离福音
Explanation:
莱德利 · 赫耶尔伊 3uritievrirjrirhruebwkwieheoo2hfjcbvi3hd
Answer:
B velocity
Explanation:
A 2.50-W beam of light of wavelength 124 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.16 eV. Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts) of this metal
Answer:
φ = 13.43 x 10⁻¹⁹ J = 8.4 eV
Explanation:
Using the Einstein's Photoelectric equation:
Energy of Photon = Work Function + Kinetic Energy of Electron
[tex]\frac{hc}{\lambda} = \phi + K.E[/tex]
where,
h = Plank's Constant = 6.625 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 124 nm = 1.24 x 10⁻⁷ m
φ = work function = ?
K.E = Kinetic Energy of Electrons = (4.16 eV)([tex]\frac{1.6\ x\ 10^{-19}\ J}{1\ eV}[/tex]) = 2.6 x 10⁻¹⁹ J
Therefore,
[tex]\frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{1.24\ x\ 10^{-7}\ m} = \phi + 2.6\ x\ 10^{-19}\\\\\phi=16.03\ x\ 10^{-19}\ J - 2.6\ x\ 10^{-19}\ J[/tex]
φ = 13.43 x 10⁻¹⁹ J = 8.4 eV
A 3.1-mole sample of an ideal gas is gently heated at constant temperature 320 K. It expands from initial volume 23 L to final volume V2. A total of 1.7 kJ of heat is added during the expansion process. What is V2? Let the ideal-gas constant R = 8.314 J/(mol • K).
From the ideal gas law,
PV = nRT ==> P = nRT/V
where P is the pressure exerted by the gas on the container. The work W done by this pressure as the volume of the gas changes from V₁ to V₂ is given by the integral,
[tex]W = \displaystyle \int_{V_1}^{V_2}P\,\mathrm dV \implies W = nRT \ln\left(\dfrac{V_2}{V_1}\right)[/tex]
and solving for V₂ gives
[tex]V_2 = V_1\exp\left(\dfrac{W}{nRT}\right)[/tex]
If you add 1.7 kJ of heat to the system, which does the aforementioned work, the gas will expand to a volume of
[tex]V_2 = (23\,\mathrm L)\exp\left(\dfrac{1.7\,\mathrm{kJ}}{(3.1\,\mathrm{mol})\left(8.314\frac{\rm J}{\mathrm{mol}\cdot\mathrm K}\right)(320\,\mathrm K)}\right) \approx \boxed{28 \,\mathrm L}[/tex]
A T-shirt is launched at an angle of 30° with an initial velocity of 25 m/s how long does it take to reach the peak? How long is it in the air for totally?
Answer:
The launched angle θ = 30 degrees, the initial velocity Vo = 20 m/s, the initial horizontal velocity Vox= ?, the initial vertical velocity Voy = ?, the time of flight t = ? the maximum height h = ?
Vox = Vo * (cos of 30 degrees)
Voy = Vo * (sin of 30 degrees)
t = 2 * (Voy / g)
h = Voy * 0.5 t - 1/2 g * (0.5t)^2
I have given the equations for you to use, just plug – in the values and then solve in a step by step manner.
Answer:
approximately 15.68 meters.
Explanation:
Here is how;
First, let's calculate the time of flight for the t-shirt. We can use the vertical motion equation:
y = y0 + v0y * t - 0.5 * g * t^2
where:
y is the vertical displacement (27.7 m)
y0 is the initial vertical position (0 m)
v0y is the vertical component of the initial velocity (v0 * sin(theta))
g is the acceleration due to gravity (9.8 m/s^2)
t is the time of flight
Plugging in the values:
27.7 = 0 + (25.8 * sin(63.6°)) * t - 0.5 * 9.8 * t^2
Simplifying the equation, we get a quadratic equation:
4.9t^2 - (25.8 * sin(63.6°))t + 27.7 = 0
Solving this quadratic equation will give us the time of flight, t. Using the quadratic formula, we find that:
t ≈ 1.23 s
Now, let's find the horizontal displacement of the t-shirt using the horizontal motion equation:
x = x0 + v0x * t
where:
x is the horizontal displacement
x0 is the initial horizontal position (0 m)
v0x is the horizontal component of the initial velocity (v0 * cos(theta))
t is the time of flight
Plugging in the values:
x = 0 + (25.8 * cos(63.6°)) * 1.23
Calculating this:
x ≈ 14.92 m
The t-shirt falls short of reaching the person by the horizontal distance of:
Shortfall = 30.6 m - 14.92 m
Calculating this:
Shortfall ≈ 15.68 m
Therefore, the t-shirt will be approximately 15.68 meters short of reaching the person.
A) In terms of electrolysis, it’s been said from multiple sources online that “Using water's density and relative atomic populations, it is estimated by a mass balance that approximately 2.38 gallons of water are consumed as a feedstock to produce 1 kg of hydrogen gas (14.13 liters), assuming no losses.”
B) However, 1 Gallon of water is said to contain approximately 4,707 liters of hydrogen.
How can both statements be correct under normal atmospheric conditions, since even with 80% efficiency of current PEM electrolyzers the first statement (A) is nowhere near the +4,000 liters of the second approximation (B)?
Answer:
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26. A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?
Answer:
Explanation:
The formula for determining the Emf induced in a loop is:
[tex]\varepsilon = \dfrac{d \phi}{dt}[/tex]
[tex]\varepsilon = \dfrac{d (B*A)}{dt}[/tex]
[tex]\varepsilon = A \times \dfrac{dB}{dt}[/tex]
[tex]\varepsilon = (side (l))^2 \times \dfrac{dB}{dt}[/tex]
where;
square area A = ( l²)
l² = 6.0 cm = 6.0 × 10⁻²
∴
[tex]\varepsilon = ( 6.0 \times 10^{-2})^2 \times 5.0 \times 10^{-3} \ T/S[/tex]
[tex]\varepsilon =18 \times 10^{6} \ V[/tex]
Recall that:
The resistivity of copper = [tex]1.68 \times 10^{-8}[/tex] ohm m
We can as well say that the length of the copper wire = perimeter of the square loop;
The perimeter of the square loop = 4L
Thus, the length of the copper wire = 4 (6.0 × 10⁻² )m
= 24× 10⁻² m
Finally, the current in the loop is determined from the formula:
V = IR
where,
V = voltage
I = current and R = resistance of the wire
Making "I" the subject:
I = V/R
where;
[tex]R = \dfrac{\rho \times l}{A}[/tex]
[tex]R = \dfrac{\rho \times l}{\pi * r^2}[/tex]
[tex]R = \dfrac{1.68 *10^{-8} \times 24*10^{-2}}{\pi * (1*10^{-3})^2}[/tex]
[tex]R = 0.001283 \ ohms[/tex]
∴
[tex]I = \dfrac{18*10^{-6}}{0.001283}[/tex]
I = 14.029 mA
A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m vertically above its starting point. a) what is the vertical component of its initial velocity? b) what is the horizontal component of velocity?
Explanation:
Given:
t = 20 seconds
x = 3000 m
y = 450 m
a) To find the vertical component of the initial velocity [tex]v_{0y}[/tex], we can use the equation
[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]
Solving for [tex]v_{0y}[/tex],
[tex]v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}[/tex]
[tex]\:\:\:\:\:\:\:=120.5\:\text{m/s}[/tex]
b) We can solve for the horizontal component of the velocity [tex]v_{0x}[/tex] as
[tex]x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}[/tex]
or
[tex]v_{0x} = 150\:\text{m/s}[/tex]
An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of 1/50 times the weight of the object is pushing the object up (weight=mg). If we assume that water resistance exerts a force on the abject that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 40 m/s? Assume that the acceleration due to gravity is 9.81 m/sec^2.
Answer:
a) Fnet = mg - Fb - Fr
b) 8.67 secs
Explanation:
mass of object = 80 kg
Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696
Proportionality constant = 10 N-sec/m
a) Calculate equation of motion of the object
Force of resistance on object due to water = Fr ∝ V
= Fr = Kv = 10 V
Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object while mg ( weight ) acts in the negative y - direction on the object.
Fnet = mg - Fb - Fr
∴ Equation of motion of the object ( Ma = mg - Fb - Fr )
b) Calculate how long before velocity of the object hits 40 m/s
Ma = mg - Fb - Fr
a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V
V = u + at ---- ( 1 )
u = 0
V = 40 m/s
a = 9.6138 - 0.125 V
back to equation 1
40 = 0 + ( 9.6138 - 0.125 (40) ) t
40 = 4.6138 t
∴ t = 40 / 4.6138 = 8.67 secs
Two streams merge to form a river. One stream has a width of 8.3 m, depth of 3.2 m, and current speed of 2.2 m/s. The other stream is 6.8 m wide and 3.2 m deep, and flows at 2.4 m/s. If the river has width 10.4 m and speed 2.8 m/s, what is its depth?
Answer:
The depth of the resulting stream is 3.8 meters.
Explanation:
Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:
[tex]\dot V_{1} + \dot V_{2} = \dot V_{3}[/tex] (1)
All volume flows are measured in cubic meters per second.
Dimensionally speaking, we can determine the depth of the resulting stream ([tex]h_{3}[/tex]), in meters, by expanding (1) in this manner:
[tex]w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}[/tex]
[tex]h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}}[/tex] (2)
[tex]v_{1}, v_{2}[/tex] - Speed of the merging streams, in meters per second.
[tex]h_{1}, h_{2}[/tex] - Depth of the merging streams, in meters.
[tex]w_{1}, w_{2}[/tex] - Width of the merging streams, in meters.
[tex]w_{3}[/tex] - Width of the resulting stream, in meters.
[tex]v_{3}[/tex] - Speed of the resulting stream, in meters per second.
If we know that [tex]w_{1} = 8.3\,m[/tex], [tex]h_{1} = 3.2\,m[/tex], [tex]v_{1} = 2.2\,\frac{m}{s}[/tex], [tex]w_{2} = 6.8\,m[/tex], [tex]h_{2} = 3.2\,m[/tex], [tex]v_{2} = 2.4\,\frac{m}{s}[/tex], [tex]w_{3} = 10.4\,m[/tex] and [tex]v_{3} = 2.8\,\frac{m}{s}[/tex], then the depth of the resulting stream is:
[tex]h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}[/tex]
[tex]h_{3} = 3.8\,m[/tex]
The depth of the resulting stream is 3.8 meters.
: A fan is placed on a horizontal track and given a slight push toward an end stop 1.80 meters away. Immediately after the push, the fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2. What is the maximum possible velocity (magnitude) the cart can have after the push so that the cart turns around just before it hits the end-stop
Answer:
The initial velocity is 1.27 m/s.
Explanation:
distance, s = 1.8 m
acceleration, a = - 0.45 m/s^2
final velocity, v = 0
let the initial velocity is u.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times 0.45\times 1.8\\\\u = 1.27 m/s[/tex]
We have that the Initial velocity is mathematically given as
u=1.27m/s
Maximum possible velocity
Question Parameters:
a slight push toward an end stop 1.80 meters away
he fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2
Generally the equation for the third equation of motion is mathematically given as
Vf^2 = Vi^2 + 2ad
Therefore
0=u^2+0.45*1.8
u=1.27m/s
For more information on velocity visit
https://brainly.com/question/17617636
A boy walks from point C to point D which is 50 m apart. Then, he walks back to point C. what is his displacement of his whole journey ?
A.25 m
B.75 m
C.50 m
D.0 m
Answer: D. 0 m
Explanation:
Concept:
Here, we need to know the concept of displacement.
Displacement is defined to be the change in position of an object.
The difference between displacement and distance is the total movement of an object without any regard to direction, while displacement is the pure change of position.
If you are still confused, please refer to the attachment below for a graphical explanation.
Solve:
STEP ONE: the boy walks from point C to point D (a distance of 50 m)
C ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ D
50 m
STEP TWO: the boy walks from point D to point C (a distance of 50 m)
D ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ C
50 m
STEP THREE: find the displacement
The boy started with point C
The boy ended with point C
He did not change his position throughout the journey.
Therefore, his displacement is 0 m.
Hope this helps!! :)
Please let me know if you have any questions
Physics question plz help ASAP
Answer:
Option D.
Explanation:
From the question given above, the following data were obtained:
Force applied (F) = 5 N
Extention (e) = 0.075 m
Spring constant (K) =?
The spring constant for the spring can be obtained as follow:
F = Ke
5 = K × 0.075
Divide both side by 0.075
K = 5 / 0.075
K = 67 N/m
Thus, the spring constant for the spring is 67 N/m
The US currently produces about 27 GW of electrical power from solar installations. Natural gas, coal, and oil powered installations produce about 740 GW of electrical power. The average intensity of electromagnetic radiation from the sun on the surface of the earth is 1000 W/m2 . If solar panels are 30% efficient at converting this incident radiation into electrical power, what is the total surface area of solar panels responsible for the 27 GW of power currently produced
Answer:
The total surface area is "90 km²".
Explanation:
Given:
Power from solar installations,
= 27 GW
Other natural installations,
= 740 GW
Intensity,
[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]
%n,
= 30%
Now,
⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]
then,
⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]
[tex]=90 \ GW[/tex]
As we know,
⇒ [tex]I=\frac{P}{A}[/tex]
by substituting the values, we get
[tex]1000=\frac{90\times 10^9}{A}[/tex]
[tex]A = \frac{90\times 10^9}{10^3}[/tex]
[tex]=90\times 10^6[/tex]
[tex]=90 \ km^2[/tex]
What Are the type's of Tidal turbines?
Answer:
Types of tidal turbines
Axial turbines.
Crossflow turbines.
Flow augmented turbines.
Oscillating devices.
Venturi effect.
Tidal kite turbines.
Turbine power.
Resource assessment.
Answer:
Axial turbines
Crossflow turbines
flow augmented turbines
Total Internal Reflection: A ray of light in glass strikes a water-glass interface at an angle of incidence equal to one-half the critical angle for that interface. The index of refraction for water is 1.33, and for the glass it is 1.43. What angle does the refracted ray in the water make with the normal
Answer:
θ₄ = 37.2º
Explanation:
For this exercise it must be solved in two parts, the first part we look for the critical angle, for this we use the law of refraction with the angle in the middle of transmission of tea = 90º
n₁ sin θ₁= n₂ sin 90
θ₁ = sin⁻¹ [tex]\frac{n_2}{n_1}[/tex]
θ₁ = sin⁻¹ (1.33 / 1.43)
θ₁ = 68.4º
They indicate that the angle of incidence is half of the critical angle
θ₃ = 68.4 / 2 = 34.2º
Let's use the law of refraction again
n₁ sin θ₃ = n₂ sin θ₄
sin θ₄ = [tex]\frac{n_1}{n_2}[/tex] sin θ₃
sin θ₄ = [tex]\frac{1.43}{1.33}[/tex] sin 34.2
θ₄ = sin⁻¹ 0.604345
θ₄ = 37.2º
Circular Motion of Charges: A charged particle is observed traveling in a circular path of radius R in a uniform magnetic field. If the particle were traveling twice as fast, the radius of the circular path would be
Answer:
V = 168 km³
Explanation:
...............
Una persona de 76 kg está siendo retirada de un edificio en llamas mientras se muestra en la figura. Calcule la tensión
en las dos cuerdas si la persona está momentáneamente inmovil.
Ayuda por favor.
Answer:
T1 = 736.6 N, T2 = 193.5 N
Explanation:
W = 76 N
The tension is T1 and T2.
By use of Lami's theorem
[tex]\frac{T_1}{Sin100}=\frac{T_2}{Sin165}=\frac{W}{Sin 95}\\\\So, \\\\T_1 = \frac{76\times 9.8\times Sin 100}{Sin 95} = 736.6 N \\And\\T_2 = \frac{76\times 9.8\times Sin 165}{Sin 95} = 193.5 N \\[/tex]
A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m>s. What is the maximum temperature increase that the water could have as a result of this event
Answer:
The rise in temperature is 0.06 K.
Explanation:
mass of bullet, m = 15 g
initial speed, u = 865 m/s
final speed, v = 534 m/s
mass of water, M = 13.5 kg
specific heat of water, c = 4200 J/kg K
The change in kinetic energy
[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]
According to the conservation of energy, the change in kinetic energy is used to heat the water.
K = m c T
where, T is the rise in temperature.
3473 = 13.5 x 4200 x T
T = 0.06 K
how much energy will be created if a man of mass of 50 kg is destroyed completely.
Answer:
the energy released or created is 4.5 x 10¹⁸ J
Explanation:
Given;
mass of the given matter (man), m = 50 kg
Apply Einstein's mass defect equation.
When a mass of any given matter is completely destroyed, the energy released is calculated as follows;
E = mc²
where;
E is the released or created
c is the speed of light = 3 x 10⁸ m/s
E = (50) x (3 x 10⁸)²
E = 4.5 x 10¹⁸ J
Therefore, the energy released or created is 4.5 x 10¹⁸ J
A green object will absorb ____________________ light and reflect ____________________ light. (ref: p.447-455)
Answer:
A green object will absorb all light except for green light and reflect blue and yellow light.
A ball of mass m is dropped from a height h above the ground. neglecting air resistance then determine the speed of the ball when it is at a height y above the ground and determine the speed of the ball at y if at the instant of release it already has an initial upward speed vi at the initial altitude h.
Answer:
Explanation:
kinematic equation (g will have a negative value if we assume UP is positive)
v² = u² + 2as
a) v = √(0² + 2(g)(y - h))
b) v = √(vi² + 2(g)(y - h))
