Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound
The question is incomplete. The complete question is :
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.
What is the frequency of the sound?
Solution :
Given :
The distance between the two loud speakers, [tex]d = 1.8 \ m[/tex]
The speaker are in phase and so the path difference is zero constructive interference occurs.
At the point [tex]D[/tex], the speakers are out of phase and so the path difference is [tex]$=\frac{\lambda}{2}$[/tex]
Therefore,
[tex]$AD-BD = \frac{\lambda}{2}[/tex]
[tex]$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$[/tex]
[tex]$\lambda = 2 \times 0.4985$[/tex]
[tex]$\lambda = 0.99714 \ m$[/tex]
Thus the frequency is :
[tex]$f=\frac{v}{\lambda}$[/tex]
[tex]$f=\frac{340}{0.99714}$[/tex]
[tex]f=340.9744[/tex] Hz
When you have a straight horizontal line on a velocity time graph, what does this tell you about the object’s motion in terms of velocity and acceleration?
Answer:
It tell you that the velocity is constant, what means that there's no acceleration
A 69.0-kg astronaut is floating in space, luckily he has his trusty 28.0-kg physics book. He throws his physics book and accelerates at 0.0130 m/s2 in the opposite direction. What is the magnitude of the acceleration of the physics book?
Answer:
0.032 [tex]m/s^2[/tex]
Explanation:
Given :
Weight of the astronaut = 69 kg
Weight of the physics book = 28 kg
Acceleration of the astronaut = 0.0130 [tex]m/s^2[/tex]
The force that is applied on the astronaut :
[tex]F=ma[/tex]
[tex]$=69 \times 0.013$[/tex]
= 0.897 N
Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N
Therefore, the acceleration of the physics book is given by :
[tex]$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$[/tex]
[tex]$a = \frac{0.897}{28}$[/tex]
a = 0.032 [tex]m/s^2[/tex]
Hence, the acceleration of the physics book is 0.032 [tex]m/s^2[/tex].
Answer:
The acceleration of astronaut is 5.27 x 10^-3 m/s^2.
Explanation:
mass of astronaut, M = 69 kg
Mass of book, m = 28 kg
acceleration of book, a = 0.013 m/s^2
Let the acceleration of astronaut is A.
According to the Newton's third law, for every action there is an equal and opposite reaction.
So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.
M A = m a
69 x A = 28 x 0.013
A = 5.27 x 10^-3 m/s^2
Cho hai mặt cầu đồng tâm O tích điện đều. Bán kính của hai mặt cầu lần lượt là R1 và R2 (R2>R1). Điện tích mặt trong là q và mặt ngoài là Q
Tính cường độ điện trường tại một điểm cách tâm O một đoạn r (biết R1 < r < R2)
Tính hiệu điện thế giữa hai mặt cầu
Answer:
you will stc ohxoyxct txxtx xigigjjgjvvixiffjz,iffzikzfjvixii. hi h ohigiogooigoh
Explanation:
k jjvhvojvovvojivivivihvi hj
Do you believe in ghost
Answer:
well its about our thinking but i do believe in ghost a little
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone subsequently falls to the ground, which is 14.5 m below the point where the stone leaves his hand.
At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
How much time is the stone in the air?
elapsed time:
Answer:
Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)
Explanation:
Time=Distance/Speed
=14.5/6.63
=2.19s(to 3s.f.)
Acceleration=Final Velocity(v)-Initial Velocity(u)/Time
9.81=v-6.63/2.19
v-6.63=21.5
v=28.1m/s
You are trying to push a 14 kg canoe across a beach to get it to a lake. Initially, the canoe is at rest, and you exert a force over a distance of 5 m until it has a speed of 2.1 m/s.
a. How much work was done on the canoe?
b. The coefficient of kinetic friction between the canoe and the beach is 0.2. How much work was done by friction on the canoe?
c. How much work did you perform on the canoe?
d. What force did you apply to the canoe?
Answer:
I hope this is correct!
Explanation:
a) work = (force)(displacement)
We know that d = 5m, now we just need to find the force
We need to calculate the acceleration first using a = v^2 - u^2 / 2d
final velocity (v) = 2.1 m/s , initial velocity (u) = 0 m/s , displacement (d) = 5m
a = 2.1^2 - 0^2 / 2(5)
= 0.441 m/s^2
Now we can find force using f = ma
f = (14kg)(0.441m/s^2)
= 6.174 newtons
Finally, you can calculate work now that you have force!
work = (6.174n)(5m)
= 30.87 J (you may need to round sig figs!)
b) Work done by friction = (coefficient of kinetic friction)(mg)(v)
m = 14kg, g = 9.8, coefficient of friction = 0.2
force due to friction = (0.2)(14kg x 9.8)(2.1)
= 57.624 J (again you may need to round to sig figs)
c) Work you perform = total work in direction of motion + work done by friction
total work = 30.87 J, work done by friction = 57.624 J
Work you perform = 30.84 + 57.624
= 88.464 J
d) Force you applied = work you performed / distance
work you performed = 88.464 / 5 m
= 17.6929 N (again you may need to round to sig figs)
Hope this helps ya! Best of luck <3
Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle
a) The mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle is 23.6 cm².
a) The mass flow rate through the nozzle can be calculated with the following equation:
[tex] \dot{m_{i}} = \rho_{i} v_{i}A_{i} [/tex]
Where:
[tex]v_{i}[/tex]: is the initial velocity = 20 m/s
[tex]A_{i}[/tex]: is the inlet area of the nozzle = 60 cm²
[tex]\rho_{i}[/tex]: is the density of entrance = 2.21 kg/m³
[tex] \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s [/tex]
Hence, the mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle can be found with the Continuity equation:
[tex] \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} [/tex]
[tex] 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} [/tex]
[tex] A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} [/tex]
Therefore, the exit area of the nozzle is 23.6 cm².
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I hope it helps you!
a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.
We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:
[tex]\dot m_{in} = \dot m_{out}[/tex] (1)
Where:
[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.
[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.
Given that air flows at constant rate, we expand (1) by dimensional analysis:
[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex] (2)
Where:
[tex]\rho_{in}, \rho_{out}[/tex] - Air density at inlet and outlet, in kilograms per cubic meter.
[tex]A_{in}, A_{out}[/tex] - Inlet and outlet area, in square meters.
[tex]v_{in}, v_{out}[/tex] - Inlet and outlet velocity, in meters per second.
a) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex] and [tex]v_{in} = 20\,\frac{m}{s}[/tex], then the mass flow rate through the nozzle is:
[tex]\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}[/tex]
[tex]\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)[/tex]
[tex]\dot m = 0.265\,\frac{kg}{s}[/tex]
The mass flow rate through the nozzle is 0.265 kilograms per second.
b) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex], [tex]v_{in} = 20\,\frac{m}{s}[/tex], [tex]\rho_{out} = 0.762\,\frac{kg}{m^{3}}[/tex] and [tex]v_{out} = 150\,\frac{m}{s}[/tex], then the exit area of the nozzle is:
[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex]
[tex]A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}[/tex]
[tex]A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}[/tex]
[tex]A_{out} = 2.320\times 10^{-3}\,m^{2}[/tex]
[tex]A_{out} = 23.202\,cm^{2}[/tex]
The exit area of the nozzle is 23.202 square centimeters.
which watch is more preferable for the measurement of time among pendulum, quartz and atomic watch
Answer:
pendulum, quartz
Explanation:
potential diffetence
Answer:
6v
Explanation:
V=IR
V= 2* 3
V= 6 volts
explain why our sweat is salty?
Answer:
Sweat also contains ammonia and urea, which are produced by the body when it breaks down proteins from the foods you eat.
Hope this helps..
.. Solve: 91
Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of
a slit of width 12x10^-5 cm when the slit illuminated by monochromatic light of wave length
6000 A
[KUET’10-11)
(a) 30°
(b) 60°
(c) 15°
(d) None of these
Solution
Explanation:
bro I have no idea fam......
State TRUE or FALSE.
1. We use muscular force to lift a bucket of water.
2. A bow uses mechanical force of the bow string to shoot an arrow.
3. The force of friction enables us to walk on earth.
4. Plants use solar energy to make their food.
5. The energy stored inside the earth is called atomic energy
Answer:
1. True
2. False
3. True
4. True
5. True
Answer:
that is pure falsereeeeeeeee
Explanation:
In order to test an intentionally weak adhesive, the bottom of the small 0.15-lb block is coated with adhesive and then the block is pressed onto the turntable with a known force. The turntable starts from rest at t = 0 and uniformly accelerates with a = 2 rad/s^2. If the adhesive fails at exactly t = 3 s, then determine:
a. the magnitude of the ultimate shear force that the adhesive supports
b. the angular displacement of the turntable at the time of failure
Answer:
answer
Explanation:
it is the answer which was presented in the year
You want to swim from one side of a river to another side. Assume your speed is three miles per hour in the west direction, with negligible water velocity. When you reach a certain point, you will encounter water flow with a velocity of 6.2 miles per hour in the north direction. What is your resultant speed and direction
Answer:
speed = 6.71 mph and angle is 71.2 degree.
Explanation:
speed of person, u = 3 miles per hour
speed of water, v = 6 miles per hour
Resultant speed
[tex]V =\sqrt{v^2 + u^2}\\\\V = \sqrt{3^2 + 6^2}\\\\V = 6.71 mph[/tex]
The angle from the west is
tan A = 6/2 = 3
A = 71.6 degree
II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4-49. ( ) If the buckets are at rest, what is the tension in each cord? ( ) If the two buckets are pulled upward with an acceleration of 1.25 m/s by the upper cord, calculate the tension in each cord
Answer:
Here , mass of bucket ,m = 3.2 Kg
Now , let the tension in upper rope is T1
the tension in the middle rope is T2
a)
For lower bucket, balancing forces in vertical direction
T2 - mg = 0
T2 = mg
T2 = 3.2 *9.8
T2 = 31.36 N
tension in the middle rope is 31.36 N
For the upper bucket , balancing forces in vertical direction
T1 - T2 - mg = 0
T1 = T2 + 3.2 *9.8
T1 = 62.72 N
the tension in the upper rope is 62.72 N
B)
for a = 1.25 m/s^2
Using second law of motion ,for both the buckets
Fnet = ma
T1 - 2mg = 2m*a
T1 = 2*3.2*(9.8 +1.25)
T1 = 70.72 N
the tension in the upper rope is 70.7 N
Now , the lower bucket
Using second law of motion,
T2 - mg = ma
T2 = 3.2 * (9.8 + 1.25)
T2 = 35.36 N
the tension in the lower rope is 35.36 N
Which of the following describes an electric conductor
The following describes an electric conductor : A material that has low resistance and allows the charges to move freely. The correct option is D.
What is conductor?A conductor is a material or metal which allows the electrons to flow through it. In other words, a conductor allows the current to pass through them.
A battery also called as the voltage source, provides sufficient voltage or energy to excite electrons in the conductor.
Opposition offered to the flow of current is called as the resistance. The electrical element used in the circuit is the resistor.
So, an electric conductor is a material that has low resistance and allows the charges to move freely.
Thus, the correct option is D.
Learn more about conductor.
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01.04 Law of Conservation of Energy
science question
Answer:
law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. Calculate the frequency of this light. Be sure to include units in your answer.
Answer:
5.71×10¹⁴ Hz
Explanation:
Applying,
v = λf................. Equation 1
Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency
make f the subject of the equation
f = v/λ............. Equation 2
From the question,
Given: λ = 525 nm = 5.25×10⁻⁷ m,
Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s
Substitute these values into equation 2
f = (3.0×10⁸)/(5.25×10⁻⁷)
f = 5.71×10¹⁴ Hz
Hence the frequency of light is 5.71×10¹⁴ Hz
Choose one. 5 points
Use the equation from week 3:
frequency =
wavespeed
wavelength
and the wavelength you found in #3 to calculate the frequency of this photon (remember the speed of
light is 3E8 m/s);
7.6E14 Hz
6.0E14 Hz
4,6E14 Hz
The frequency is 4,6E14 Hz.
What is the frequency?
Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.
Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.
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Two different galvanometers G1 and G2, have internal resistances r1and r2. The galvanometers G1 and G2 require the same current IC1=IC2 for a full-scale deflection of their pointers. These galvanometers G1 and G2 are used to build lab-made ammeters A1 and A2 . Both ammeters A1 and A2 have the same maximum scale reading Imax1=Imax2=Imax. To build A1 ,shunt resistor of resistance Rsh1is used and to build A2 , shunt resistor of resistance Rsh2 is used. The value of these shunt resistor resistances are such that: Rsh1=3Rsh2. What is the ratio oftheir internal resistances: r1:r2?
Answer:
there are 3 photos attached. so check
Explanation:
If the horizontal range and the max height of a body projected at an angle titre to the horizontal is K and Q respectively. Show that the muzzle velocity Vo is given by: Vo=√[2gQ + K²/8Q]
Explanation:
Recall that
[tex]K = \dfrac{v_0^2\sin2\theta}{g}\:\:\:\:\:\:\:\:\:(1)[/tex]
and
[tex]Q = \dfrac{v_0^2\sin^2\theta}{2g}\:\:\:\:\:\:\:\:\:(2)[/tex]
From Eqn(2), we can write
[tex]\sin\theta = \sqrt{\dfrac{2gQ}{v_0^2}}\:\:\:\:\:\:\:\:\:(3)[/tex]
Using the identity [tex]\sin\theta = 2\sin\theta \cos\theta[/tex], we can rewrite Eqn(1) as
[tex]\dfrac{gK}{2v_0^2} = \sin\theta \cos\theta[/tex]
Squaring the above equation, we get
[tex]\dfrac{g^2K^2}{4v_0^4} = \sin^2\theta \cos^2\theta[/tex]
[tex]\:\:\:\:\:\:\:\:\:=\sin^2\theta(1 - \sin^2\theta)\:\:\:\:\:\:\:(4)[/tex]
Use Eqn(3) on Eqn(4) and we will get the following:
[tex]\dfrac{g^2K^2}{4v_0^4} = \dfrac{2gQ}{v_0^2}(1 - \dfrac{2gQ}{v_0^2})[/tex]
This simplifies to
[tex]\dfrac{gK^2}{8v_0^2Q} = 1 - \dfrac{2gQ}{v_0^2}[/tex]
Rearranging this further, we get
[tex]1 = \dfrac{2gQ}{v_0^2} + \dfrac{gK^2}{8v_0^2Q}[/tex]
Putting [tex]v_0^2[/tex] to the left side, we get
[tex]v_0^2 = 2qQ + \dfrac{gK^2}{8Q}[/tex]
Finally, taking the square root of the equation above, we get the expression for the muzzle velocity [tex]v_0[/tex] as
[tex]v_0 = \sqrt{2gQ + \dfrac{gK^2}{8Q}}[/tex]
A student of mass 50kg takes 15seconds to run up a flight of 50 steps. If each step is 20cm, calculate the potential energy of the student at the maximum height
Answer:
the answer is 49000 joules at the maximum height
Explanation:
we know the mass (50kg)
we know the acceleration due to gravity(9.8m/s²)
we know the height too(maximum height meaning the 50th step so we multiply 50 with 20cm as each step is 20 cm and we get 1000 cm, convert to m it is 100 m
the formula is potential energy=mgh
m for mass
g for acceleration due to gravity
h for height
multiply them
50x9.8x100
we get 49000
the unit of potential energy is joules so the answer is
49000 joules
Answer:
49000 joules
Explanation:
hope it helpss
A wire long and with mass is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude . What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards
The question is incomplete. The complete question is :
A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.
Solution :
Given :
Length of the wire, L = 0.6 m
Mass of the wire length, m = 11 g
= [tex]11 \times 10^{-3}[/tex] kg
Magnetic field , B = 0.4 T
Know we know that :
ILB = mg
or [tex]$I=\frac{mg}{BL}$[/tex]
[tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]
[tex]I=0.44963\ A[/tex]
[tex]I = 449.63 \ mA[/tex]
Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum uncertainty in the electron's momentum
Answer:
Uncertainty in position (∆x) = 24 × 10⁻¹⁵ mUncertainty in momentum (∆P) = ?Planck's constant (h) = 6.26 × 10⁻³⁴ Js[tex]\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi} [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} } [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} } [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21} [/tex]
[tex]\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}[/tex]
two object A and B vertically thrown upward with velocities of 80m/s and 100m/s at two seconds interview where and when will the two object meet.
Answer:
THIS IS YOUR ANSWER:
☺✍️HOPE IT HELPS YOU ✍️☺
In a photoelectric effect experiment, it is observed that violet light does not eject electrons from a particular metal. Next, red light with the same intensity is incident on the same metal. Which result is possible
Answer:
No ejection of photo electron takes place.
Explanation:
When a photon of suitable energy falls on cathode, then the photoelectrons is emitted from the cathode. This phenomenon is called photo electric effect.
The minimum energy required to just eject an electron is called work function.
The photo electric equation is
E = W + KE
where, E is the incident energy, W is the work function and KE is the kinetic energy.
W = h f
where. h is the Plank's constant and f is the threshold frequency.
Now, when the violet light is falling, no electrons is ejected. When the red light is falling, whose frequency is less than the violet light, then again no photo electron is ejected from the metal surface.
how much heat is produced in one hour by an electric iron which draws 2.5ampere when connected to a 100V supply
Explanation:
I=2.5 Ampere ; V=100V ;t = 1 hour=60secs
We know Heat = VIt
H=100×2.5×60=15,000J
A rabbit is moving in the positive x-direction at 2.70 m/s when it spots a predator and accelerates to a velocity of 13.3 m/s along the positive y-axis, all in 2.40 s. Determine the x-component and the y-component of the rabbit's acceleration.
Answer:
the answer is nearly 5.655 [tex]ms^{-2}[/tex]
Explanation:
Given,
[tex]v_{x}=2.7 ms^{-1}[/tex]
[tex]v_{y}=13.3 ms^{-1}[/tex]
[tex]t=2.4 s[/tex]
[tex]a_{x}=\frac{2.7}{2.4}=1.125 ms^{-2}[/tex] (as [tex]a=\frac{v-u}{t}[/tex])
[tex]a_{y}=\frac{13.3}{2.4}=5.542 ms^{-2}[/tex]
[tex]a=\sqrt{a_{x}^{2}+a_{y}^{2} }[/tex]
[tex]=\sqrt{1.125^{2}+5.542^{2} }[/tex]
[tex]=5.655 ms^{-2}[/tex]
hope you have understood this...
pls mark my answer as the brainliest
(c) It takes you hours to to bring the turkey from to . During that time, the electrical grid transfers a constant Watts of power into the the oven. Take the turkey and the air in the oven to be your system. What was the thermal transfer of energy between the system and the surroundings
Complete Question
(c) It takes you 5 hours to to bring the turkey from 10.0°C to 75.0 °C. During that time, the electrical grid transfers a constant 2500.0 Watts of power into the the oven. Take the turkey and the air in the oven to be your system. What was the thermal transfer of energy between the system and the surroundings?
Answer:
[tex]Q=4.50 *10^7J[/tex]
Explanation:
From the question we are told that:
Time [tex]t=5hours[/tex]
Temperature rise [tex]dT= 65\textdegree[/tex]
Power [tex]P=2500.0 Watts[/tex]
Generally, the equation for Power is mathematically given by
[tex]P=\frac{Q}{t}[/tex]
Therefore
[tex]Q=2500*5*360[/tex]
[tex]Q=4.50 *10^7J[/tex]
