Answer:
The right solution is "4.8° east of north".
Explanation:
Given:
Distance,
= 500 km
Speed,
[tex]\vec{v}=120 \ m/s[/tex]
Wind (towards west),
[tex]v_0=10 \ m/s[/tex]
According to the question, we get
The angle will be:
⇒ [tex]\Theta=Cos^{-1}(\frac{v_0}{v_1} )[/tex]
[tex]=Cos^{-1}(\frac{10}{120} )[/tex]
[tex]=85.21[/tex] (north of east)
hence,
The direction must be:
⇒ [tex]\Theta'=90-85.21[/tex]
[tex]=4.79^{\circ}[/tex]
or,
[tex]=4.8^{\circ}[/tex] (east of north)
Choose the CORRECT statements. The superposition of two waves.
I. refers to the effects of waves at great distances.
Il. refers to how displacements of the two waves add together.
Ill. results into constructive interference and destructive interference
IV. results into minimum amplitude when crest meets trough.
V. results into destructive interference and the waves stop propagating.
A. I and II
B. II and III
C. I, II and III
D. II, III and IV
E. III, IV and V
F. II, III, IV and V
Answer:
A
Explanation:
I guess not that much confidential!
A balloon is filled with 80 liters of gas on a day where the temperature was 34 degrees at sea level which is 101.3 kPa and released. As the balloon rises to a certain altitude, the temperature drops to 0 degrees celsius and the balloon doubles in volume. What is the atmospheric pressure at that altitude?
Answer:
0.444atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (
P2 = final pressure (
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question,
P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm
P2 = ?
V1 = 80L
V2 = 160L (double of V1)
T1 = 34°C = 34 + 273 = 307K
T2 = 0°C = 0 + 273 = 273K
Using P1V1/T1 = P2V2/T2
0.999 × 80/307 = P2 × 160/273
79.92/307 = 160P2/273
Cross multiply
307 × 160P2 = 79.92 × 273
49120P2 = 21818.16
P2 = 21818.16 ÷ 49120
P2 = 0.444
P2 = 0.444atm
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2
Answer: The weight of the object is 29.4 N
Explanation:
To calculate the weight of the object, we use the equation:
[tex]W=m\times g[/tex]
where,
m = mass of the object = 3 kg
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]
Hence, the weight of the object is 29.4 N
Một học sinh làm thí nghiệm sóng dừng trên dây cao su dài L với hai đầu A và B cố định . Xét điểm M trên dây sao cho khi sợi dây duỗi thẳng thì M cách B một khoảng a < L/2 . Khi tần số sóng là f = f1 = 60 Hz thì trên dây có sóng dừng và lúc này M là một điểm bụng . Tiếp tục tăng dần tần số thì lần tiếp theo có sóng dừng ứng với f = f2=72 Hz và lúc này M không phải là điểm bụng cũng không phải điểm nút . Thay đổi tần số trong phạm vi từ 73 Hz đến 180 Hz , người ta nhận thấy với f = fo thì trên dây có sóng dừng và lúc này M là điểm nút . Lúc đó , tính từ B ( không tính nút tại B ) thì M có thể là nút thứ ?
A 3-kg projectile is launched at an angle of 45o above the horizontal. The projectile explodes at the peak of its flight into two pieces. A 2-kg piece falls directly down and lands exactly 50 m from the launch point. Determine the horizontal distance from the launch point where the 1-kg piece lands.
1517.4 m
Step-by-step explanation:
Since the projectile broke up at the peak of its flight, it already traveled half its initial range so we can find its initial launch velocity [tex]v_0[/tex] from the equation
[tex]\frac{1}{2}R= \dfrac{1}{2} \left(\dfrac{v_0^2}{g}\sin 2\theta_0 \right)[/tex]
where [tex]\theta_0 = 45°[/tex] and [tex]\frac{1}{2}R = 50\:\text{m}[/tex] so we will get [tex]v_0=31.3\:\text{m/s}[/tex]. Next, we can use the equation
[tex]v_y = v_0y - gt = v_0 \sin 45 - gt[/tex]
and since [tex]v_y=0[/tex] at its peak, we get t = 22.1 s. Let's set this aside for a moment and we'll use it later.
At the top of its peak, we can use the conservation law of linear momentum. Let M be the mass if of the original projectile, [tex]m_1[/tex] be the mass of the larger fragment (2 kg) and [tex]m_2[/tex] be the mass of the smaller fragment (1 kg). We can write the conservation law as
[tex]Mv_0x = m_1V_1 + m_2V_2[/tex]
where [tex]V_1\:\text{and}\:V_2[/tex] are the velocities of the fragments immediately after the break up. But we also know that [tex]V_1=0[/tex] so the velocity of [tex]m_2[/tex] can be calculated from the conservation law as
[tex]Mv_0 \cos 45° = m_2V_2[/tex]
or
[tex]V_2 = \dfrac{M}{m_2}v_0 \cos 45° = 66.4\:\text{m/s}[/tex]
Now we can calculate the horizontal distance the smaller fragment traveled after the break up. Recall that the amount of time for it to go up is also the amount of time to get down so the horizontal distance x is
[tex]x = V_2 t = (66.4\:\text{m/s})(22.1\:\text{s})= 1467.4\:\text{m}[/tex]
Therefore, the total distance traveled from the launch point is
[tex]D = 50\:\text{m} + 1467.4\:\text{m}=1517.4\:\text{m}[/tex]
two factor of a number are 5 and 6 .what is the number show working
Answer:
30
Explanation:
since [tex]\frac{30}{5}[/tex]=6
[tex]\frac{30}{6}[/tex]=5
then both 5 and 6 are factors of 30
Have a nice day
A 10,000J battery is depleted in 2h. What power consumption is this? *
A) 5000W
B) 3W
C) 1.4W
D) 20000W
show your work please
Answer:
P = 1.4 W
Explanation:
Given that,
The work done or the energy of the battery, E = 10,000 J
Time, t = 2 h
We need to find the power consumption. Let it is P. Power is the rate of doing work. So,
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{10,000}{2\times 3600}\\\\P=1.38\ W[/tex]
or
P = 1.4 W
So, the power of the battery is 1.4 W.
If a rock displaces 7 ml of water, what is the volume of the rock?
Answer:
if i am not mistaken the volume is 7, because it only took that much space
A string of length 3m and total mass of 12g is under a tension of 160N. A transverse harmonic wave with wavelength 25cm and amplitude 2cm travels to the right along the string. It is observed that the displacement at x=0 at t=0 is 0.87cm. a) What is the wave? b) Wrote the wave function, y(x,t) c) Find the particle velocity at position x=0 at time t=10s. What is the maximum particle velocity?
Answer:
What is the answer bro idont now
At a playground, Maryam a 3-year old girl and Zahirah a 6-year old girl are playing with the swings. Maryam is sitting while Zahirah is standing on the swing. Both of them were given the same push by their mother. Choose the CORRECT statements:
A. Maryam is swinging faster than Zahirah.
B. Zahirah is swinging faster than Maryam.
C. Both swings at the same pace.
D. Maryam is swinging faster since she is younger.
E. Zahirah is swing faster since she is older.
Answer:
both swings at the same place
Explanation:
because there mother is giving same amount of force to both.
If the car falls down the side of the cliff, what is happening to the gravitational potential energy of the falling car? (Assume the bottom of the cliff is zero)
Answer:
Sentences with many clauses and phrases are difficult to understand because the clauses and phrases typically _____.
modify other clauses and phrases in the sentence
refer to other sentences in the passage
make it hard to determine where the sentence ends
change the intended meaning of the sentence
Explanation:
A man pulls his dog (m=20kg) on a sled with a force of 100N at a 60° angle from the horizontal. What is the horizontal component of the force?
A) 100N
B) 196N
C) 50N
D) 86N
show your work please
Answer:
the horizontal component of the force is 50 N
Explanation:
Given;
force applied by the man, F = 100 N
angle of inclination of the force, θ = 60⁰
mass of the dog, m = 20 kg
The horizontal component of the force is calculated as;
[tex]F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N[/tex]
Therefore, the horizontal component of the force is 50 N
What is the magnitude of a vector that has the following components: x = 32 m y = -59 m
Answer:
Explanation:
Since the x and y components are given
The vectors Magnitude = √32²+(-59)²
=67.12m
If R1 and R2 are in parallel and R3 is in series with them then equivalent resistance will be
Answer:
Refer to the attachment!~
A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?
16s
300s
15s
23s
The boy pushed the sled for 16 seconds.
We have a boy who pushes his little brother on a sled.
We have to determine for how long time does boy push the sled.
State Work - Energy Theorem.The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
According to the question -
The sled is initially at rest → initial velocity (u) = 0.
Final velocity (v) = 4 m/s
Mass of boy and sled (M) = 40 kg
Power developed (P) = 20 W = 20 Joules/sec
According to work - energy theorem -
Work done (W) = Δ E(K) = E(f) - E(i)
Therefore -
W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule
Now, Power is defined as the rate of doing work -
P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]
20 = [tex]\frac{320}{t}[/tex]
t = 16 seconds
Hence, the boy pushed the sled for 16 seconds.
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A heat engine with 0.100 mol of a monatomic ideal gas initially fills a 3000 cm3 cylinder at 800 K. The gas goes through the following closed cycle Isothermal expansion to 5000 cm3 ?
Part A How much work does this engine do per cycle? Express your answer with the appropriate units. sochoric cooling to 200 K -Isothermal compression to 3000 cm3. - Isochoric heating to 800 K Value Units
Part B What is its thermal efficiency? Express your answer with the appropriate units.
Answer:
below
Explanation:
Part A) This engine works per cycle is 254.9 J.
Part B) The thermal efficiency is 23.42%
What is the thermal efficiency?The thermal efficiency of any heat engine is represented in percentage of heat energy converted into work.
For isothermal expansion, work done is
W₁ =nRT₁ x ln(V₂/V₁)
W₁ = 0.1 x 8.314 x 800 x ln(5000/3000)
W₁ = 339.8 J =Q₁
For isochoric cooling ,
W₂ =0
Q₂ =nCvdT = 0.1 x 3R/2 x (T₂-T₁)
Q₂ = -748.3 J
For isothermal compression,
W₃ =nRT₂ ln (V₄/V₃)
W₃ = 0.1 x 8.314 x 200 x ln(3000/5000)
W₃ = -84.9J
For isochoric heating
W₄ =0
Q₄ =nCvdT = 0.1 x 3R/2 x (800-200)
Q₄ = -748.3 J
Total work done in all the process W = W₁ +W₂ +W₃ +W₄
W =254.9 J
Thus, the work done is 254.9 J
Thermal efficiency = Work done/Heat taken
η = W/ Q₁ +Q₄
η = [254.9 / 339.8 +748.3 ] x 100 %
η = 0.2342 x 100 %
η = 23.42%
Thus, the thermal efficiency is 23.42%
Learn more about thermal efficiency.
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Suppose there are 3 molecules in a container. If each molecule has a 1-in-2 chance of being in the left half of the container, what is the probability that there are exactly 2 molecules in the left half of the container?
Answer:
Total probability = 3/8
Explanation:
Below is the calculation:
Number of molecules in the container = 3
The probability of one molecule in the left half, P = 3 / 2 = 1.5 or 1/2
The probability of second molecule in the left half, P1 = (3/4)
Total probability = P x P1
Total probability = (1/2) x (3/4)
Total probability = 3/8
HELP ME PLEASE!!!
Which 2 statements are true about this chemical reaction that forms acid rain?
Answer:
B.
Explanation:
HNO2 is less stable thus dissociates easily to HNO3 + NO + H2O while HNO3 is a strong acid. Thus when they react with H2O they form acid rain
Answer:
B
Explanation:
dont have one just trust me
A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that is perpendicular to the platform at its center. The angular speed of the merry-go-round is constant, and a child at a distance of 1.4 m from the axis has a tangential speed of 2.2 m/s. What is the tangential speed of another child, who is located at a distance of 2.1 m from the axis?
(a) 1.5 m/s
(b) 3.3 m/s
(c) 2.2 m/s
(d) 5.0 m/s
(e) 0.98 m/s
Answer:
[tex]V_2=3.3m/s[/tex]
Explanation:
From the question we are told that:
Distance [tex]d_1=1.4m[/tex]
Tangential speed [tex]V=2.2m/s[/tex]
Distance 2 [tex]d_2=2.1m[/tex]
Generally the equation for Angular velocity is mathematically given by
[tex]w=\frac{v}{r}[/tex]
Therefore
[tex]\frac{v_1}{r_1}=\frac{v_2}{r_2}[/tex]
[tex]V_2=\frac{2.2*2.1}{1.4}[/tex]
[tex]V_2=3.3m/s[/tex]
The wavelength of visible light range of 400 to 750mm .what is the corresponding range of photon energies for visible light
Answer:
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
Explanation:
The energy of a photon is calculated using the following equation;
E = hf
where;
h is Planck's constant = 6.63 x 10⁻³⁴ Js
f is frequency of the photon
[tex]E = h \frac{c}{\lambda} \\\\where;\\\\\lambda \ is \ the \ wavelength\\\\c \ is \ the \ speed \ of \ light \ = 3\times 10^8 \ m/s\\\\When \ \lambda = 400 \ mm = 400 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{400 \times 10^{-3}} \\\\E = 4.973 \times 10^{-25} \ J[/tex]
[tex]When \ \lambda = 750 \ mm = 750 \ \times 10^{-3} \ m\\\\E = \frac{(6.63 \times 10^{-34})(3\times 10^8)}{750 \times 10^{-3}} \\\\E = 2.652 \times 10^{-25} \ J[/tex]
The range of the photon energies is between:
2.652 x 10⁻²⁵ J to 4.973 x 10⁻²⁵ J
Determine the values of m and n when the following average magnetic field strength of the Earth is written in scientific notation: 0.0000451 T. Enter m and n, separated by commas.
Answer:
B = 4.51×10⁻⁵ T
Explanation:
Given that,
The average magnetic field strength of the Earth is 0.0000451 T.
We need to write the value in the form of scientific notation. Any number in scientific notation is written as follows :
N=a×bⁿ
Where
n is any integer and a is a real no
So,
0.0000451 = 4.51×10⁻⁵ T
So, the required answer is equal to 4.51×10⁻⁵ T.
A 620 N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is 870 kg. As the elevator starts moving, the scale reads 450 N.
Required:
a. Find the acceleration of the elevator (magnitude and direction).
b. What is the acceleration if the scale reads 670 N?
c. If the scale reads zero, should the student worry? Explain.
d. What is the tension in the cable in parts (a) and (c)?
Answer:
(a) 9.28 m/s2
(b) 9.03 m/s2
(c) 9.8 m/s2
(d) 450 N, 670 N
Explanation:
mass of elevator + student, m = 870 kg
Reading of scale, R = 450 N
(a) When the elevator goes down, the weight decreases.
Let the acceleration is a.
By the Newton's second law
m g - R = m a
870 x 9.8 - 450 = 870 a
a = 9.28 m/s2
(b) R = 670 N
Let the acceleration is a.
870 x 9.8 - 670 = 870 a
a = 9.03 m/s2
(c) If the scale reads zero, it mean the elevator is falling freely. The acceleration is downwards and its value is 9.8 m/s2.
(d) Tension in cable is 450 N and 670 N.
A rock is thrown from the edge of the top of a 51 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 74 m from the base of the building 8 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.
Answer:
The speed of projection is 34 m/s.
Explanation:
Height of building, h = 51 m
horizontal distance, d = 74 m
time, t = 8 s
Let the angle is A and the speed is u.
d = u cos A x t
74 = u cos A x 8
u cos A = 9.25 .... (1)
Use second equation of motion
[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]
Squaring and adding both the equations
[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]
Một vật chuyển động tròn đều có chu kì T = 0,25 s. Tính tần số chuyển động f của vật?
Explanatio
omega=2pi/T
Answer:
0
0000
Explanation:
a soap bubble was slowly enlarged from radius 4cm to 6cm and amount of work necessary for enlargement is 1.5 *10 calculate the surface tension of soap bubble joules
Answer:
The surface tension is 190.2 N/m.
Explanation:
Initial radius, r = 4 cm
final radius, r' = 6 cm
Work doen, W = 15 J
Let the surface tension is T.
The work done is given by
W = Surface Tension x change in surface area
[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]
You take your pulse and observe 80 heartbeats in a minute. What is the period of your heartbeat? What is the frequency of your heartbeat?
Answer:
120 beats per minute.
Explanation:
If I take your pulse and observe 80 heartbeats in a minute. Then the period of your heartbeat is 0.8 s and frequency is 1.3Hz.
What is Heartbeat ?A pulse is the term used in medicine to describe the tactile arterial palpation of the cardiac cycle (heartbeat) by skilled fingertips. Any location where an artery can be compressed close to the surface of the body, such as the carotid artery in the neck, the radial artery in the wrist, the femoral artery in the groyne, the popliteal artery behind the knee, the posterior tibial artery near the ankle joint, and on the foot, can be used to palpate the pulse (dorsalis pedis artery). Heart rate may be determined by monitoring pulse, or the number of arterial pulses per minute. Auscultation, which is the process of counting the heartbeats while listening to the heart using a stethoscope, is another way to determine the heart rate. Typically, three fingers are used to gauge the radial pulse.
Given,
heart beat = 80 beats/min = 1.3 beats/s
Frequency is nothing but how much beats is heart having in one second and that is 1.3 beats/s. Hence frequency of heart is 1.3Hz.
The Period is reciprocal of frequency,
T = 1/f = 0.8 s
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How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?
Answer:
Explanation:
From the given information:
the car's momentum = momentum of the truck
∴
(a) 816 kg × v = 2650 kg × 16.0 km/h
v = (2650 kg × 16.0 km/h) / 816 kg
v = 51.96 km/hr
(b) 816 kg × v = 9080 kg × 16.0 km/h
v = (9080 kg × 16.0 km/h) / 816 kg
v = 178.04 km/hr
