Explanation:
Displacement of a body moving in circular motion is called uniform circular motion.
hope it is helpful to you
Answer:
A constantly moving object with consistent circular movement. However, for its change in direction, it is accelerating
Explanation:
Uniform circular motion in a circle at constant rate can be described as the motion of the object. When an object moves in a circle, it changes its direction constantly. The object moves tangently to the circle at all times. As the velocity vector direction is the same as the object motion direction, the velocity vector is tangent to the circle. This is shown in the animation on the right by a vector arrow.
An item is accelerating that moves in a circle. Objects that accelerate are subjects that change their speed – either the velocity (i.e. the vecteur magnitude) or the direction. An object with consistent circular movement moves at a constant speed. However, because of its change in direction, it is accelerating. The acceleration direction is inside. The animation on the right shows this through a vector arrow
For an object with only a uniform circular movement, the final motion is the net force. The The net force acting on this object is directed to the middle of the circle. The net force is an inner or centripetal force. Without such a deepest force, an object would continue in a straight direction, never deviating. Regrettably, with the inward net force, perpendicular to the vector, the object changes the direction and is accelerated internally.
A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which one will fall faster
Answer:
on the moon, they will fall at the timeon earth, the coin will fall faster to the groundExplanation:
A coin and feather dropped in a moon experience the same acceleration due to gravity as small as 1.625 m/s², and because of the absence of air resistance both will fall at the same rate to the ground.
If the same coin and feather are dropped in the earth, they will experience the same acceleration due to gravity of 9.81 m/s² and because of the presence of air resistance, the heavier object (coin) will be pulled faster to the ground by gravity than the lighter object (feather).
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Answer:
I DON'T UNDERSTAND
Explanation:
GUESS A MISUNDERSTANDING PLZ PUT A UNDERSTANDABLE QUESTION.
A submarine has a "crush depth" (that is, the depth at which
water pressure will crush the submarine) of 400 m. What is
the approximate pressure (water plus atmospheric) at this
depth? (Recall that the density of seawater is 1025 kg/m3, g=
9.81 m/s2, and 1 kg/(m-s2) = 1 Pa = 9.8692 x 10-6 atm.)
Answer:
P =40.69 atm
Explanation:
We need to find the approximate pressure at a depth of 400 m.
It can be calculated as follows :
P = Patm + ρgh
Put all the values,
[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]
So, the approximate pressure is equal to 40.69 atm.
Astronauts in space move a toolbox from its initial position ????????→=<15,14,−8>m to its final position ????????→=<17,14,−1>m. The two astronauts each push on the box with a constant force. Astronaut 1 exerts a force ????1→=<18,7,−12>???? and astronaut 2 exerts a force ????2→=<16,−10,16>????.
Required:
What is the total work performed on the toolbox?
If both forces are measured in Newtons, then the net force is
F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N
The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector
d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m
The total work done by the astronauts on the toolbox is then
F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J
The work done by the two astronauts is equal to 96 J.
What is work done?work done?Work done is defined as the product of force applied and the distance moved by the force.
Work done = Force × DistanceThe forces applied = 18+16 N, 7+ -10 N, and -12 + 16N
Forces = 34 N, -3 N, and 4N
Distances = (17 - 15, 14 - 14, -1 - - 8) m
Distances = 2, 0, 7
Work done = 34 × 2 + -3 × 0 + 4 × 7
Work done = 96 J
Therefore, the work done by the two astronauts is equal to 96 J.
Learn more about work done at: https://brainly.com/question/25573309
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what is the time taken by moving body with acceleration 0.1m/s2 if the initial or finak velocities are 20m/s and 30m/s respectively?
Answer:
t= 100s
Explanation:
use v=v0+at
plug in givens and solve for t
30=20+0.1*t
t= 100s
An airplane increases its speed at the average rate of 15 m/s2. How much time does it take to increase its speed from 100 m/s to 160 m/s
Answer:
4 s
Explanation:
Acceleration (a) = 15 m/s²Initial velocity (u) = 100 m/sFinal velocity (v) = 160 m/sWe are asked to calculate time taken (t).
By using the first equation of motion,
[tex]\longrightarrow[/tex] v = u + at
[tex]\longrightarrow[/tex] 160 = 100 + 15t
[tex]\longrightarrow[/tex] 160 - 100 = 15t
[tex]\longrightarrow[/tex] 60 = 15t
[tex]\longrightarrow[/tex] 60 ÷ 15 = t
[tex]\longrightarrow[/tex] 4 s = t
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.71 s. He breaks the tackle and runs straight forward another 24.0 m in 5.20 s. Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
Answer:
Average Velocity = 3.63 m/s
Explanation:
First, we will calculate the total displacement of the quarterback, taking forward direction as positive:
Total Displacement = 15 m - 3 m + 24 m = 36 m
Now, we will calculate the total time taken for this displacement:
Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s
Therefore, the average velocity will be:
[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]
Average Velocity = 3.63 m/s
write down the following units in the ascending of their value A) mm nm cm um B) 1m 1cm 1km 1mm. convert the following units into SI without changing their values? A)3500g B)2.5km C)2h
Answer:
A) nm, um, mm, cm
B) 1mm, 1cm, 1m, 1km
A) 3500g, B) 2500m, C) 7200 seconds
Thermometers and Temperature Scales
While traveling outside the United States, you feel sick. A companion gets you a thermometer, which says your temperature is 40.9. What scale is that on? What is your Fahrenheit temperature? Should you seek medical help?
Answer:
105.62°F
Explanation:
When the body temperature having fever is measured to be 40.9 on a scale then it must be a Celsius scale thermometer because 37°C is the normal temperature of a healthy human. In case of fever the given temperature is measured on a standard Celsius scale.
The relation between Fahrenheit and Celsius scale is:
[tex]\frac{C}{5}=\frac{F-32}{9}[/tex]
[tex]F=\frac{9C}{5} +32[/tex]
[tex]F=105.62^{o}F[/tex]
It is a high fever and an immediate medical help must be taken.
During 57 seconds of use, 330 C of charge flow through a microwave oven. Compute the size of the electric current.
Answer:
5.78amps
Explanation:
Given data
Time t= 57 seconds
Charge Q= 330C
Current I= ??
The expression for the electric current is given as
Q= It
Substituting we have
330= I*57
I= 330/57
I=5.78 amps
Hence the current is 5.78amps
The electric potential ( relative to infinity ) due to a single point charge Q is 400 V at a point that is 0.6 m to the right of Q. The electric potential (relative to infinity) at a point that is 0.90 m to the left of 0 is:_____.
A. + 400 V.
B. -400 V.
C. + 200 V.
Answer:
The potential at a distance of 0.9 m is 266.67 V.
Explanation:
Charge = Q
Potential is 400 V at a distance 0.6 m .
Let the potential is V at a distance 0.9 m.
Use the formula of potential.
[tex]V = \frac{Kq}{r}\\\\\frac{V}{400}=\frac{0.6}{0.9}\\\\V = 266.67 V[/tex]
Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their waveform repeats every time interval T.
a. transverse
b. longitudinal
c. periodic
d. sinusoidal
Answer:
periodic
Explanation:
As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.
Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is
[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]
where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
[tex]A=2.01×10^{16}\:\text{nuclei}[/tex]
Explanation:
Given:
[tex]\lambda = 4.96×10^3 s[/tex]
[tex]A_0 = 3.21x10^{17}[/tex] nuclei
t = 1.98×10^4 s
[tex]A=A_02^{-\frac{t}{\lambda}}[/tex]
[tex]A=(3.21×10^{17}\:\text{nuclei}) \left(2^{-\frac{1.98×10^4}{4.96×10^3}} \right)[/tex]
[tex]\:\:\:\:\:\:\:=2.01×10^{16}\:\text{nuclei}[/tex]
I need help on this physics problem.
Answer:
the speed of the nerve impulse in miles per hour is 201.59 mi/hr
Explanation:
Given;
the speed of the nerve impulse, v = 90.1 m/s
To convert this speed in meters per second to miles per hour, we use the following method;
1,609 meter = 1 mile
3,600 s = 1 hour
[tex]v(mi/h) = 90.1 \ \frac{m}{s} \times \frac{1 \ mile}{1,609 \ m} \times \frac{3,600 \ s}{1 \ hour} = (\frac{90.1 \times 3,600}{1,609} )\frac{mi}{hr} = 201.59 \ mi/hr[/tex]
Therefore, the speed of the nerve impulse in miles per hour is 201.59 mi/hr
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 40.0 mph and half the distance at 60.0 mph . On her return trip, she drives half the time at 40.0 mph and half the time at 60.0 mph.
Required:
a. What is Julie's average speed on the way to grandmother's house?
b. What is her average speed in the return trip?
Answer:
a. The average speed on her way to Grandmother's house is 48.08 mph
b. The average speed in the return trip is 50 mph.
Explanation:
The average speed (S) can be calculated as follows:
[tex] S = \frac{D}{T} [/tex]
Where:
D: is the total distance
T: is the total time
a. To find the total distance in her way to Grandmother's house, we need to find the total time:
[tex]T_{i} = t_{1_{i}} + t_{2_{i}} = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}}[/tex]
Where v is for velocity
[tex] T = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}} = \frac{(100/2) mi}{40.0 mph} + \frac{(100/2) mi}{60.0 mph} = 1.25 h + 0.83 h = 2.08 [/tex]
Hence, the average speed on her way to Grandmother's house is:
[tex]S_{i} = \frac{D}{T_{i}} = \frac{100 mi}{2.08 h} = 48.08 mph[/tex]
b. Now, to calculate the average speed of the return trip we need to calculate the total time:
[tex]D = v_{1_{f}}\frac{T_{f}}{2} + v_{2_{f}}\frac{T_{f}}{2} = \frac{T_{f}}{2}(v_{1_{f}} + v_{2_{f}})[/tex]
[tex]100 mi = \frac{T_{f}}{2}(40 mph + 60 mph)[/tex]
[tex] T_{f} = \frac{200 mi}{40 mph + 60 mph} = 2 h [/tex]
Therefore, the average speed of the return trip is:
[tex]S_{f} = \frac{D}{T_{f}} = \frac{100 mi}{2 h} = 50 mph[/tex]
I hope it helps you!
Solids diffuse because the particles cannot move.
A. Can
B. Not enough info
C. Cannot
D. Sometimes will
Solids cannot diffuse.
Why don’t you see tides ( like those of the ocean ) in your swimming pool ?
A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 800-N firefighter has climbed 4.00 m along the ladder from the bottom. (b) If the ladder is just on the verge of slipping when the firefighter is 9.00 m from the bottom, what is the coefficient of static friction between ladder and ground
Answer:
a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36
Explanation:
a) This is a balancing act.
Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive
-w x - W x₂ + R y = 0 (1)
usemso trigonometry to find distances
cos 60.08 = x / 7.5
x = 7.5 cos 60.08
x = 3.74 m
fireman
cos 60.08 = x₂ / 4
x2 = 4 cos 60
x2 = 2 m
wall support
sin 60.08 = y / 15
y = 15 are 60.08
y = 13 m
we substitute in equation 1
R y = w x + W x2
R = (w x + W x2) / y
R = (500 3.74 +800 2) / 13
R = 266.92 N
now let's write the expressions for the translational equilibrium
X axis
R -fr = 0
R = fr
fr = 266.92 N
Y Axis
Fy - w-W = 0
fy = 500 + 800
fy = 1300 N
b) ask the friction coefficient
the firefighter's distance is
cos 60.08 = x₃ / 9.00
x₃ = 9 cos 60
x₃ = 5.28 m
from equation 1
R = (w x + W x₃) / y
R = 500 3.74 + 800 5.28) / 13
R = 468.769 N
we saw that
fr = R = 468.769
The expression for the friction force is
fr = μ N
in this case the normal is the ratio to pesos
N = Fy
N = 1300 N
μ N = fr
μ = fr / N
μ = 468,769 / 1300
μ = 0.36
Find the amount og work done
Answer:
100j
Explanation:
Question 8 a-e plz
Answer:
(a) t = 0 s
(b) t = 0 s, 30 s, 55 s
(c) t = 40 s to t = 60 s
(d) t = 10 s to t = 15 s
(e) a = 6 m/s^2
Explanation:
(a) The car is at starting position at t = 0 s and v = 0 m/s.
(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.
(c) from t = 40 s to 60 s the car is moving in the negative direction.
(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.
(e) The slope of the velocity time graph gives acceleration.
a = (60 - 0) / (10 - 0) = 6 m/s^2
Water is falling on the blades of a turbine at a rate of 100 kg/s from a certain spring. If the height of spring be 100m, then the power transferred to the turbine will be: a) 100 KW b) 10 KW c) 1 KW d) 100 W
Answer:
Natae Si Jordan Kaya Sya Napaihe
Explanation:
haha
What has a wind speed of 240 kph or greater?
Answer:
SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.
Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the spring to a distance of 0.200 mm , how far up the slope will an identical ice cube travel before reversing directions
Answer:
The correct answer will bs "2.41 m".
Explanation:
According to the question,
M = 50 g
or,
= 0.050 kg
[tex]\Theta = 25^{\circ}[/tex]
k = 25.9 N/m
Δx = 0.200 m
Let the traveled distance be "x".
By using trigonometry, the height will be:
⇒ [tex]h = l Sin \Theta[/tex]
hence,
⇒ [tex]Potential \ energy \ at \ the \ top=Spring \ potential \ energy[/tex]
[tex]Mgh=\frac{1}{2} k(\Delta x)^2[/tex]
By putting the values, we get
[tex]0.050\times 9.8\times lSin 25^{\circ}=\frac{1}{2}\times 25.0\times (0.200)^2[/tex]
[tex]l=2.41 \ m[/tex]
IS ANYONE THERE..??!
Answer:
hmmmmmmmmmmmmmmmmmmmmmmmm y
What would the radius (in mm) of the Earth have to be in order for the escape speed of the Earth to equal (1/21) times the speed of light (300000000 m/s)? You may ignore all other gravitational interactions for the rocket and assume that the Earth-rocket system is isolated. Hint: the mass of the Earth is 5.94 x 1024kg and G=6.67×10−11Jmkg2G=6.67\times10^{-11}\frac{Jm}{kg^2}G=6.67×10−11kg2Jm
Answer:
The expected radius of the Earth is 3.883 meters.
Explanation:
The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:
[tex]U = K[/tex] (1)
Where:
[tex]U[/tex] - Gravitational potential energy, in joules.
[tex]K[/tex] - Translational kinetic energy, in joules.
Then, we expand the formula by definitions of potential and kinetic energy:
[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the Earth. in kilograms.
[tex]m[/tex] - Mass of the rocket, in kilograms.
[tex]r[/tex] - Radius of the Earth, in meters.
[tex]v[/tex] - Escape velocity, in meters per second.
Then, we derive an expression for the escape velocity by clearing it within (2):
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)
If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]
[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]
[tex]r = 3.883\,m[/tex]
The expected radius of the Earth is 3.883 meters.
Assume that I = E/(R + r), prove that 1/1 = R/E + r/E
[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \frac{1}{I} = \frac{R}{E} + \frac{r}{E} }}}}}}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]
[tex]I = \frac{ E}{ R + r} \\[/tex]
[tex] ➺\:\frac{I}{1} = \frac{E}{R + r} \\[/tex]
Since [tex]\frac{a}{b} = \frac{c}{d} [/tex] can be written as [tex]ad = bc[/tex], we have
[tex]➺ \: I \: (R + r) = E \times 1[/tex]
[tex]➺ \: \frac{1}{I} = \frac{R + r}{E} \\ [/tex]
[tex]➺ \: \frac{1}{I} = \frac{R}{E} + \frac{r}{E} \\ [/tex]
[tex]\boxed{ Hence\:proved. }[/tex]
[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}[/tex]
When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?
Answer:
because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.
Need ur help,,, :-[ :-{
...... ............ .. ..
Answer:
Graph B express the magnetic relationship of magnetic flux and electronic flow
what is the dimensional formula of force and torque
Answer:
Units. Torque has the dimension of force times distance, symbolically T−2L2M. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit newton metre (N⋅m) and never the joule. The unit newton metre is properly denoted N⋅m.
Dimension: M L2T−2
In SI base units: kg⋅m2⋅s−2
Other units: pound-force-feet, lbf⋅inch, ozf⋅in
Answer:
hope it is helpful to you
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