Explanation:
The acceleration of an object is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
Dimension of velocity is, [tex][v]=[LT^{-1}][/tex]
Dimension of time is, [tex][t]=[T][/tex]
Dimension of acceleration is, [tex][a]=[LT^{-2}][/tex]
Option 1.
[tex]\dfrac{v^2}{x}=[LT^{-2}][/tex]
Option 2.
[tex]xt^2=[LT^2][/tex]
Option 3.
[tex]\dfrac{x}{t^2}=[LT^{-2}][/tex]
Option 4.
[tex]\dfrac{v}{t}=[LT^{-2}][/tex]
So, from above calculations, it is clear that option (1),(3) and (4) have the dimensions of acceleration. Hence, this is the required solution.
Answer:
a)[tex]\dfrac{v^2}{x}[/tex]
c)[tex]\dfrac{x}{t^2}[/tex]
d)[tex]\dfrac{v}{t}[/tex]
Explanation:
Acceleration :
Acceleration is the rate of change of velocity of the particle.
The unit of acceleration is m/s².
In mathematical form,
[tex]a=\dfrac{dv}{dt}[/tex]
We will check the all option on base of unit
(a). [tex]\dfrac{v^2}{x}[/tex]
Where, v = velocity
x = position
The unit of velocity and position are m/s and m.
The dimension formula of velocity and position
[tex]v = LT^{-1}[/tex]
[tex]x=L[/tex]
Put the unit in the given equations
[tex]\dfrac{v^2}{x}=\dfrac{L^2T^{-2}}{L} =\dfrac{L}{T^2}[/tex]
(b). [tex]xt^2=L\times T^2[/tex]
(c). [tex] \dfrac{x}{t^2}=\dfrac{L}{T^2}[/tex]
(d). [tex]\dfrac{v}{t}=\dfrac{LT^{-1}}{T}=\dfrac{L}{T^2}[/tex]
Hence, This is the required solution.
If the initial velocity is 100 m/s at 42.8 ◦ above the horizontal and the initial height of the target is 95.5 m , how high above the ground does the collision take place? The acceleration due to gravity is 9.8 m/s 2 .
Answer:
Explanation:
Since it only mentions the height we only have to concern ourselves with the y direction, up and down, so that's nice. Also, acceleration due to gravity is -9.8 m/s^2
First we want the formula for the position of the object, for that we need the initial speed. We have an initial speed but we need the speed in the y direction. To do that we break the speed we have into its x and y components. Imagine a right triangle where the hypotenuse is 100 and the angle between that and the horizontal leg is 42.8 degrees. Notice that the legs of the right triangle are vertical and horizontal, they represent the y and x components of the speed respectively. Anyway, we want the y component, so the vertical. Relative to the given angle that is the "opposite" side, when we use SOH CAH TOA. Let me know if this doesn't make sense and I can draw it out.
Now, using trig, we can find the y component with SOH. so sin(42.8)=y/100. Make sure your calculator is in degree mode of course. we get
y = 100*sin(42.8) = 67.94 so the vertical speed is 67.94 m/s.
With that now we can use the displacement formula s = vi*t + .5a*t^2 if you don't remember this formula you definitely should have some kind of paper or something with all your physics formula. Anyway, we can plug in all values except for t, which is our variable.
s = 95.5
vi=67.94
a= -9.8
95.5 = 67.94t + .5(-9.8)t^2
Then if you subtract 95.5 from both sides you have a traditional quadratic equation you can solve with the quadratic formula, or completing the square or however you solve them. If you need help with this step let me know, or if something else didn't make sense.
High‑speed ultracentrifuges are useful devices to sediment materials quickly or to separate materials. An ultracentrifuge spins a small tube in a circle of radius 13.1 cm at 50000.0 rev/min. What is the centripetal force experienced by a sample that has a mass of 0.00310 kg?
The force needed to move a body in a curved way is understood as centripetal force. The centripetal force experienced by a sample will be 7.14 N.
What is centripetal force?The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.
The direction of centripetal force is always in the path of the center of the course.
The given data in the problem;
r is the radius =13.1 cm=0.13 m
n is the rpm = 50000.0 rev/min
To find velocity first we have to go to angular velocity
[tex]\omega = \frac{2\pi n}{60} \\\\ \omega = \frac{2\times3.14\times50000}{60} \\\\\omega = 132.63\; rad/sec.[/tex]
The centripetal force is given as
[tex]\rm F_C= m\omega^{2} r\\\\F_C= 0.00310(132.63)^2 \times 0.131\\\\\rm{F_C=7.14 N[/tex]
Hence The centripetal force experienced by a sample will be 7.14 N.
To learn more about the centripetal force refer to the link;
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77. Two forces act on a parachutist falling in air: the force of gravity and air resistance. If the fall is steady, with no gain or loss of speed, then the parachutist is in dynamic equilibrium. How do the magnitudes of the gravitational force and air resistance compare?
Answer:
Explanation:
When a parachutist jump from an aeroplane, there are two forces acting on the parachutist.
One is the force called gravitational force which is acting downwards.
Other is the force of air resistance which is acting upwards.
When the parachute is not opened, the gravitational force is more than the air resistance force due to which the parachutist experiences a force with which he is coming down and thus , there is an acceleration in the body and the velocity of the body goes on increasing.
When the parachute is opened, the gravitational force is balanced by the air resistance force due to which there is no net force acting on the body of parachutist and thus, he comes down with uniform velocity as acceleration in the body is zero.
In this condition, the magnitude of gravitational force is equal to the air resistance force.
A model of a helicopter rotor has four blades, each of length 3.80 m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 450 rev/min.a. What is the linear speed of the blade tip?b. What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity?
Answer:
a)V= 179.056 m/s
b)[tex]a_{r}=843.71 g \ m/s^2[/tex]
Explanation:
Given that
Length of blade = 3.8 m
Rotational speed N= 450 rev/min
We know that
[tex]\omega =\dfrac{2\pi N}{60}\ \frac{rad}{s}[/tex]
[tex]\omega =\dfrac{2\pi \times 450}{60}\ \frac{rad}{s}[/tex]
ω=47.12 rad/s
Linear velocity
We know that linear velocity V = ω x r
Here r = 3.8 m
So by putting the values
V = ω x r
V = 47.12 x 3.8 m/s
V= 179.056 m/s
Radial acceleration
[tex]a_{r}=\omega ^2r\ m/s^2[/tex]
[tex]a_{r}=47.12^2\times 3.8 \ m/s^2[/tex]
[tex]a_{r}=8437.12 \ m/s^2[/tex]
[tex]a_{r}=843.71 g \ m/s^2[/tex]
The linear speed of the blade tip is approximately 178.6 m/s. The radial acceleration of the blade tip is approximately 864 times the acceleration of gravity.
Explanation:The two parts of this question relate to basic concepts in physics, specifically relating to circular motion and the relationship between linear speed, radial acceleration, and the acceleration of gravity. The model of a helicopter rotor in question has four blades, each of length 3.80 m, and rotates at a speed of 450 rev/min.
a. What is the linear speed of the blade tip? The linear velocity (v) of the tip of the blade can be calculated using the formula v = rω, where r is the radial distance (half the length of the blade) and ω is the angular velocity. The angular velocity can be converted to rad/s by multiplying the rotational speed (450 rev/min) by 2π rad/rev and then by 1/60 min/s, giving approximately 47 rad/s. Then, v = (3.8 m) * (47 rad/s) gives a linear velocity of approximately 178.6 m/s.
b. What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity? The radial (or centripetal) acceleration (a_r) can be calculated using the formula a_r = ω²r. Substituting the known values gives a_r = (47 rad/s)² * 3.8 m = approximately 8460 m/s². As the acceleration of gravity is approximately 9.8 m/s², the radial acceleration is therefore about 864 times the acceleration of gravity.
Learn more about Circular motion here:https://brainly.com/question/2285236
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Mike's car, which weighs 1,000 kg, is out of gas. Mike is trying to push the car to a gas station, and he makes the car go 0.05 m/s2. Using Newton's Second Law, you can compute how much force Mike is applying to the car.
Answer:
The answer to your question is: F = 50 N
Explanation:
Data
mass = 1000 kg
acceleration = 0.05m/s2
F = ?
Formula
F = m x a
Substitution
F = 1000 kg x 0.05 m/s2 = 50 kgm/s2 = 50 N
Mike is applying a force of 50 N to the car.
The magnitude of applied force on the car by Mike is 50 N.
Given data:
The mass of car is, m = 1000 kg.
The magnitude of acceleration of car is, [tex]a = 0.05 \;\rm m/s^{2}[/tex].
According to Newton's second law of motion, the force applied on the object is expressed as the product of mass of object and magnitude of acceleration caused by the applied force on the object.
Therefore,
[tex]F = m \times a[/tex]
Here, F is magnitude of applied force on car.
Solving as,
[tex]F = 1000 \times 0.05\\F = 50 \;\rm N[/tex]
Thus, we can conclude that Mike is applying 50 N of force on his car.
learn more about the Newton's second law here:
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An uncharged molecule of DNA (deoxyribonucleic acid) is 2.14 µm long. The ends of the molecule become singly ionized so that there is a charge of −1.6 × 10−19 C on one end and +1.6 × 10−19 C on the other. The helical molecule acts like a spring and compresses 1.4% upon becoming charged. Find the effective spring constant of the molecule. The value of Coulomb’s constant is 8.98755 × 109 N · m2 /C 2 and the acceleration due to gravity is 9.8 m/s 2 . Answer in units of N/m.
Answer:
K = 1.72 *10^{-11} N/m
Explanation:
given data:
L= 2.14*10^{-6} m
[tex]\Delta = 1.4%* of L[/tex]
[tex]= \frac{1.4}{100} *2.14*10^{-6}[/tex]
= 2.99*10^{-8} m
[tex]L' = L - \Delta L[/tex]
[tex]L ' = 2.14*10^{-6} -2.99*10^{-8}[/tex]
[tex]L = 2.11*10^{-6} m[/tex]
electrostatic force = spring force
[tex]\frac{ kq^2}{L'^2} = K \Delta L[/tex]
putting all equation to get spring constant K value
[tex]\frac{8.98*10^9 *1.6*10^{-19}}{2.11*10^{-6}} = K 2.99*10^{-8}[/tex]
K = 1.72 *10^{-11} N/m
The flywheel of a steam engine runs with a constant angular speed of 158 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 2.6 h. What is the magnitude of the constant angular acceleration of the wheel in rev/min^2? Do not enter the units.
Answer:
α=1.013rev/min^2
Explanation:
Wo=initial angular speed=158rev/min
t=time=2.6h=156min
Wf=final angular speed=0rev /min , this happens because in the end the wheel stops
this is a circular motion with uniform acceleration therefore we can use the following equation
α=angular aceleration=(Wf-Wo)/t
α=(0-158)/156=1.013rev/min^2
Vectors Have
A.magnitude only.
B.magnitude and direction.
C.direction.
D.neither magnitude nor direction.
Answer:
B.magnitude and direction
Explanation:
Vector is moved from one place to another.
Example: acceleration
only scalar have magnitude only
Example: speed , temperature
The correct answer is B. Magnitude and direction
Explanation:
In physics, vectors are used to represent motion, forces acting on a body, among other common concepts. A vector is usually represented as a line that connects to different points and has a specific direction. Due to this, all vectors have a magnitude (specific size or length representing a concept) and a direction (position of forces, movements, etc). For example, in physics, it is common to use vectors to represent the forces that act on a body such as gravity or friction and each of these ave specific magnitudes represented with the length of the vector and specific directions depending on the position of the vector. Thus, vectors have both magnitude and direction.
A soccer ball is kicked from the ground with an initial speed of 20.1 m/s at an upward angle of 47.8˚. A player 53.0 m away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? Neglect air resistance.
Answer:3.95 m/s
Explanation:
Given
Initial velocity(u)[tex]=20.1 m/s[/tex]
launch angle[tex]=47.8^{\circ}[/tex]
Player is 53 m away
Range of projectile
[tex]R=\frac{u^2sin2\theta }{g}[/tex]
[tex]R=\frac{20.1^2\times sin95.6}{9.8}[/tex]
R=41.02 m
so he need to run 53-41.02=11.98 m
Time of flight of projectile [tex]=\frac{2usin\theta }{g}[/tex]
T=3.03 s
thus average speed of boy [tex]s_{avg}=\frac{11.98}{3.03}=3.95 m/s[/tex]
A body of mass m moves in a horizontal direction such that at time t its position is given byx(t)=at4+bt3+ct,where a, b, and c are constants. (a) What is the acceleration of the body? (b) What is the time-dependent force acting on the body?
Answer:
The acceleration is given by de second derivative of x(t) which is equal to [tex]\frac{d^f{2}f(x) }{d^{2}x }=12at^{2} + 6bt[/tex] m/s^2
Explanation:
a) We have the equation x(t)=at^4+bt^3+ct which is the position of the body of mass m at a time t
Where a, b and c are constants
From the rules of differenciation we have that the first derivative of the position is the velocity and the second derivative is the acceleration.
Hence the first derivative of the function is equal to [tex]4at^{3} +3bt^{2}+c[/tex][/tex] m/s
Don´t forget to write down the unities
Then we have to derivate again this equation, so we have
[tex][tex]\frac{d^{2}f(x) }{d^{2}x }=12at^{2} + 6bt[/tex] m/s^2[/tex]
b) Remembering the Newton´s laws we know that
[tex]F=ma[/tex]
where:
F is the force
m is the mass
and a is the acceleration
From the first part we know the value of the acceleration which is
[tex]\frac{d^{2}f(x) }{d^{2}x }=12at^{2} + 6bt[/tex] m/s^2
So using the second law formula and replacing the values we have that
F=m([tex]\frac{d^{2}f(x) }{d^{2}x }=12at^{2} + 6bt[/tex] ) N
Remember the that N= Newton which is kg*m/s^2
Which of the following have quantized values? Explain your selections. a. the pitch of a note played on a slide trombone b. the pitch of a note played on a flute c. the wavelengths of light produced by the heating elements in a toaster d. the wind speed at the top of Mt. Everest
Answer:
Option b and c
Explanation:
Quantized values are discrete in nature and are not continuous.
These values can be expressed as integral multiples or integers.
Therefore,
The pitch of notes played on trombone are of continuous form and are not quantized.The pitch of the notes that are played on a flute are dependent on the number of open and closed holes on the instrument, clearly discrete in nature thus quantized.The wavelength of light is produced as a result of transition of electron in between energy levels which occurs in discrete steps thus quantized.The wind speed at Mt. Everest's top is continuous and thus not quantized.Technician A says a power stroke in a modern common rail engine only uses a single injection of fuel. Tecnician B says a late-model common rail diesel has multiple injection events during a combustion cycle. Who is correct?
Answer:
B
Explanation:
In a common rail system, the fuel is distributed to the injectors from the rail where there is high pressure accumulation.High pressure fuel pump feds the rail and with the help of the start and end signals the injector is activated.For example in diesel common rail direct injection systems, the rail is connected to injectors using individual pipes in that the injectors work hand in hand with the fuel pump which ensures fuel injection timing and amount. Comparing this with early models, the fuel pump was responsible for timing, quantity and pressure.So power stroke in a modern common rail system does not only rely on single injection of pump but also has multiple injection events during combution process.
A student at the top of building of height h throws one ball upward with the initial speed V and then throws a second ball downward with the same initial speed. How do the final speeds of the balls compare when they reach the ground?
Answer:
They are the same (assuming there is no air friction)
Explanation:
Take a look at the picture.
When the first ball (the one thrown upward) gets to the point marked as A, the speed will has the exact same value V but the velocity will now point downward (just like the second ball).
So if you think about it, the first ball, from point A to the ground, will behave exactly like the second ball (same initial speed, same height).
That is why the speeds will be the same when they reach the ground.
Answer:
Velocity is same
Explanation:
Case I:
When the ball throws upwards
Let the velocity of the ball as it hits the ground is V'.
Initial velocity, u = V
Final velocity, v = V'
height = h
acceleration due to gravity = g
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
By substituting the values
[tex]V'^{2}=V^{2}+2(-g)(-h)[/tex]
[tex]V'=\sqrt{V^{2}+2gh}[/tex] .... (1)
Case II:
When the ball throws downwards
Let the velocity of the ball as it hits the ground is V''.
Initial velocity, u = V
Final velocity, v = V''
height = h
acceleration due to gravity = g
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
By substituting the values
[tex]V''^{2}=V^{2}+2(-g)(-h)[/tex]
[tex]V''=\sqrt{V^{2}+2gh}[/tex] .... (2)
By comparing the equation (1) and equation (2), we get
V' = V''
Thus, the velocity of balls in both the cases is same as they strikes the ground.
A motorboat maintained a constant speed of 21 miles per hour relative to the water in going 14 miles upstream and then returning. The total time for the trip was 1.5 hours. Use this information to find the speed of the current.
a) 3 miles per hour
b) 5 miles per hour
c) 7 miles per hour
d) 4 miles per hour
e) 2 miles per hour
Answer:
c)7 mph
Explanation:
Going downstream the speed of the boat is added to the speed of the water, on the way back it is opposite to it, then the speed of the water is substracted to that of the boat, and we have the total time of the trip, then we can write down a system of equations as follows:
x:=Speed of water that is being asked
t1:= time going downstream
t2:= time going upstream
(21+x)*t1=14 (1)
(21-x)*t2=14 (2)
t1+t2=1.5 (3)
then t1=1.5-t2, replacing this in (1) we have (21+x)*(1.5-t2)=14, then t2=1.5-(14/(21+x)) (4). Doing the same in (2) we get t2=14/(21-x) (5), this way we can use (4) and (5) and get that:
1.5-(14/(21+x))=14/(21-x), doing the operations we get that 588/(441-x^2)=1.5 or equivalently 1.5x^2-73.5=0 or x^2-49=0 or (x-7)(x+7)=0, then the possible answers are x=7 or x= -7, here we consider the positive answer, that in magnitude is the same as the other, because we already considered the direction of the water into the equations. Then the answer is c) 7mph.
Final answer:
The speed of the current is 3 miles per hour, obtained by setting up an equation based on the total time taken for an upstream and downstream trip and solving for the current's speed. So the correct option is a.
Explanation:
To find the speed of the current, we can set up an equation considering the speed of the boat in still water (which we know is 21 miles per hour) and the speed of the current, which we will call 'c'.
When the boat goes upstream, the effective speed is reduced by the speed of the current, so it's 21 - c miles per hour. Similarly, when the boat goes downstream, the effective speed is increased by the speed of the current, making it 21 + c miles per hour.
Let's calculate the time taken to travel upstream and downstream using the distance and the effective speed. For upstream, it is 14 / (21 - c) hours, and for downstream, it is 14 / (21 + c) hours. The total time for the trip is given as 1.5 hours, so we sum the upstream and downstream times to set up the equation:
14 / (21 - c) + 14 / (21 + c) = 1.5
By solving this equation for 'c', we find that the speed of the current is 3 miles per hour, which corresponds to option (a).
Starting from home, you bicycle 25 km north in 2.4 h and then turn around and pedal straight home in 1.4 h . What are your (a) displacement at the end of the first 2.4 h , (b) average velocity over the first 2.4 h , (c) average velocity for the homeward leg of the trip, (d) displacement for the entire trip, and (e) average velocity for the entire trip
Answer:
Part a)
[tex]d = 25 km[/tex]
Part b)
[tex]v_{avg} = 10.42 km/h[/tex]
Part c)
[tex]v_{avg} = 17.86 km/h[/tex]
Part d)
Displacement for entire trip = 0
Part e)
Average velocity for entire trip will be zero
Explanation:
Part a)
Displacement after t = 2.4 hours is the straight line distance between initial and final positions
so we have
[tex]d = 25 km[/tex]
Part b)
Average velocity is defined as
[tex]v_{avg} = \frac{displacement}{time}[/tex]
[tex]v_{avg} = \frac{25 km}{2.4 h}[/tex]
[tex]v_{avg} = 10.42 km/h[/tex]
Part c)
During his return journey the displacement will be same
[tex]displacement = 25 km[/tex]
[tex]time = 1.4 h[/tex]
so average velocity is defined as
[tex]v_{avg} = \frac{25 km}{1.4 h}[/tex]
[tex]v_{avg} = 17.86 km/h[/tex]
Part d)
Displacement for entire trip = 0
as initial and final position will be same
Part e)
Average velocity for entire trip will be zero
Final answer:
The displacement after 2.4 hours of cycling is 25 km north. The average velocity is 10.42 km/h north for the first 2.4 hours, 17.86 km/h south for the return trip, and the average velocity for the entire trip is zero because the displacement is zero.
Explanation:
To address the questions posed by the student, let's examine each part of the cycling trip:
(a) Displacement at the end of the first 2.4 h
The displacement after the first 2.4 hours is the straight-line distance from the starting point to the turnaround point, which would be 25 km directly north. Displacement is a vector quantity that refers to the change in position of an object.
(b) Average velocity over the first 2.4 h
The average velocity is the displacement divided by the time taken. Since the cyclist traveled 25 km north in 2.4 hours, we calculate the average velocity as 25 km / 2.4 h = 10.42 km/h north.
(c) Average velocity for the homeward leg of the trip
On the way back, the cyclist travels 25 km to the south (back home) in 1.4 h. The average velocity for this would be 25 km / 1.4 h = 17.86 km/h south.
(d) Displacement for the entire trip
Since the cyclist returns straight home, the displacement for the entire trip is zero, because there is no net change in position from the starting point.
(e) Average velocity for the entire trip
The total distance travelled is 25 km out and 25 km back, for a total of 50 km. The total time taken is 2.4 h for the first leg and 1.4 h for the return, which adds up to 3.8 h. However, the displacement is zero, therefore the average velocity for the entire trip is zero.
The electrical resistance of dry skin is 100 kΩ, but can be lowered to 20 Ω if electrode contact area is large and conducting gel is used on the skin. If Vmax of the defibrillatror 500 V and lasts 0.01 s, what is the maximum possible current delivered to the heart during defibrillation?
Answer:
25 Ampers
Explanation:
As the current in a defibrillation is DC (Direct current), we can use simply the Ohm's law which shows:
V=I*R
Then if we want to know the current, we can follow the next step for I:
I=V/R
Using the information, and asumming electrode contact area is large and conducting gel is used on the skin (as is normal) , we can solve:
Imax=Vmax/R
Imax= 500 V / 20 ohms
Imax= 25 Ampers
*Fact: The time in this case doesn't make any difference in the result.
A baseball is thrown at an angle of 20° relative to the ground at a speed of 25 m/s if the ball was caught 50 m from the thrower how long was it in the air ?
2.1 s
0.5 s
10 s
5 s
Answer:
2.1 s
Explanation:
The motion of the ball is a projectile motion. We know that the horizontal range of the ball is
[tex]d = 50 m[/tex]
And that the initial speed of the ball is
[tex]u=25 m/s[/tex]
at an angle of
[tex]\theta=20^{\circ}[/tex]
So, the horizontal speed of the ball (which is constant during the entire motion) is
[tex]u_x = u cos \theta = 25 \cdot cos 20^{\circ} = 23.5 m/s[/tex]
And since the horizontal range is 50 m, the time taken for the ball to cover this distance was
[tex]t=\frac{d}{u_x}=\frac{50}{23.5}=2.1 s[/tex]
which is the time the ball spent in air.
A flask weights 45.4 g when it is empty and 121.8 g when filled with water. When the same flask is filled with another liquid, the mass is 91.39 g. What is the density of the second liquid?
Answer:
The answer to your question is: 1.19 gr/ml
Explanation:
empty flask weighs = 45.4 g
filled with water = 121.8 g
filled with a different substance = 91.39 g
density of the second liquid =
First we must calculate the mass of water in the flask by subtracting the flask filled with water and the flask empty.
mass = flask filled with water - the flask empty.
mass = 121.8 - 45.4 gr
mass = 76.4 gr
The density of water is always 1 gr/ml
and density is mass/volume
from here we calculate the volume = mass/density
volume = 76.4/1 = 76.4 ml
Now we calculate the density of the other liquid
density = 91.39/76.4 = 1.19 g/ml
what is the mass of a dog that weighs 382 N?(unit=kg)
Explanation:
If I assume you are talking about on Earth, then by using the equation
Weight Force=mg
382=(10)m
m=382/10
m=38.2 kg
Note that regardless of whether you are on Earth or any other planet or body the mass of something does not change. Only the weight of it changes as the gravitational acceleration (g) varies from planet to planet.
Answer : The mass of a dog that weighs 382 N is, 38.93 kg
Explanation :
Formula used :
[tex]F=m\times g[/tex]
where,
F = force = 382 N
m = mass of a dog = ?
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Now put all the given values in the above formula, we get:
[tex]382N=m\times 9.8m/s^2[/tex]
[tex]382\frac{kg.m}{s^2}=m\times 9.8m/s^2[/tex]
[tex]m=38.98kg[/tex]
Thus, the mass of a dog that weighs 382 N is, 38.93 kg
If a child pulls a sled through the snow on a level path with a force of 60 N exerted at an angle of 39° above the horizontal, find the horizontal and vertical components of the force.
Explanation:
Given that,
Force with which a child pulls a sled, F = 60 N
It is at an angle of 39° above the horizontal. We need to find the horizontal and vertical components of the force.
The horizontal component is given by :
[tex]F_x=30\ cos(39)=23.31\ N[/tex]
The vertical component is given by :
[tex]F_y=30\ sin(39)=18.87\ N[/tex]
So, the horizontal and vertical components of the force are 23.31 N and 18.87 N. Hence, this is the required solution.
To find the horizontal and vertical components of the force, use the formulas Fx = F * cos(angle) and Fy = F * sin(angle), respectively.
Explanation:To find the horizontal and vertical components of the force, we can use trigonometry. The horizontal component of the force can be found using the formula: Fx = F * cos(angle), where F is the magnitude of the force and angle is the angle above the horizontal. Plugging in the values, we get Fx = 60 N * cos(39°) = 45.85 N.
The vertical component of the force can be found using the formula: Fy = F * sin(angle), where F is the magnitude of the force and angle is the angle above the horizontal. Plugging in the values, we get Fy = 60 N * sin(39°) = 36.34 N.
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Your companion on a train ride through Illinois notices that telephone poles near the tracks appear to be passing by very quickly, while telephone poles in the distance are passing by much more slowly. This is an example of______________.
Answer:
Motion Parallax
Explanation:
According to my research on different scientific anomalies or terminology, I can say that based on the information provided within the question this is an example of Motion Parallax. This term refers to a depth cue where an individual views objects that are closer to us as moving faster than those objects further from us. Which is what the companion in this situation is describing.
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
A straight, nonconducting plastic wire 8.00 cm long carries a charge density of 100 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop.
(a) Find the magnitude and direction of the Electric Field that is produced 6.0cm directly above its midpoint.(b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the Electric field on a point 6.0cm directly above its center.
To find the electric field at a point directly above the midpoint of the plastic wire, use the formula for the electric field due to a charged line. When the wire is bent into a circle, the electric field at a point directly above its center is zero.
Explanation:To find the electric field at a point directly above the midpoint of the plastic wire, you can use the formula for the electric field due to a charged line. The formula is given by E = (k * λ) / r, where E is the electric field, k is the Coulomb's constant (9 * 10^9 Nm²/C²), λ is the charge density, and r is the distance from the wire.
In this case, the charge density is 100nC/m and the distance is 6.0cm. Plug in the values to calculate the electric field. Since the wire is nonconducting, it does not affect the direction of the electric field, so the magnitude is the only value you need to find.
(b) When the wire is bent into a circle lying flat on the table, the electric field at a point directly above its center is zero. This is because the electric field due to each small segment of the wire cancels out by the symmetry of the circular shape.
A car accelerates at a constant rate from zero to 33.7 m/s in 10 seconds and then slows to 17.6 m/s in 5 seconds. What is its average acceleration to the nearest tenth of a m/s2 during the 15 seconds?
Answer:
[tex]a_{avg} = 1.17 m/s^{2}[/tex]
Given:
initial velocity, u = 0
final velocity, v = 33.7 m/s
t = 10 s
final velocity, v' = 17.6 m/s
t' = 5 s
Total time, T = 10 + 5 = 15 s
Solution:
The rate of change of velocity of an object is referred to as the acceleration of that object.
Average accelaeration, [tex]a_{avg} = \frac{\Deta v}{\Delta t}[/tex]
Now,
Initial acceleration of the body, [tex]a = \frac{v - u}{t}[/tex]
[tex]a = \frac{33.7 - 0}{10} = 3.37 m/s^{2}[/tex]
Now, average acceleration during the 15 seconds:
[tex]a = \frac{v' - u}{T}[/tex]
[tex]a = \frac{17.6 - 0}{15} = 1.17 m/s^{2}[/tex]
A solid cylinder with a radius of 5.08 cm starts from rest at the top of a 12.0 meter long ramp inclined 20.3° above the horizontal. When it reaches the bottom of the ramp 3.25 seconds later the cylinder has a final linear velocity of 7.38 m/s .What was the average angular velocity of the cylinder?
Final answer:
The average angular velocity of the cylinder is calculated using the relationship between linear velocity and radius, with the formula ω = v / r. With a final linear velocity of 7.38 m/s and a radius of 5.08 cm, the average angular velocity is about 145.27 rad/s.
Explanation:
To calculate the average angular velocity of the cylinder, we must first understand the relationship between linear velocity and angular velocity. Angular velocity (ω) is related to linear velocity (v) through the radius of the object according to the equation ω = v / r, where r is the radius of the cylinder. Given that the final linear velocity is 7.38 m/s and the radius is 5.08 cm (or 0.0508 meters), we can calculate the average angular velocity.
Calculating the average angular velocity:
Convert the radius to meters: 5.08 cm = 0.0508 m.Use the relationship ω = v / r to find the average angular velocity.ω = 7.38 m/s / 0.0508 m = 145.27 rad/s.Therefore, the average angular velocity of the cylinder is approximately 145.27 rad/s.
diver uses a sonar device to determine the distance to her diving partner. The device sends a sound wave and records the time it takes for the wave to reach the diving partner and return to the device. Suppose the wave travels at a rate of about 4000 feet per second. The wave returns 0.8 second(s) after it was sent. How far away is the diving partner?
Answer:
Assuming the divers are not moving. The diving partner is at 1600 feet.
Explanation:
The sound wave have to go to the diving partner and return, so its traveling the double of the total distance.
Been X= distance, V= speed, and t= time
[tex]x=v*t\\[/tex]
[tex]WaveTravelDistance=4000*0.8\\\\Distance to the partner=WaveTravelDistance/2\\\\WaveTravelDistance=3200feet\\\\Distance to the partner=1600feet[/tex]
The driver of a car traveling at a speed of 20.0 m/s slams on the brakes and comes to a stop in 5.8 s. If we assume that the car's speed changed at a constant rate (constant net force), what was theaverage speed during this 5s interval?
average speed = m/s
Answer:
10 m/s
Explanation:
u = 20 m/s
v = 0 m/s
t = 5.8 s
Let a be the acceleration and s be the distance traveled by the car in time t.
use first equation of motion
v = u + a t
0 = 20 + a x 5.8
a = -3.45 m/s^2
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]0^{2}=20^{2}-2 \times3.45 \times s[/tex]
s = 58 m
The average speed is defined as the ratio of distance traveled to the time taken.
[tex]Average speed = \frac{58}{5.8} = 10 m/s[/tex]
thus, the average speed of the car is 10 m/s.
Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 30 m?
Answer:
[tex]\frac{dA}{dt} = 188.5 m^2/s[/tex]
Explanation:
As we know that area of the circle at any instant of time is given as
[tex]A = \pi r^2[/tex]
now in order to find the rate of change in area we will have
[tex]\frac{dA}{dt} = 2\pi r\frac{dr}{dt}[/tex]
here we know that
rate of change of radius is given as
[tex]\frac{dr}{dt}= 1 m/s[/tex]
radius of the circle is given as
[tex]r = 30 m[/tex]
now we have
[tex]\frac{dA}{dt} = 2\pi (30)(1)[/tex]
[tex]\frac{dA}{dt} = 60\pi[/tex]
[tex]\frac{dA}{dt} = 188.5 m^2/s[/tex]
A 4.33 kg cat has 41.7 J of KE How fast is the cat moving?
Answer:
The answer to your question is:
Explanation:
Data
mass = 4.33 kg
E = 41.7 J
v = ?
Formula
Ke = (1/2)mv²
Clear v from the equation
v = √2ke/m
Substitution
v = √2(41.7)/4.33
v = 19.26 m/s Result
If an automobile had a 100%-efficient engine, transferring all of the fuel's energy to work, would the engine be warm to your touch? Would its exhaust heat the surrounding air? Would it make any noise? Would it vibrate? Would any of its fuel go unused? Discuss.
Answer:
The engine would be warm to touch, and the exhaust gases would be at ambient temperature. The engine would not vibrate nor make any noise. None of the fuel entering the engine would go unused.
Explanation:
In this ideal engine, none of these events would happen due to the nature of the efficiency.
We can define efficiency as the ratio between the used energy and the potential generable energy in the fuel.
n=W, total/(E, available).
However, in real engines the energy generated in the combustion of the fuel transforms into heat (which heates the exhost gases, and the engine therefore transfering some of this heat to the environment). Also, there are some mechanical energy loss due to vibrations and sound, which are also energy that comes from the fuel combustion.
Final answer:
This answer addresses the implications of a hypothetical 100%-efficient engine in an automobile, discussing its heat production, exhaust impact, noise and vibration levels, and fuel usage.
Explanation:
If an automobile had a 100%-efficient engine:
Would the engine be warm to your touch? - Yes, the engine would still produce heat as a byproduct of its operation, even with 100% efficiency, due to thermodynamics.Would its exhaust heat the surrounding air? - The exhaust gases would still carry thermal energy and heat the air around.Would it make any noise or vibration? - While noise and vibration are not related to engine efficiency, they stem from mechanical components and combustion processes.Would any of its fuel go unused? - With 100% efficiency, all the fuel's energy would convert to work, leaving no unused fuel.A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 241 torr. Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature.
What is the mole fraction of hexane? (Assume ideal behavior.)
Answer:
The mole fraction of hexane is 0.67
Explanation:
This problem can be solved using Dalton´s law and Raoult´s law.
The vapor pressure of a mixture of gases is the sum of the partial pressures of each gas (Dalton´s law).
Example:
In a mixture of gases A and B
Pt = PA + PB
where:
Pt = total pressure
PA = partial pressure of A
PB = partial pressure of B
In an ideal solution, the vapor pressure of each component is equal to the vapor pressure of the pure component times the mole fraction of the component in the solution (Raoult´s law).
Example:
In a solution containing A and B, the vapor pressure of A in the solution will be:
PA = P(pure A) * Xa
Where
PA = vapor pressure of A in the solution
P(pure A) = vapor pressure of pure A
Xa = mole fraction of A
In our problem, we have that the vapor pressure of the solution is 241 torr.
Then, using Dalton´s law:
Pt = P(hexane) + P(Pentane)
Using Raoult´s law:
P(hexane) = P(pure hexane) * X(hexane)
P(pentane) = P(pure pentane) * X(pentane)
We also know that the sum of the molar fractions of each component in a solution equals 1:
X(hexane) + X(pentane) = 1
X(pentane) = 1 - X(hexane)
Replacing in the Dalton´s law in terms of X(hexane):
Pt = P(pure hexane) * X(hexane) + P(pure pentane) * (1 - X(hexane))
Solving for X(hexane):
Pt = P(pure hex) * X(hex) + P(pure pent) - P(pure pent) * X(hex)
Replacing with the data:
241 torr = 151 torr * X(hex) + 425 torr - 425 torr*X(hex)
-184 torr = -274 torr * X(hex)
-184torr/-274 torr = X(hex)
X(hex) = 0.67
The mole fraction of hexane in the solution is 1.59.
Explanation:To find the mole fraction of hexane in the solution, we can use Raoult's law, which states that the partial pressure of a component in a solution is equal to the product of the mole fraction of that component and its vapor pressure.
Let's calculate the mole fraction of hexane:
Partial pressure of hexane = mole fraction of hexane * vapor pressure of hexane
241 torr = x * 151 torr
241 torr / 151 torr = x = 1.59
The mole fraction of hexane in the solution is 1.59.