A 40kg mass is pulled along a surface by a horizontal force of 300N. Friction on the mass is 30N. What is the acceleration of the mass???
Answer:
6.75 m/s²
Explanation:
The following data were obtained from the question:
Mass (m) of object = 40 Kg
Force applied (Fₐ) = 300 N
Force of friction (Fբ) = 30 N
Acceleration (a) =?
Next, we shall determine the net force acting on the mass. This can be obtained as follow:
Force applied (Fₐ) = 300 N
Force of friction (Fբ) = 30 N
Net force (Fₙ) =?
Fₙ = Fₐ – Fբ
Fₙ = 300 – 30
Fₙ = 270 N
Thus, the net force acting on the mass is 270 N.
Finally, we shall determine acceleration of the mass. This can be obtained as follow:
Mass (m) of object = 40 Kg
Net force (Fₙ) = 270 N
Acceleration (a) =?
Net force = mass × acceleration
Fₙ = m × a
270 = 40 × a
Divide both side by 40
a = 270 / 40
a = 6.75 m/s²
Therefore, the acceleration of the mass is 6.75 m/s²
Revolution + tilt = what
what is the force of gravitational attraction between an object with a mass of 0.5kg and another object that had a mass of 0.33kg and a distance between them of 0.002m
Answer:
1.1x10^-4 N
Explanation:
F = G[(m_1*m_2)/r^2]
G = Gravitational constant 6.67433x10^-11 (N*m^2)/(kg^2)
m_1 = mass 1
m_2 = mass 2
r = radius between objects
F = G[(0.5kg*0.33kg)/(0.001m)^2]
F = 1.1x10^-4 N
Explain why there are missing harmonics for standing waves on a string with a loose (free) end. (Hint: Is there a pattern to how much of a wavelength fits between the ends of the string with a loose end
Answer:
It is because there are a odd number of wavelengths fitting onto the string
Explanation:
The general formula for the harmonics on a string loose at one end is f = nv/2L where n = odd integer, v = speed of wave in string and L = length of string.
For a string loose at one end, there are a odd number of wavelengths which can fit onto the string. Since this is the case, we only have harmonics of odd numbers and thus we have even number harmonics missing.
What is the average power developed by a motor as it lifts a 400.-kilogram mass at constant speed through a vertical distance of 10.0 meters in 8.0 seconds
Answer:
The average power developed by a motor is 4905 Watts.
Explanation:
The average power is given by:
[tex] P = \frac{W}{t} [/tex]
Where:
W: is the work
t: is the time = 8.0 s
First, let's find the work:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the displacement = 10.0 m
The force is in the vertical motion, and since the movement of the mass is at constant speed the force is:
[tex] F = mg = 400 kg*9.81 m/s^{2} = 3924 N [/tex]
Hence, the average power is:
[tex] P = \frac{F*d}{t} = \frac{3924 N*10.0 m}{8.0 s} = 4905 J/s = 4905 W [/tex]
Therefore, the average power developed by a motor is 4905 Watts.
I hope it helps you!
The average power developed by a motor as it lifts the mass is 4,900 W.
The given parameters:
mass lifted, m = 400 kgdistance traveled, d = 10 mtime of motion, t = 8.0 sThe average power developed by a motor as it lifts the mass is calculated as follows;
[tex]P = \frac{E}{t} \\\\P = \frac{mgh}{t} \\\\P = \frac{400 \times 9.8 \times 10}{8} \\\\P = 4,900 \ J/s[/tex]
Thus, the average power developed by a motor as it lifts the mass is 4,900 W.
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A steel cable of diameter 3.0 cm supports a load of 2.0kN. What is the fractional length increase of the cable compared with the length when there is no load if Y=2.0 x 10^11 Pa?
Answer:
1.429*10^-5 m
Explanation:
From the question, we are given that
Diameter of the cable, d = 3 cm = 0.03 m
Force on the cable, F = 2 kN
Young Modulus, Y = 2*10^11 Pa
Area of the cable = πd²/4 = (3.142 * 0.03²) / 4 = 0.0028 / 4 = 0.0007 m²
The fractional length = Δl/l
Δl/l = F/AY
Δl/l = 2000 / 0.0007 * 2*10^11
Δl/l = 2000 / 1.4*10^8
Δl/l = 1.429*10^-5 m
Therefore, the fractional length is 1.429*10^-5 m long
A tired squirrel (mass of approximately 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm in order to do a mere 0.50 Joule of work. If the tired squirrel does all this work in 2 seconds, then determine its power.
Answer:
Power = 0.25 Watts
Explanation:
Given:
Work done = 0.50 Joule of work
Time taken = 2.0 second
Find:
Power
Computation:
Power = Work done / Time taken
Power = 0.50 J / 2
Power = 0.25 Watts
If the tired squirrel does all this work in 2 seconds with work of 0.5 J the power is 0.25 Watts.
Given:
Mass, m = 1 kg
Work, W = 0.5 J
Time, t = 2 s
The concept of power, which measures the pace at which work is done or the rate at which energy is transmitted or converted, is crucial to physics. It measures how quickly energy is produced or utilized in a system, to put it simply. In the International System of Units (SI), power is a scalar quantity and is measured in watts (W).
The power is given by:
Power = Work ÷ Time
P = 0.50 Joules ÷ 2 seconds
P = 0.25 Watts
Hence, the power exerted by the tired squirrel is 0.25 Watts.
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Scenario
Andrea is driving her car and getting on Interstate 70. She is traveling 10 m/s at the beginning of the
highway on-ramp and accelerates to 30 m/s at the end of the on-ramp. It takes her 4 seconds to get to
full speed
What is Andrea's acceleration to get on the on-ramp ?
Answer:
5m/s/s
Explanation:
The units for velocity is m/s.
The units for acceleration is m/s/s
So, Determine the change in velocity and divide it by the time that change took place.
(30m/s - 10m/s)/4s = 20m/s/(4s) = 5m/s/s
When energy changes form, energy is...
A) lost.
B) created.
C) not lost or created.
Answer:
C
Explanation:
Energy is neither created or destroyed, but it can change forms...
what is the force on a 1000 kg elevator that is falling freely at 9.8m/s2
Answer:
Since it is falling freely, the only force on it is its weight, w. w = m ⋅ g = 1000kg ⋅ 9.8m s2 = 9800N To draw a Free Body Diagram, draw an elevator cage (I am sure you would get lots of points for drawing it with intricate detail) with a downward force of 9800 N. I hope this helps,
Explanation:
Explanation:
❀ [tex] \underline {{\underline{ \text{Given} }}}: [/tex]
Mass ( m ) = 1000 kgAcceleration ( a ) = 9.8 m/s²❀ [tex] \underline{ \underline{ \text{To \: find}}} : [/tex]
Force ( F )❀ [tex] \underline{ \underline{ \text{Solution}}} : [/tex]
[tex] \boxed{ \sf{force = mass \times acceleration}}[/tex]
Plug the known values :
⟶[tex] \sf{1000 \times 9.8}[/tex]
⟶ [tex] \sf{9800 \: N}[/tex]
[tex] \red{\boxed{ \boxed{ \tt{⟿ \: Our \: final \: answer : 9800 \: N}}}}[/tex]
Hope I helped !♡
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spaceship of mass m travels from the Earth to the Moon along a line that passes through the center of the Earth and the center of the Moon. (a) At what distance from the center of the Earth is the force due to the Earth twice the magnitude of the force due to the Moon
Answer:
the correct result is r = 3.71 10⁸ m
Explanation:
For this exercise we will use the law of universal gravitation
F = [tex]- \frac{m_{1} m_{2} }{r^2}[/tex]
We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m
rocket force -Earth
F₁ = - \frac{m' M }{r^2}
rocket force - Moon
F₂ = - \frac{m' m }{(d-r)^2}
in the problem ask for what point the force has the relation
2 F₁ = F₂
let's substitute
2 [tex]2 \frac{M}{r^2} = \frac{m}{(d-r)^2}[/tex]
(d-r) ² = [tex]\frac{m}{2M}[/tex] r²
d² - 2rd + r² = \frac{m}{2M} r²
r² (1 -\frac{m}{2M}) - 2rd + d² = 0
Let's solve this quadratic equation to find the distance r, let's call
a = 1 - \frac{m}{2M}
a = 1 - [tex]\frac{7.36 10^{22} }{2 \ 5398 10^{24}}[/tex] = 1 - 6.15 10⁻³
a = 0.99385
a r² - 2d r + d² = 0
r = [tex]\frac {2d \frac{+}{-} \sqrt{4d^2 - 4 a d^2}} {2a}[/tex]
r = [2d ± 2d [tex]\sqrt{1-a}[/tex]] / 2a
r = [tex]\frac{d}{a}[/tex] (1 ± √ (1.65 10⁻³)) = [tex]\frac{d}{a}[/tex] (1 ± 0.04)
r₁ = \frac{d}{a} 1.04
r₂ = \frac{d}{a} 0.96
let's calculate
r₁ = [tex]\frac{3.84 10^8}{0.99385}[/tex] 1.04
r₁ = 401.8 10⁸ m
r₂ = \frac{3.84 10^8}{0.99385} 0.96
r₂ = 3.71 10⁸ m
therefore the correct result is r = 3.71 10⁸ m
Is it possible to stand backwards on a flight of stairs?
Answer:
yes
Explanation:
if your destination is on the top floor, but you are facing towards the bottom floor, you are facing the wrong way. It's all relative. Honestly, I hate this question, because then it turns into, is it possible to go backwards anywhere. Backwards and forwards, right and left, it's all relative to which way you are facing and where you want to go. If something is on the left and you turn 180 degrees, then that thing is on the right. If you are going forward towards your house and you turn around, you are now going backwards, relative to your house. But if your perspective is relative to that tree in the opposite direction of your house, you are now facing the right way. See, it's all relative to where you are going.
Example
A 1.8 m tallman stand in an elevator accelerating upward at 12 m/s?,
what is the blood pressure in the brain and foot.
Take the height difference between the heart and the brain to be
0.35 m?
13.3 x 103 Pa] & [Pblood = 1060 kg.m-3]
Note: [: P Heart
Answer:
Explanation:
From the given information; the diagram below shows a clearer understanding.
The blood pressure in the brain [tex]P_{brain} = P_{heart} - \delta ( a - g ) h[/tex]
= 13300 - 1060 (12-9.81) 0.35
= 13300 - 1060 (2.19) 0.35
= 13300 - 812.49
= 12487.51 Pa
The blood pressure in the feet [tex]P_{feet} = P_{heart} + \delta (a + g) h[/tex]
= 13300 + 1060 (12 + 9.81) 1.45
= 13300 + 1060( 21.81 ) 1.45
= 13300 + 33521.97
= 46821.97 Pa
Answer:
The blood pressure in the brain = [tex]12487.51 pa[/tex]The blood pressure in the feet = [tex]46821.97pa[/tex]Explanation:
[tex]P_brain = P_heart - y(a - g)h\\\\P_brain = 13300 - 1060 (12-9.81)0.35\\\\P_brain = 12487.51 pa\\\\[/tex]
[tex]P_feet = P_heart + y(a+g)*h_r\\\\P_feet = 13300 + 1060(12+9.81)*(1.8-0.35)\\\\P_feet = 46821.97pa[/tex]
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is c20h12 organic or not organic
Answer:
.
Explanation:
A motorcycle skids to a stop on a road.
What is the equal and opposite force for the force of the motorcycle's friction pushing on the road as
described by Newton's third law?
Choose 1 answer:
Normal force of the road pushing up on the motorcycle
Earth's force of gravity pulling down on the motorcycle
Motorcycle's force of gravity pulling up on the Earth
Friction of the road on the motorcycle in the opposite direction
Answer:
Friction of the road on the motorcycle in the opposite direction
Explanation:
Khanacademy
Which property describes if a mineral breaks down into flat pieces? this is science
A Cleavage
B Color
C Fracture
D Luster
A stationary 12.5 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.50 and of kinetic friction is 0.30. Show all work including units.
A horizontal force of 15 N is applied to the object.
a. Draw a free body diagram with the forces to scale.
b. Determine the force of friction.
c. Determine the acceleration of the object.
When the object is at rest, there is a zero net force due the cancellation of the object's weight w with the normal force n of the table pushing up on the object, so that by Newton's second law,
∑ F = n - w = 0 → n = w = mg = 112.5 N ≈ 113 N
where m = 12.5 kg and g = 9.80 m/s².
The minimum force F needed to overcome maximum static friction f and get the object moving is
F > f = 0.50 n = 61.25 N ≈ 61.3 N
which means a push of F = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.
So:
(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.
(b) Friction has a magnitude of 15 N because it balances the pushing force.
(c) The object is in equilibrium and not moving, so the acceleration is zero.
Homeostasis refers to the ability of the body to maintain a stable internal environment despite changes in external conditions.
True
False
Answer:
True
Explanation:
A redecor travelling of 94 m/s s lows at a anstant
race to a velocity of 22m Is over Ils How
ar dies it move during this time?
Answer:
638 m.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 94 m/s
Final velocity (v) = 22 m/s
Time (t) = 11 s
Distance (s) =?
We can obtain the distance travelled by using the following formula:
s = (u + v) t /2
s = (94 + 22) × 11 /2
s = 116 × 11 /2
s = 1276 /2
s = 638 m
Thus, the distance travelled is 638 m.
Problem 1. A tugboat exerts a constant force of 4000N toward the right on a ship, moving it
a distance of 15m. What work is done?
Answer:
75,000 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 5000 × 15
We have the final answer as
75,000 JHope this helps you
This is about net force
So the question is what is the net force on the rope? Be sure to include units. So the answer has to be like this ex: 15N left. This isn’t the answer for the first one. And then say which team won so pls help me. Please please help
Answer:
(a) [tex]Net\ Force = 2N[/tex] towards the right
(b) The green team wins
Explanation:
Given
[tex]Blue = 40N[/tex]
[tex]Green = 42N[/tex]
Solving (a): The net force
Because the force in the opposite directions, the net force is the subtraction of the smaller force from the bigger one.
So, we have:
[tex]Net\ Force = Green - Blue[/tex]
Substitute values for Blue and Green
[tex]Net\ Force = 42N- 40N[/tex]
[tex]Net\ Force = 2N[/tex]
Solving (b): Who wins the round
The body will move toward the Green Team because the force larger [tex](42N > 40N)[/tex]
So, the green team will win
EASY!!
A 5000kg truck and a 50 kg person have the same momentum. How is this possible?
Explanation:
The mass of a truck = 5000 kg
Mass of a person = 50 kg
The momentum of a body is given by :
p = mv
m is mass
If momentum of the truck and the person is same. Let v₁ and v₂ are velocity of the truck and the person.
[tex]5000\times v_1=50\times v_2\\\\\dfrac{v_1}{v_2}=\dfrac{50}{5000}\\\\\dfrac{v_1}{v_2}=\dfrac{1}{100}\\\\v_2=100v_1[/tex]
So, it can only possible if the velocity of the person is 100 times of the velocity of the truck.
Sunita uses a lever of length 5m to lift a piece of stone weighing 1000N.
If the distance of the load from the fulcrum is Im, how much effort does she
apply to lift the piece of stone
Answer:
F = 250 [N] (Force exerted by Sunita)
Explanation:
First we must understand the use of the lever, in the attached picture we see that it is fulcrum and other features.
The moment that the load exerts with respect to the fulcrum should be calculated, as the distance is 1 [m].
[tex]M=F*d\\M=1000*1\\M=1000[N*m][/tex]
Now Sunita must exert the same moment using the remaining distance (4 [m]).
[tex]1000=F_{sunita}*4\\F_{sunita}=1000/4\\F_{sunita}=250 [N][/tex]
The matter between stars is called?
A. asteroid belt
B. space
C. interstellar medium
D. Kuiper Belt
Answer:
your answer is C
Explanation:
Which of the following best describes the charge of the nucleus of an atom?
A. The nucleus can have a positive, neutral, or negative charge.
B. The nucleus always has a positive charge.
c. Jahe nucleus always has a neutral charge.
D. The nucleus always has a negative charge.
SUBMIT
On moving a charge of 20 coulombs by 2 cm, 2J of work is done then the potential difference between the points is
Answer:
The potential difference between the points is: 0.1 V
Explanation:
Given
Work W = 2JCharge q₀ = 20 CTo determine
The potential difference between the points is
We can use the formula to determine the potential difference between the points
[tex]V_A-V_B=\frac{W}{q_0}[/tex]
where W is done by moving charge q₀ from point A and point B
substituting W = 2J, and q₀ = 20 C
[tex]=\frac{2}{20}[/tex]
[tex]= 0.1[/tex] V
Therefore, the potential difference between the points is: 0.1 V
Two masses are precisely 1 m apart from each other. The gravitational force each exerts on the other is exactly 1 N. If the masses are identical, what is each mass?
Answer:
122444 Kg
Explanation:
Using the formula for gravitational force;
F= Gm1m2/r^2
Given that
G = constant of gravitation = 6.67 × 10−11 newton-metre2-kilogram−2.
m1 and m2 = the masses
r= distance of separation
Note that the question specifically mentioned that the masses are identical
so m1=m2 = m^2
F= Gm^2/r^2
Fr^2 = Gm^2
m^2 = Fr^2/G
m^2 = 1 × 1^2/6.67 × 10−11:
m =√ 6.67 × 10−11
m= 122444 Kg
What does a light-year measure? A. volume B. weight C. circumference D. distance
Answer:
Distance. A light-year is the distance a single photon of light can travel in empty space in a year.
Answer:
D-Distance
Explanation:
A football player kicks a ball with a mass of .55 kg. The average acceleration of the
football was 14.9 m/s2. How much force did the kicker supply to the football?
A.8.2 m/s2
B.)8.2 N
C.)35.24 N
D.)35.24 m/s2
Answer:
8.2N
Explanation:
I took the test and got it right :)
Length of table is 1.0 metre, 1.00 metre and 1.000 metre. Which one is more accurate?
Answer:
1.000 metre
Explanation:
I hope it's helpful