Hi there!
Since the crate is being slid at a constant speed, the forces sum to 0 N. In this instance, the following forces occur in the axis of interest:
Wsinθ = downward acceleration along incline due to gravity (N)
Fκ = kinetic friction force along incline (N)
A = applied force (N)
The acceleration due to gravity and friction force act in the same direction, so:
Wsinθ + Fκ = A
Solve for sinθ using right triangle trigonometry:
sinθ = O/H = 3/6 = 0.5
Rearrange the equation for the force of kinetic friction and solve:
Fκ = A - 0.5W
Fκ = 30.4 - 20 = 10.4 N
Now, recall that:
Work = Force × displacement (W = F × d)
Since the box's displacement is in the same axis as the force but OPPOSITE direction, we must use:
W = Fdcosθ
Angle between displacement and friction force is 180°.
cos(180) = -1
Work done by friction = -Fd = -10.4(6) = -62.4 J
In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.
Find the central force acting on the electron as it revolves in a circular orbit of radius 4.63 x 10-11 m.
Answer in units of N.
The central force acting on the electron as it revolves in a circular orbit is [tex]9.52 \times 10^{-8} \ N[/tex].
The given parameters;
speed of electron, v = 2.2 x 10⁶ m/sradius of the circle, r = 4.63 x 10⁻¹¹ mThe central force acting on the electron as it revolves in a circular orbit is calculated as follows;
[tex]F = \frac{M_e v^2}{r} \\\\[/tex]
where;
[tex]M_e[/tex] is mass of electron = 9.11 x 10⁻³¹ kg
[tex]F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N[/tex]
Thus, the central force acting on the electron as it revolves in a circular orbit is [tex]9.52 \times 10^{-8} \ N[/tex].
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Does the coefficient of kinetic friction depend on the speed?
Answer:
yes because if the speed/K.E. of body increases the friction will also apply more instantly
A father places his daughter in a swing that is 0.60\,\text{m}0.60m0, point, 60, start text, m, end text above ground. Then he raises the swing to a height of 1.3\,\text{m}1.3m1, point, 3, start text, m, end text and lets go. The girl and the swing have a combined mass of 14\,\text{kg}14kg14, start text, k, g, end text. Assume friction is negligible and use g = 9.8\,\text{m/s}^2g=9.8m/s
2
g, equals, 9, point, 8, start text, m, slash, s, end text, squared.
Calculate the girl’s fastest speed.
This question involves the concepts of the law of conservation of energy and kinetic energy.
The girl's fastest speed is "3.7 m/s".
According to the law of conservation of energy, the girl will have the fastest speed at mean position, which will be calculated as follows:
Loss in Potential Energy = Gain in Kinetic Energy
[tex]mg\Delta h=\frac{1}{2}mv^2\\\\v=\sqrt{2g\Delta h}[/tex]
where,
v = maximum speed = ?
g = acceleration due to gravity = 9.81 m/s²
Δh = change in height = 1.3 m - 0.6 m = 0.7 m
Therefore,
[tex]v=\sqrt{2(9.81\ m/s^2)(0.7\ m)}[/tex]
v = 3.7 m/s
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What would be the difference in your welght if you were four times farther from the center of Earth than you are now?
you would die.
the core of the earth is as hot as the sun surface so you would be dead
I need help asap! Will give brainliest
[tex]\\ \tt\hookrightarrow \dfrac{(3\times 10^4kg)(4\times 10^4m)}{6\times 10^4s)}[/tex]
[tex]\\ \tt\hookrightarrow \dfrac{12\times 10^8kgm}{6\times 10^4s}[/tex]
[tex]\\ \tt\hookrightarrow 2kgm/s[/tex]
We have calculated momentum
PLEASE HELP!! 10 points!!!! brainliest!!!
Why is a sound wave considered to be a mechanical wave? What properties of sound
waves are responsible for pitch and volume
Answer:
Mechanical Sound Waves - A sound wave moves through air by displacing air particles in a chain reaction. ... Sound energy, or energy associated with the vibrations created by a vibrating source, requires a medium to travel, which makes sound energy a mechanical wave. For pitch the higher the frequency, the higher the pitch. Frequency is measured in hertz, or cycles per second. Frequency also affects loudness, with higher-pitched sounds being perceived as louder. Amplitude and frequency of sound waves interact to produce the experiences of loudness and pitch.
Hi, please help! •_•
One workman is measured as having a power of 528W. His weight is 800N. He can develop the same power climbing a ladder, whose rungs are 30cm apart. How many rungs can he climb in 5s?
Answer:
He can climb 11 rungs
Explanation:
Power = 528W, Force = 800N, Time= 5s Total Distance moved = 30x cm or 0.3x m, where x is the number of rungs he climbed
Power = Work done/time
Work done = Power × time
= 528 × 5
= 2640Joules
Also Work done = Force × total distance moved
Total Distance moved = Work done/force
0.3x = 2640/800
0.3x = 33
x = 3.3/0.3
x = 11
Please help if you can
Answer:
The answer is B
Explanation:
Because when the both sides aren't balanced one side has to cause motion. (fall down)
Do two indentical objects become statically charged when you rub them together? Explain why they do or do not.
Answer:
Yes, they do. When two different materials are rubbed together, there is a transfer of electrons from one material to the other material. This causes one object to become positively charged and the other object to become negatively charged.
Explanation:
hope this helps
free tingsssssssssss
Explanation:
yayyy thx u soo muchihuhuhu
Answer:
thx
Explanation:
Physicists call any change in energy an impulse true or false?
who has the best answer for this
what percentage more water is used to provide us with electricity vs for irrigation
Answer:
electricity because it has more percentage nd energy
Explanation:
mark me brainliest pl
What did Ernest Rutherford expect to happen when he aimed a beam of particles at a thin gold foil
Answer:
he expected a fire to happen I think
Answer:
When Ernest Rutherford aimed a beam of particles at a thin gold foil he expected that the particles would be deflected slightly after passing through the foil.
Explanation:
hope it helped
Someone answer? Please
write of the following does not represent electrical power in a circuit ?
(a) I²R
(b) IR²
(c)VI
(d) V²/R
Help me pls both answers i’ll give brainly
A new moon is discovered orbiting Neptune with an orbital speed of 9.3 x103 m/s. Neptune's mass is 1.0 x1026 kg. A) What is the radius of the new moon's orbit? B) What is the orbital period? Assume that the orbit is circular. (G = 6.673 x10-11 N-m²/kg?
I am unique even though there are billions of me no two are the same what am I
AND IT'S NOT A FINGER-PRINT
Since every ice crystal takes a different route to the ground. They will float through various clouds with various temperatures and moisture contents, which will cause the ice crystal to develop in a distinctive manner.
What make snowflakes unique?A snowflake's shape changes as it flies through the atmosphere, thus no two are ever the same. Even two flakes floating side by side will each be blown through various vapour and humidity levels to produce a genuinely unique shape.
One of the most recognizable representations of winter weather are snowflakes. There are an unlimited amount of conceivable shapes for snowflakes, and each one is distinct.
Therefore, snowflakes unique even though there are billions.Even two floating flakes will each be exposed to different vapour and humidity conditions, creating a really distinctive form.
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Give proof that mass is constant and weight keeps changing.
will mark brainliest if answer
I won't touch the object because there is a warning sign and that means you stay out if it because it is a danger sign that High Resistance Material that means it is dangerous.
How to find final velocity and initial velocity and acceleration ? :(
Explanation:
Final velocity=Initial velocity+(acceleration×time)
4 ways to find initial velocity:
1) Initial velocity=Final velocity-(acceleration×time)
2) Initial velocity=(Distance/Time)-((acceleration×time)/2)
3) Initial velocity=√Final velocity-(2×(acceleration×distance))
4) Initial velocity=2(distance/time)-Final velocity
Total force = Mass×Acceleration
(F=ma)
What two things related to sedimentary rocks do you see at the Grand Canyon ?
Answer:
sandstone and mudstone
Explanation:
A weightlifter lifts a 350 N set of weights over his head a vertical distance of 2.5 meters above the floor. Calculate the work the athlete does on the weights, assuming he lifts them at a constant speed.
A. 525 J
B. 750 J
C. 1005 J
D. 700 N•m
E. 875 J
A 2-kg object is pulled to the left with a force of 30 N and to the right with a force of 9 N. What is the acceleration of the object?
An object that weighs 75 N is pulled horizontally to the right with a force of 50 N. The force of friction on this object is 30 N to the left. What is the acceleration of the object?
The force on the object is 50 N to the right and 30 N to the left.
⇒ The net force on the object is, =50−30=20 N to the right.
The weight of the object is, =75 N.
⇒ The mass of the object is ==759.8 kg.
⇒ The acceleration of the object is ==20759.8=2.613 m/s 2.
The acceleration of the object is 10 [tex]m/s^2[/tex].
What is Acceleration?Acceleration is defined as the rate of change of velocity of an object with respect to time which are vector quantities i.e. both magnitude and direction. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.
It can be expressed as:
a= v/t
where, a is the acceleration
v is the velocity
t is the time taken
It can also be expressed as F=ma
where, F= force applied
m= mass of the object
So, for above given information
The force on the object is 9N to the right and 30N to the left. The net force on the object is 21N to the left
The mass of the object is 2kg
So, a= 21N/ 2 kg = 10.5 [tex]m/s^2[/tex]
Thus, the acceleration of the object is 10 [tex]m/s^2[/tex].
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Convert 8 m/s to miles per hour
Answer:
17.8954903 or 17.9 mph
Explanation:
m/s is meters per second. 1609.344 meters is one mile. There are 3,600 seconds in an hour. 8 meters *3600 seconds= 28,800. 28,800/1609.344 = 17.8954903. Simplify to 17.9
What net force is needed to increase the velocity of a 5 kg object by 3 m/s over the course of 2.5 seconds?
Hi there!
Recall Newton's Second Law:
[tex]\large\boxed{\Sigma F = ma}[/tex]
∑F = net force (N)
m = mass (kg)
a = acceleration (m/s²)
We must begin by solving for the acceleration using the following:
a = Δv/t
In this instance:
Δv = 3 m/s
t = 2.5 sec
a = 3/2.5 = 1.2 m/s²
Now, plug this value along with the mass into the equation for net force:
[tex]\Sigma F = 5(1.2) = \boxed{6 N}}[/tex]
Answer:
[tex]\boxed {\boxed {\sf 6 \ Newtons}}[/tex]
Explanation:
According to Newton's Second Law of Motion, force is the product of mass and acceleration. The mass of the object is 5 kilograms, but the acceleration is unknown.
We can find acceleration using the following formula:
[tex]a= \frac{ \Delta v}{t}[/tex]
The velocity of the object increases by 3 meters per second. It accelerates in 2.5 seconds.
Δv= 3 m/s t= 2.5 sSubstitute the values into the formula.
[tex]a= \frac{3 \ m/s}{2.5 \ s}[/tex]
[tex]a= 1.2 \ m/s/s = 1.2 \ m/s^2[/tex]
Now we know the mass and acceleration.
m= 5 kg a= 1.2 m/s²Substitute the values into the force formula.
[tex]F=ma[/tex]
[tex]F= (5 \ kg)(1.2 \ m/s^2)[/tex]
[tex]F= 6 \ kg*m/s^2[/tex]
Convert the units. 1 kilogram meter per second squared is equal to 1 Newton. Our answer of 6 kilogram meters per second squared is equal to 6 Newtons.
[tex]F= 6 \ N[/tex]
I need some help with this.
[tex]I = \frac{V}{R} \\ [/tex]
Thus :
[tex]I = \frac{160}{2} \\ [/tex]
[tex]I = 80 \: \: A[/tex]
So :
[tex]V = 160[/tex]
[tex]I = 80[/tex]
[tex]R = 2[/tex]
(a) One medium is air, in which the speed of light is 3.00 x 108 m/s. The other
medium is ice, in which light has a speed of 2.29 x 108 m/s. Use this information to
identify medium A and medium B. Explain
Answer:
WARNING WRONG SPELLING.that is a ternadow and very dangeruse air .Consider a car of mass m moving with an initial speed Vo on a straight, flat road. At time t=0, the driver fully applies the breaks to avoid colliding with debris in the road in front of the car. The car’s wheels lock, causing the car to slide on the roadway until the car stops, before running over the debris. The distance that the car slides is D. The coefficient of kinetic friction between the car’s tires and the roadway is a constant value of Uk.
A) Sketch a free body diagram for the car sliding to rest and write Fnet equations
A) does the value D increase or decrease with increasing initial speed?
B) does the value D increase or decrease with increasing the coefficient of friction?
C) derive an equation for the acceleration of the car in terms of Uk and physical consonants
If an input force of 202N is applied to the handles of the wheelbarrow in the sample problem how large is the output force that just lifts the load?
For this case, the relationship between the two forces is given by:
F1 = nF2
Where,
F1: output strengthF2: input forcen: mechanical advantageThen, replacing the values we have:
F1 = (2.2)(202)
Having the calculations we have:
F1 = 444.4N
Answer: The output force that only lifts the load is F1 = 444.4N.