The average force exerted by the floor on the ball during the collision is zero.
Explanation:To find the average force exerted by the floor on the ball during the collision, we can use the impulse-momentum principle. The impulse experienced by the ball is equal to the change in momentum. Since the ball rebounds back at the same speed, the change in momentum is zero. Therefore, the average force exerted by the floor on the ball is also zero.
Final answer:
The average force exerted by the floor on a 0.5 kg ball bouncing off the floor with a change in velocity from -4 m/s to 4 m/s, over a time of 0.05 s, is 40 N directed upwards.
Explanation:
To calculate the average force exerted by the floor on the ball, you can use the concept of impulse, which is the change in momentum. The change in momentum is equal to the final momentum minus the initial momentum. Since the speed of the ball is the same before and after the bounce, but the direction has changed, the change in velocity is twice the speed of the ball.
The formula for impulse (I) is:
I = Δp = m(v_f - v_i)
Where:
m is the mass of the ball
v_i is the initial velocity of the ball (before the bounce)
v_f is the final velocity of the ball (after the bounce)
Δp is the change in momentum
The impulse given by the floor can also be described by the average force (F_avg) multiplied by the time (t) the force was applied:
I = F_avg ⋅ t
Therefore, we can equate the two:
m(v_f - v_i) = F_avg ⋅ t
By plugging in the values:
(0.5 kg)(4 m/s + 4 m/s) = F_avg ⋅ (0.05 s)
Solving for F_avg gives:
F_avg = ±8.0 N
The positive value indicates that the average force direction is upwards.
An object is dropped from a tower, 400 ft above the ground. The object's height above ground t seconds after the fall is s(t)equals400 minus 16 t squared. Determine the velocity and acceleration of the object the moment it reaches the ground. The velocity of the object the moment it reaches the ground is nothing ft/s.
Answer: v= 160ft/s
a=32ft/s^2 constant
Explanation:
s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.
0=400-16t^2
400=16t^2
25=t^2
t=5s
ds/dt=v(t)=0-32t
dv/dt=a(t)=-32 constant(gravity)
v(t)=-32(5s)= -160ft/s negative sign is only showing direction
The velocity of the object the moment it reaches the ground is 160 ft/s.
What is Velocity?Velocity is defined as the directional speed of an object.How to solve this problem?The problem can be solved by following steps.
s(t) = 400-16t^2 (given)
We know that derivative of position is Velocity v(t) and the derivate of velocity is acceleration a(t)Let s(t)=0
Therefore
0= 400-16t^2
400=16t^2
25=t^2
Therefore
t = 5sec
Now as we know that the derivative of the position is Velocity
so v(t) = ds/dt = -32t
where t = 5sec
substitute the value of t in v(t)
Therefore, v(t) = -32(5) = -160
The direction is negative
Hence the velocity is 160ft/s
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A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6 m, y = 8.5 m, and has velocity ~vo = (9 m/s) ˆı + (−2.5 m/s) ˆ . The acceleration is given by ~a = (4.5 m/s 2 ) ˆı + (3 m/s 2 ) ˆ . What is the x component of velocity after 3.5 s? Answer in units of m/s.
The x component of velocity after 3.5 seconds is 24.75 m/s, calculated using the equation v = v0 + at with the given initial velocity and constant acceleration.
To calculate the x component of velocity after 3.5 seconds for a particle under constant acceleration, we use the equation:
v = v0 + at
Here v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time elapsed. Given v0 = 9 m/s in the x direction, and acceleration a = 4.5 m/[tex]s^2[/tex] in the x direction, and time t = 3.5 s, the calculation is:
v = 9 m/s + (4.5 x (3.5))
The x component of velocity after 3.5 seconds is:
v = 9 m/s + 15.75 m/s
v = 24.75 m/s
A 1.5-kg object has a velocity of 5j m/s at t = 0. It is accelerated at a constant rate for five seconds after which it has a velocity of (6i + 12j ) m/s. What is the magnitude of the resultant force acting on the object during this time interval?
Answer:
2.76 N
Explanation:
m = mass of the object = 1.5 kg
v₀ = initial velocity at t = 0, = 0 i + 5 j
v = final velocity of the object at t = 5, = 6 i + 12 j
t = time interval = 5 sec
a = acceleration of the object = ?
Acceleration of the object is given as
[tex]a = \frac{v - v_{o}}{t}[/tex]
inserting the values
a = ((6 i + 12 j) - (0 i + 5 j))/5
a = (6 i + 7 j)/5
a = 1.2 i + 1.4 j
magnitude of the acceleration is given as
|a| = √((1.2)² + (1.4)²)
|a| = 1.84 m/s²
magnitude of the resultant force is given as
|F| = m |a|
|F| = (1.5) (1.84)
|F| = 2.76 N
To find the magnitude of the resultant force, we use Newton's second law of motion. Evaluating the acceleration at 2.0s gives a magnitude of 24.8 m/s^2. Using the formula F = ma, the magnitude of the resultant force is 37.2 N.
Explanation:To find the magnitude of the resultant force acting on the object during the time interval, we need to use Newton's second law of motion, which states that the force is equal to the mass of the object multiplied by its acceleration.
First, we need to find the acceleration of the object. We can use the formula:
a(t) = 5.0i + 2.0tj - 6.0t^2 km/s^2
By evaluating a(2.0 s), we get a magnitude of 24.8 m/s^2.
Now, we can use Newton's second law:
F = ma
Substituting the values, we get:
F = 1.5 kg * 24.8 m/s^2
F = 37.2 N
Therefore, the magnitude of the resultant force acting on the object during this time interval is 37.2 N.
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If you drop an object from a height of 1.4 m, it will hit the ground in 0.53 s. If you throw a baseball horizontally with an initial speed of 35 m/s from the same height, how long will it take the ball to hit the ground?
Answer:
The ball to hit the ground in 0.53 s.
Explanation:
Given that,
Height = 1.4 m
Time t = 0.53 s
Initial speed = 35 m/s
We need to calculate the time when the ball to hit the ground
Using equation of motion
[tex]s_{y}=u_{y}t-\dfrac{1}{2}gt^2+h_{0}[/tex]
Where, s= vertical height
u= vertical velocity
t = time
h = height
Put the value in equation
[tex]0=0-\dfrac{1}{2}\times9.8\times t^2+1.4[/tex]
[tex]t^2=\dfrac{1.4}{4.9}[/tex]
[tex]t=\sqrt{\dfrac{1.4}{4.9}}[/tex]
[tex]t=0.53\ s[/tex]
Hence, The ball to hit the ground in 0.53 s.
Final answer:
A baseball thrown horizontally with an initial speed will take the same amount of time to fall as a dropped object from the same height, which is 0.53 seconds, because the horizontal speed does not affect the vertical fall time.
Explanation:
The time it takes for an object to fall from a height solely depends on the force of gravity and the initial vertical speed. Since the baseball is thrown horizontally with an initial speed of 35 m/s, this speed does not affect the vertical fall time. The ball will hit the ground in the same duration as any object dropped from the same height without any initial vertical velocity, provided that air resistance is negligible. The previously stated object took 0.53 s to fall from a height of 1.4 m, therefore the baseball will also take 0.53 seconds to hit the ground.
An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end. scaffold weighs 500-N and the worker stands 1.0 m from one end, what is the tension in the rope farther from the worker? (a) 1300 N (b) 1800 N (c) 450 N (d) None of these.
Explanation and answer:
This problem is best answered by drawing a figure as a first step.
ABC is the scaffold.
A downward force of 500N is applied downwards at 1m from end A.
The weight of 800N is exerted by the scaffold uniformly distributed between A & C.
At A and C, ropes are attached to support the load.
Let Fc=tension in rope passing through C.
Take moments about A:
Fc = (500N * 1m +800N*(3+1)/2m / 4m
= (500 Nm + 1600Nm) / 4m
= 2100 Nm / 4m
= 525 N
The magnetic field produced by a long straight current-carrying wire is A) inversely proportional to the current in the wire and proportional to the distance from the wire. B) inversely proportional to both the current in the wire and the distance from the wire. C) proportional to both the current in the wire and the distance from the wire D) proportional to the current in the wire and inversely proportional to the distance from the wire. E) independent of both the current in the wire and the distance from the wire.
Answer:
The magnetic field produced by a long straight current-carrying wire is directly proportional to the current in the wire and inversely proportional to the distance from the wire.
Explanation:
The magnetic field produced by a long straight current-carrying wire is given by :
[tex]B=\dfrac{\mu_0I}{2\pi d}[/tex]............(1)
Where
[tex]\mu_o[/tex] = permeability of free space, [tex]\mu_o=4\pi\times 10^{-7}\ T-m/A[/tex]
I = current flowing in the wire
d = distance from wire
From equation (1), it is clear that the magnetic field produced by a long straight current-carrying wire is directly proportional to the current flowing and inversely proportional to the distance from the wire. So, the correct option is (d).
The magnetic field produced by a long straight current-carrying wire is proportional to the current in the wire and inversely proportional to the distance from the wire, demonstrated by Ampere's Law and the right-hand rule.
Explanation:The magnetic field produced by a long straight current-carrying wire is D) proportional to the current in the wire and inversely proportional to the distance from the wire. This relationship is described by Ampere's Law and the formula B = μI/2πr, where B is the magnetic field strength, μ is the permeability of free space, I is the current in the wire, and r is the distance from the wire. This suggests that as the current increases, the magnetic field strength increases, and as the distance from the wire increases, the magnetic field strength decreases.
Furthermore, the direction of the magnetic field is given by the right-hand rule. If you point the thumb of your right hand in the direction of the current, your fingers will curl in the direction of the magnetic field loops, which form concentric circles around the wire.
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Imagine you had to physically add electrons, one at a time, to a previously neutral conductor. You add one electron very easily, but the second electron requires more work. In your initial post to the discussion, explain why this is. Also, what happens to the work needed to add the third, fourth, fifth, and subsequent electrons
Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of light to get from the Sun to the Earth (in s) is written in scientific notation. Note: the speed of light is approximately 3.0×108 m/s.
The time it takes a beam of light to get from the Sun to the Earth is [tex]4.99\times 10^2secs[/tex]
The formula for calculating the average speed is expressed according to the formula:
[tex]Speed= \frac{distance}{time}[/tex]
Given the speed of light as [tex]3.0 \times 10^8m/s[/tex]
Distance from earth to the sum is [tex]149.6\times 10^9m[/tex]
Substitute the given parameters into the formula to get the time "t"
[tex]t=\frac{d}{t} \\t=\frac{149.6\times10^9m}{3.0\times10^8}\\t=49.87\times 10\\t =498.7secs[/tex]
Hence the time it takes a beam of light to get from the Sun to the Earth is [tex]4.99\times 10^2secs[/tex]
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An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m/s, it then flies a further distance of 40100 m and afterwards its velocity is 47.5 m/s. Find the airplane\'s acceleration and calculate how much time elapses while the airplane covers those 40100 m.
Explanation:
We'll need two equations.
v² = v₀² + 2a(x - x₀)
where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.
x = x₀ + ½ (v + v₀)t
where t is time.
Given:
v = 47.5 m/s
v₀ = 34.3 m/s
x - x₀ = 40100 m
Find: a and t
(47.5)² = (34.3)² + 2a(40100)
a = 0.0135 m/s²
40100 = ½ (47.5 + 34.3)t
t = 980 s
How many electrons does it take to make up 4.33 C of charge?
Answer:
Number of electrons, [tex]n=2.7\times 10^{19}[/tex]
Explanation:
It is given that,
Charge, q = 4.33 C
We need to find the number of electrons that make 4.33 C of charge. According to quantization of charge as :
[tex]q=ne[/tex]
n = number of electrons
e = electron's charge
[tex]n=\dfrac{q}{e}[/tex]
[tex]n=\dfrac{4.33\ C}{1.6\times 10^{-19}\ C}[/tex]
[tex]n=2.7\times 10^{19}[/tex]
So, the number of electrons are [tex]2.7\times 10^{19}[/tex] Hence, this is the required solution.
what is periodicity?
Hello There!
"Periodicity" are types of trends that are seen in element properties.
This is the same thing as periodic trends. These are patterns that are present in the periodic table. These trends show different properties of elements and how characteristics increase or decrease.
An electron traveling horizontally to the right enters a region where a uniform electric field is directed downward. What is the direction of the electric force exerted on the electron once it has entered the electric field?
Answer:
Upward
Explanation:
For charged particles immersed in an electric field:
- if the particle is positively charged, the direction of the force is the same as the direction of the electric field
- if the particle is negatively charged, the direction of the force is opposite to the direction of the electric field
In this problem, we have an electron - so a negatively charged particle - so the direction of the force is opposite to that of the electric field.
Since the electric field is directed downward, therefore, the electric force on the electron will be upward.
A spring has a spring constant of 81 N · m−1. What is the force (in N) required to do the following? (Enter the magnitude.) (a) compress the spring by 6 cm N (b) expand the spring by 17 cm N
Explanation:
It is given that,
Spring constant, k = 81 N/m
We need to find the force required to :
(a) Compress the spring by 6 cm i.e. x₁ = 6 cm = -0.06 m
It can be calculated using Hooke's law as :
F = - k(-x₁)
[tex]F=81\ N/m\times 0.06\ m[/tex]
F = 4.86 N
(b) Expand the spring by 17 cm i.e. x₂ = 17 cm = +0.17 m
So, F = -kx₂
[tex]F=-81\ N/m\times 0.17\ m[/tex]
F = -13.77 N
Hence, this is the required solution.
Final answer:
The force required to compress a spring with a spring constant of 81 N/m by 6 cm is 4.86 N, and the force required to expand the same spring by 17 cm is 13.77 N.
Explanation:
The force required to compress or expand a spring can be determined by Hooke's Law, which states that the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position, and is given by the equation F = kx, where k is the spring constant.
To calculate the force required to:
Compress the spring by 6 cm:
First convert 6 cm to meters (6 cm = 0.06 m). Then apply Hooke's Law: F = kx = 81 N/m times 0.06 m = 4.86 N.
Expand the spring by 17 cm:
First convert 17 cm to meters (17 cm = 0.17 m). Then apply Hooke's Law: F = kx = 81 N/m times 0.17 m = 13.77 N.
In both cases, the magnitude of force is reported, as the question specifies, ignoring the sign which indicates the direction of the force.
How much time would a force of 646 N need to stop a 82 kg object moving at 41 m/s (1 decimal place and no spaces between answer and units).
Answer:
The time is 5.21 s.
Explanation:
Given that,
Force F = 646 N
Mass m = 82 kg
Velocity v = 41 m/s
We need to calculate the acceleration
Using formula of force
F= ma
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{646}{82}[/tex]
[tex]a =7.87\ m/s^2[/tex]
We need to calculate the time
Now, using equation of motion
[tex]v = u+at[/tex]
[tex]41=0+7.87 t[/tex]
[tex]t = \dfrac{41}{7.87}[/tex]
[tex]t = 5.21\ s[/tex]
Hence, The time is 5.21 s.
Choose the letter for the acceleration of the ball during the upward motion after it is released. a) The acceleration is in the negative direction and constant. b) The acceleration is in the negative direction and increasing. c) The acceleration is in the negative direction and decreasing. d) The acceleration is zero. e) The acceleration is in the positive direction and decreasing. ea) The acceleration is in the positive direction and increasing. ea) The acceleration is in the positive direction and constant
Answer:
Option (a)
Explanation:
If a body is thrown upwards, it's velocity goes on decreasing with constant rate. It is because an acceleration is acting on the body which is equal to acceleration due to gravity and acting downwards. The value of acceleration due to gravity is constant and always acting downwards.
A pendulum with a length of 1.5 meters is released from an angle of 20 degrees.What is the period and frequency of this pendulum?
Answer:
The period and frequency of this pendulum are 2.457 s and 0.407 Hz.
Explanation:
Given that,
Length = 1.5 m
Angle = 20°
We need to calculate the period
Using formula of period
[tex]T = 2\pi\sqrt{\dfrac{L}{g}}[/tex]
Where, T = time period
g = acceleration due to gravity
l = length
Put the value into the formula
[tex]T=2\times3.14\sqrt{\dfrac{1.5}{9.8}}[/tex]
[tex]T=2.457\ sec[/tex]
We need to calculate the frequency
[tex]T = \dfrac{1}{f}[/tex]
[tex]f=\dfrac{1}{T}[/tex]
Put the value of T
[tex]f=\dfrac{1}{2.457}[/tex]
[tex]f =0.407\ Hz[/tex]
Hence, The period and frequency of this pendulum are 2.457 s and 0.407 Hz.
A 6.00 V battery has an internal resistance of 0.8322 What is the terminal voltage if it is connected in series to a circuit with a total resistance of 7380 O 5.89 V O 591V 5.87V O 5.99
Answer:
The terminal voltage will be 5.99 volt.
(d) is correct option.
Explanation:
Given that,
Voltage = 6.00
Internal r= 0.8322 ohm
Total resistance R =7380 ohm
We need to calculate the current
Using current formula
[tex]I=\dfrac{V}{R+r}[/tex]
Put the value into the formula
[tex]I = \dfrac{6}{7380+0.8322}[/tex]
[tex]I=0.000812\ A[/tex]
We need to calculate the voltage drop due to internal resistance
[tex]V' = Ir[/tex]
[tex]V'=0.000812\times0.8322[/tex]
[tex]V'=0.00067\ volt[/tex]
Now, The terminal voltage will be
[tex]V''=6-V'[/tex]
[tex]V''=6-0.00067[/tex]
[tex]V''=5.99\ volt[/tex]
Hence, The terminal voltage will be 5.99 volt,
Ammonia can be synthesized according to the equilibrium reaction shown below. If the concentrations of the reactants and products were measured and found to be 0.50 M (N2), 3.00 M (H2), and 1.98 M (NH3), what is the value of the reaction quotient? N2(g) + 3H2(g) --> 2NH3(g) Kc = 0.291 (this reaction is reversible and undergoes equilbrium)
The reaction quotient (Q) for the given reaction, calculated using initial concentrations of the reactants and the product, is approximately 0.092. This value suggests the reaction will move forward, producing more NH3 to reach equilibrium.
Explanation:The reaction quotient, commonly referred to as 'Q', is a value used to determine the direction in which a reaction will proceed. It is calculated similarly to the equilibrium constant but uses the initial concentrations instead. For this reaction, the equation for Q would be [NH3]^2 / ([N2] * [H2]^3) based on the balanced chemical equation.
The initial concentrations given in the question are 0.50 M for N2, 3.00 M for H2, and 1.98 M for NH3. To find Qc, substitute these concentrations into our Q equation to get (1.98)^2 / (0.50 * 3.00^3), which simplifies to approximately 0.092.
If Qc < Kc, the reaction will proceed in the forward direction to reach equilibrium, so in this case, since our Qc (0.092) is less than Kc (0.291), the reaction will produce more NH3 to reach equilibrium.
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A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water. Remember, the density of water is 1.0 g/m
Answer:
335°C
Explanation:
Heat gained or lost is:
q = m C ΔT
where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Heat gained by the water = heat lost by the copper
mw Cw ΔTw = mc Cc ΔTc
The water and copper reach the same final temperature, so:
mw Cw (T - Tw) = mc Cc (Tc - T)
Given:
mw = 390 g
Cw = 4.186 J/g/°C
Tw = 22.6°C
mc = 248 g
Cc = 0.386 J/g/°C
T = 39.9°C
Find: Tc
(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)
Tc = 335
Data obtained from the question
Mass of copper (M꜀) = 248 g
Volume of water = 390 mL
Density of water = 1 g/mL
Initial temperature of water (Tᵥᵥ) = 22.6 °C
Equilibrium temperature (Tₑ) = 39.9 °C
Initial temperature of copper (T꜀) =? Step 2:Determination of the mass of water
Volume of water = 390 mL
Density of water = 1 g/mL
Mass of water =?[tex]Density = \frac{mass}{volume}\\\\1 = \frac{mass}{390}[/tex]
Cross multiply
[tex]Mass = 1 * 390[/tex]
Mass of water = 390 gStep 3:Determination the initial temperature of the copper.
Mass of copper (M꜀) = 248 g
Mass of water (Mᵥᵥ) = 390 g
Initial temperature of water (Tᵥᵥ) = 22.6 °C
Equilibrium temperature (Tₑ) = 39.9 °C
Initial temperature of copper (T꜀) =? NOTE:1. Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
2. Specific heat capacity of copper (C꜀) = 0.385 J/gºC
Heat lost by copper = heat gained by water
[tex]Q_{c} = Q_{w} \\ \ M_{c} C_{c}(T_{c}-T_{e}) = M_{w} C_{w}(T_{e}-T_{w})\\248* 0.385(T_{c}-39.9) = 390*4.184(39.9-22.6)\\95.48(T_{c}-39.9) = 1631.76*17.3\\95.48(T_{c}-39.9) = 28229.448[/tex]
Divide both side by 95.48
[tex]T_{c} - 39.9 = \frac{28229.448}{95.48}\\T_{c} - 39.9 = 295.658[/tex]
Collect like terms
[tex]T_{c} = 295.658 + 39.9[/tex]
T꜀ = 335.6 °CTherefore, the initial temperature of the piece of copper is 335.6 °C.
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A 320.9 ng sample of an unknown radioactive substance was placed in storage and its mass measured periodically. After 47 days the amount of radioactive substance had decreased to 40.11 ng. How many half‑lives of the unknown radioactive substance have occurred?
Answer:
3 half life of the unknown radioactive substance have occurred.
Explanation:
Mass of sample = 320.9 ng
Mass after 1 half life = 0.5 x 320.9 = 160.45 ng
Mass after 2 half life = 0.5 x 160.45 = 80.225 ng
Mass after 3 half life = 0.5 x 80.225 =40.11 ng
So 3 half life of the unknown radioactive substance have occurred.
The unknown radioactive substance went through approximately 3 half-lives in the 47 days period.
Explanation:The subject of your question is related to radioactive decay, particularly the concept of a half-life, which is a term in physics used to describe the time it takes for half the atoms in a sample to decay. In your case, the initial mass of your substance was 320.9 ng and it decreased to 40.11 ng after 47 days. To figure out how many half-lives have occurred, we need to understand that with each half-life, the quantity of the substance halve its original mass.
A useful method to answer your question is to divide the final amount by the starting amount and then take the logarithm base 2 of the result. This will give us the number of times the amount halved, which is the number of half-lives. In your case, the calculation is like this: number of half lives = log2(320.9 ng / 40.11 ng) = approx. 3 half-lives.
This means that in 47 days, approximately 3 half‑lives of the unknown radioactive substance have occurred, based on the given data. It's important to note that because this is an approximate result, the true half-life of the substance might be slightly smaller or larger.
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If you fell 90ft while sitting down strapped in an elevator (think tower of terror ride) would you die when the elevator hit the floor? Would the power be enough? Not for school just curious
Probably not. The elevator has security system just like any device. It will automatically trigger breaks in case it detects some sort of free fall.
Person inside free falling elevator could experience 0G or no gravity effect something similar to what astronauts experience in earth's orbit.
The only damage that could happen to you is when the breaks are released you won't feel 0G anymore and fall about 4ft to the floor of the elevator.
Hope this helps.
r3t40
How long is a string under 240 N of tension whose mass is 0.086 kg if a wave travels through it at a speed of 12 m/s?
Answer:
The length of the string is 0.051 meters
Explanation:
It is given that,
Tension in the string, T = 240 N
Mass of the string, m = 0.086 kg
Speed of the wave, v = 12 m/s
The speed of the wave on the string is given by :
[tex]v=\sqrt{\dfrac{T}{M}}[/tex]
M is the mass per unit length of the string i.e. M = m/l.......(1)
So, [tex]M=\dfrac{T}{v^2}[/tex]
[tex]M=\dfrac{240\ N}{(12\ m/s)^2}[/tex]
M = 1.67 kg/m
The length of the string can be calculated using equation (1) :
[tex]l=\dfrac{m}{M}[/tex]
[tex]l=\dfrac{0.086\ kg}{1.67\ kg/m}[/tex]
l = 0.051 m
So, the length of the string is 0.051 meters. Hence, this is the required solution.
Which of the following is NOT an example of an assembly process? a) Handling b) Fitting c) Automated robot d) Orientation
Answer: I think it is D
Explanation: It is the only one that does not make sense
Final answer:
The correct answer is 'c) Automated robot' because it is a tool used within the assembly process, not a process itself like handling, fitting, or orientation.
Explanation:
The question is asking which of the provided options is not an example of an assembly process. An assembly process involves the steps required to put together parts to make a complete product. The options given are:
Handling - the act of manipulating components in preparation for assembly.
Fitting - the process of putting parts together, which is certainly part of assembly.
Automated robot - often used in assembly to perform repetitive tasks more efficiently.
Orientation - this typically refers to aligning parts in the correct position for assembly.
Based on these definitions, automated robot is not an example of an assembly process but is rather a tool that might be used in the process. Therefore, the correct answer is 'c) Automated robot'
In a super-heater (A) pressure rises, temperature drops (B) pressure rises, temperature remains constant (C) pressure remains constant and temperature rises (D) both pressure and temperature remains constant
Answer:
i believe that it is d
Explanation:
In a super heater, the temperature of the steam rises while the pressure remains constant. This process helps to remove the last traces of moisture from the saturated steam.
Explanation:In a super heater, the conclusion is that option (C) pressure remains constant and temperature rises is the correct choice. A super heater is a device used in a steam power plant to increase the temperature of the steam, above its saturation temperature. The function of the super heater is to remove the last traces of moisture (1 to 2%) from the saturated steam and to increase its temperature above the saturation temperature. The pressure, however, remains constant during this process because the super heater operates at the same pressure as the boiler.
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A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vertical.
(a) How far does the mass fall (y-displacement) before reaching its lowest point?
(b) How much work is done by gravity as it falls to its lowest point?
(c) How much work is done by the string tension as it falls to its lowest point?
(a) -0.211 m
At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is
[tex]\theta=65.4^{\circ}[/tex]
The projection of the length of the pendulum along the vertical direction is
[tex]L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm[/tex]
the full length of the pendulum when the mass is at the lowest position is
L = 36.1 cm
So the y-displacement of the mass is
[tex]\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m[/tex]
(b) 0.347 J
The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to
[tex]\Delta U = mg \Delta y[/tex]
where we have
m = 168 g = 0.168 kg is the mass of the pendulum
g = 9.8 m/s^2 is the acceleration due to gravity
[tex]\Delta y = 0.211 m[/tex] is the vertical displacement of the pendulum
So, the work done by gravity is
[tex]W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J[/tex]
And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.
(c) Zero
The work done by a force is:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and the displacement
In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula
[tex]\theta=90^{\circ}, cos \theta = 0[/tex]
and so the work done is zero.
The pendulum falls 0.212 m, the work done by gravity is 0.349 J, and the work done by string tension is 0 J.
Explanation:The first part of the question asks for the vertical displacement (y-displacement) of the pendulum. The length of the pendulum is the hypotenuse of a right triangle, and the vertical displacement is the adjacent side, so we can use the cosine function to solve: y = L*cos(θ). Plugging in the given values: y = 0.361 m * cos(65.4 degrees) = 0.149 m. So the fall is the length of the pendulum minus this displacement: 0.361 m - 0.149 m = 0.212 m.
The second part of the question asks for the work done by gravity. The work done by gravity is equal to the weight of the pendulum times the vertical distance it falls (Work = m*g*y), or 0.168 kg * 9.8 m/s² * 0.212 m = 0.349 J.
The final part of the question asks for the work done by string tension. The tension force always acts perpendicular to the direction of displacement, meaning it does no work on the pendulum, as work is defined as force times the displacement in the direction of the force. Therefore, the work done by the tension in the string is 0 J.
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A sample of gas in a balloon has an initial temperature of 23 ∘C and a volume of 1.09×103 L . If the temperature changes to 59 ∘C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?
Answer:
[tex]1.22\cdot 10^3 L[/tex]
Explanation:
We can solve the problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas is directly proportional to its temperature:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where here we have
[tex]V_1 = 1.09\cdot 10^3 L[/tex] is the initial volume
[tex]T_1 = 23^{\circ}+ 273 = 296 K[/tex] is the initial temperature
[tex]V_2[/tex] is the final volume
[tex]T_2 = 59^{\circ}+ 273 =332 K[/tex] is the final temperature
Solving for V2, we find
[tex]V_2 = \frac{V_1 T_2}{T_1}=\frac{(1.09 \cdot 10^3 L)(332 K)}{296 K}=1.22\cdot 10^3 L[/tex]
An electric heater is rated at 1400 W, a toaster is rated at 1150 W, and an electric grill is rated at 1560 W. The three appliances are connected in parallel across a 112 V emf source. Find the current in the heater.
Answer:
The current in the heater is 12.5 A
Explanation:
It is given that,
Power of electric heater, P₁ = 1400 W
Power of toaster, P₂ = 1150 W
Power of electric grill, P₃ = 1560 W
All three appliances are connected in parallel across a 112 V emf source. We need to find the current in the heater. We know that in parallel combination of resistors the current flowing in every branch of resistor divides while the voltage is same.
Electric power, [tex]P_1=V\times I_1[/tex]
[tex]I_1=\dfrac{P_1}{V}[/tex]
[tex]I_1=\dfrac{1400\ W}{112\ V}[/tex]
[tex]I_1=12.5\ A[/tex]
So, the current in the heater is 12.5 A. Hence, this is the required solution.
An electrical device draws 4.68 A at 220 V. (a) If the voltage drops by 31%, what will be the current, assuming the resistance doesn't change?
Answer:
The current will be 3.23 A.
Explanation:
Given that,
Current I = 4.68 A
Voltage V = 220 volt
We need to calculate the resistance
Using ohm's law
[tex]V = I R[/tex]
[tex]R = \dfrac{V}{I}[/tex]
Where,
V = voltage
I = current
R = resistance
Put the value into the formula
[tex]R = \dfrac{220}{4.68}[/tex]
[tex]R = 47\ \Omega[/tex]
We need to calculate the current
If the voltage drops by 31%
Voltage will be
[tex]V'=V-V\times31%[/tex]
[tex]V'=220-220\times\dfrac{31}{100}[/tex]
[tex]V'=151.8\ volt[/tex]
Now, the current will be
[tex]I = \dfrac{151.8}{47}[/tex]
[tex]I=3.23\ A[/tex]
Hence, The current will be 3.23 A.
A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.1 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 6 ft from the wall.)
Explanation:
If the distance between the bottom of the ladder and the wall is x, then:
cos θ = x / 10
Taking derivative with respect to time:
-sin θ dθ/dt = 1/10 dx/dt
Substituting for θ:
-sin (acos(x / 10)) dθ/dt = 1/10 dx/dt
Given that x = 6 and dx/dt = 1.1:
-sin (acos(6/10)) dθ/dt = 1/10 (1.1)
-0.8 dθ/dt = 0.11
dθ/dt = -0.1375
The angle is decreasing at 0.1375 rad/s.
What is the power of a motor that can accelerate a 1700 kg car from rest to 30.0 m S in 5.0 seconds?
Answer:
The power of the motor is 153000 watts.
Explanation:
It is given that,
Mass of the car, m = 1700 kg
Initially, it is at rest, u = 0
Final velocity of the car, v = 30 m/s
Time taken, t = 5 s
We need to find the power of a motor. Work done per unit time is called power of the motor. We know that the change in kinetic energy is equal to the work done i.e.
[tex]P=\dfrac{W}{t}=\dfrac{\Delta E}{t}[/tex]
[tex]P=\dfrac{\dfrac{1}{2}mv^2}{t}[/tex]
[tex]P=\dfrac{\dfrac{1}{2}\times 1700\ kg\times (30\ m/s)^2}{5\ s}[/tex]
P = 153000 watts
So, the power of the motor is 153000 watts. Hence, this is the required solution.