Answer:
α = 1930.2 rad/s²
Explanation:
The angular acceleration can be found by using the third equation of motion:
[tex]2\alpha \theta=\omega_f^2-\omega_i^2[/tex]
where,
α = angular acceleration = ?
θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s
Therefore,
[tex]2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}[/tex]
α = - 1930.2 rad/s²
negative sign shows deceleration
A gymnast weighs 450 N. She stands on a balance beam of uniform construction which weighs 250 N. The balance beam is 3.0 m long and is supported at each end. If the support force at the right end is four times the force at the left end, how far from the right end is the gymnast
Answer:
x = 9.32 cm
Explanation:
For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation
Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar
- W l / 2 - W_{child} x + N₂ l = 0
x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex] 1)
now let's use the expression for translational equilibrium
N₁ - W - W_(child) + N₂ = 0
indicate that N₂ = 4 N₁
we substitute
N₁ - W - W_child + 4 N₁ = 0
5 N₁ -W - W_{child} = 0
N₁ = ( W + W_{child}) / 5
we calculate
N₁ = (450 + 250) / 5
N₁ = 140 N
we calculate with equation 1
x = -250 1.50 + 4 140 3) / 140
x = 9.32 cm
An object moving with a constant
acceleration changes its velocity from
10ms' to 20 ms' in five seconds. What is the
distance travelled in five seconds
Answer:
Acceleration:
[tex]{ \tt{a = \frac{v - u}{t} }} \\ { \tt{a = \frac{20 - 10}{5} }} \\ { \tt{a = 2 \: m {s}^{ - 2} }}[/tex]
From third equation:
[tex]{ \bf{ {v}^{2} = {u}^{2} + 2as}} \\ { \tt{s = \frac{ {20}^{2} - {10}^{2} }{2 \times 2} }} \\ = { \tt{s = 75 \: m}}[/tex]
Answer:
Formula = m/s
Explanation:
The answer is 10 m / 5 seconds = 2 meters distance
The answer is 20 m / 5 seconds = 4 meters distance
A cycle track is 500 metres long. A cyclist completes 10 laps (that is, he rides completely round the track 10 times).
a) How many kilometres has the cyclist travelled?
b) On average it took the cyclist 50 second to complete one lap (that it, to ride round just one).
(i) What was the average speed of the cyclist?
(ii) How long in minutes and seconds dit it take the cyclist to complete the 10 laps?
c) Near the end of the run the cyclist put on a spurt. During this spurt it took the cyclist 2 seconds to increase speed from 8 m/s to 12 m/s. What was the cyclist's acceleration during this spurt?
Explanation:
a) D_t = 500m*10laps
D_t = 5000m or 5km
b)
I) v = d/t
v = 500m/50s
v = 10m/s
ii) 10m/s = 5000m/t
t = 5000m/10m/s
t = (500s)*(1min/60s)
t = 8'20" or 8 mins and 20 sec
c) v_f = v_0 + a*t
v_f-v_0 = a*t
a = (v_f-v_0)/t
a = (12m/s-8m/s)/2s
a = (4m/s)/2s
a = 2m/s²
Part(a),
The distance travelled by the cyclist is 5 km.
Part(b),
(i) The average speed is 10 m/s
(ii) The time taken to cover 10 laps is 8 minutes and 20 seconds.
Part(c),
The acceleration is 2 m/s²
What is speed?Speed is defined as the ratio of the time distance travelled by the body to the time taken by the body to cover the distance. Speed is the ratio of the distance travelled by time. The unit of speed in miles per hour.
a) The distance will be calculated as
D = 500m*10laps
D = 5000m or 5km
b) The average speed is calculated as,
I) v = d/t
v = 500m/the 50s
v = 10m/s
The time will be calculated as,
ii) 10m/s = 5000m/t
t = 5000m/10m/s
t = (500s)*(1min/60s)
t = 8'20" or 8 mins and 20 sec
The acceleration is calculated as,
c) vf = v0 + at
vf-v0 = at
a = (vf-v0)/t
a = (12m/s-8m/s)/2s
a = (4m/s)/2s
a = 2m/s²
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1 hallar el trabajo mecanico de un cuerpo que tiene una fuerza de 250 newton y recorre 750 metros
2 hallar la potencia necesaria para levantar un transformador de masa 2500kg,una altura de 4 metros en un tiempo de 30 segundos
porfa es para hoy
Answer: TRACK
Explanation:
what is meant by specific latent heat of vaporization of water is -2.26mjkg^-1 or -2.26mj/kg?
Answer:
The specific latent heat of a substance is the amount of energy required to change the state of one kilo of the substance without change in it temperature.The latent heat of vaporization or evaporation is the heat given to some mass to convert if from the liquid to the vapor phase.
Liquid plastic is frozen in a physical change that increases its volume. What can be known about the plastic after the change?
(A) Its mass will increase.
(B) Its density will increase.
(C) Its mass will remain the same.
(D) Its density will remain the same.
Answer:
c
Explanation:
Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is option C.
What is the matter?Anything which has mass and occupies space is known as matter ,mainly there are four states of matter solid liquid gases, and plasma.
These different states of matter have different characteristics according to which they vary their volume and shape.
It is known about plastic that its mass will remain the same when liquid plastic is frozen, by increasing its volume.
Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is C.
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00
Use base units to check whether the following equations are balance
(a) pressure = depth x density gravitational field strength,
(b) energy = mass x (speed of light).
dod to molt solid
Answer:
In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.
The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid, or p = p 0 + ρ h g , p = p 0 + ρ h g , 14.4
Granite: 2.70 × 10 32.70 × 10 3
Lead: 1.13 × 10 41.13 × 10 4
Iron: 7.86 × 10 37.86 × 10 3
Oak: 7.10 × 10 27.10 × 10 2
If the mass of an object is 15 kg and the velocity is -4 m/s, what is the momentum?
momentum p= m x v = 15 x -4 = -60 N.s
A bicyclist moves along a straight line with an initial velocity vo and slows downs. Which of the following the best describes the signs set for the initial position, initial velocity and the acceleration ?
The sign set after the slowdown of the bicycle will be positive for the position, negative for velocity, and negative for acceleration.
What is velocity?The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.
According to Que, when a bicyclist moves in a straight line and slows down, then the velocity decrease as displacement is decreasing, and the acceleration also decreases only displacement increases.
Therefore, the sign set for the position is +ve, for velocity it is -ve, and for acceleration also -ve
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A biker slows down after traveling in a long, straight line at initial velocity v0. Which of the following the best \sdescribes the signs set for the initial position, initial velocity and the acceleration? Initial position Initial velocity Acceleration
A. Positive Negative Negative
B. Positive Positive Negative
C. Negative Positive Negative
D. Negative Negative Positive
E. Negative Negative Negative
Flag question
Consider the pressure and force acting on the
dam retaining a reservoir of water. Suppose the
dam is 500-m wide
and the water is 80.0-m
deep at the dam, as illustrated below. What is
the average pressure on the dam due to the
water?
Answer:
P = density (p) * g * h
P = 1000 kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2
since kg m / s^2 = Newtons
The average pressure is 1/2 (pressure at 0m + pressure 80 m) for liquid of uniform density
what additional load will be required to cause the extension of 2.0cm when an elastic wire extend by 1.0cm when a load of 20g range from it
Answer:
The additional load is 20g
given A=4i-10j and B= 7i+5j find b such that A+bB is a vector pointing along the x-axis (i.e has no y component)
Answer:
-4/7
Explanation:
Given the following
A=4i-10j and B= 7i+5j
A+ bB = 4i-10j + (7i+5j)b
A+ bB = 4i-10j + 7ib+5jb
A+ bB =
The vector along the x-axis is expressed as i + 0j
If the vector A+ bB is pointing in the direction of the x-axis then;
[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]
Hence the value of b is -4/7
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
According to the statement, we have following system of vectorial equations:
[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)
[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)
[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)
By applying (1) and (2) in (3):
[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]
[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]
And we get two scalar equations after analyzing each component:
[tex]4+7\cdot \beta = c[/tex] (4)
[tex]-10+5\cdot \beta = 0[/tex] (5)
We solve for [tex]\beta[/tex] in (5):
[tex]\beta = 2[/tex]
And for [tex]c[/tex] in (4):
[tex]c = 4+7\cdot (2)[/tex]
[tex]c = 18[/tex]
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
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27. The part of the Earth where life exists .
Mesosphere
Stratosphere
Troposphere
Biosphere
Answer:
Biosphere is the part of the earth where life exists.
Physics question plz help ASAP
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88°C I am not 100% sure this is right but I am 98% sure this IS right
Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.
Compute the PE when the spring is compressed by 0.50 m.
Answer
E - 1/2 K x^2 potential energy of compressed spring
E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m
A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 40.0 N directed at an angle of 35.0deg below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
Answer:
N = 107.94 N
Explanation:
For this exercise we must use Newton's second law.
Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical
X axis
Fₓ = ma
ej and
N -F_y - W = 0
let's use trigonometry to decompose the applied force
cos -35 = Fₓ / F
sin -35 = F_y / F
Fₓ = F cos -35
F_y = F sin -35
Fₓ = 40.0 cos -35 = 32.766 N
F_y = 40.0 sin -35 = -22.94 N
we substitute
N = Fy + W
N = 22.94 + 85
N = 107.94 N
a coach is travelling east wards at 12.6 m/s after 12 second its velocity is 9.5 m/s in the same direction. what is the acceleration and direction of its acceleration?
pls do it with the formula
thx mates :)
[tex]\\ \rm\longmapsto a=\dfrac{v-u}{t}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{12.6-9.5}{12}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{3.1}{12}[/tex]
[tex]\\ \rm\longmapsto \overrightarrow{a}=0.25m/s^2[/tex]
Where is the center of mass of homogeneous body which has a regular
Following the definition of the center of mass, "In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero."
(see explanation below)
Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 2.0×1092.0×109 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.
Use two significant figures in your answer, expressed in kilometers.
Answer:
26 km
Explanation:
Let's say our "cable" has a cross section of 1 m²
Then each meter of cable would weight 7900(9.8) = 77420 N
A Pascal is a Newton per square meter
2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles
Megan accelerates her skateboard from 0 m/s to 8 m/s in 2 seconds. What is the magnitude of the acceleration of the skateboard?
O 8 m/s^2
O 16 m/s^2
O 2 m/s^2
O 4 m/s^2
Answer:
chk picture for eqn
Explanation:
A 55kg bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion . Find the ( magnitude and direction ) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.1m/s2
Answer:
148.5 N
Explanation:
Given that,
The mass of a bungee jumper, m = 55 kg
The downward acceleration, a = 7.1 m/s²
We need to find the net force acting on the jumper. As it is moving in downward direction, net force is :
T = m(g-a)
Put all the values,
T = 55(9.8 - 7.1)
= 148.5 N
So, the force exerted on the bungee cord is 148.5 N.
Answer:
The downward force is 148.5 N.
Explanation:
mass, m = 55 kg
downwards acceleration, a = 7.1 m/s^2
Let the force is F.
According to the newton's second law
m g - F = m a
F = m (g - a)
F = 55 (9.8 - 7.1)
F = 148.5 N
what is the value of x if x-36=5?
Answer:
Therefore, the value of x is 41
Explanation:
x=5+36
x=41
A 10 n force is applied horizontally on a box to move it 10 m across a frictionless surface. How much work was done to move the box?
Answer:
[tex]\boxed {\boxed {\sf 100 \ J}}[/tex]
Explanation:
We are asked to calculate the work done to move a box.
Work is the product of force and distance or displacement.
[tex]W= F*d[/tex]
A 10 Newton force is applied horizontally on the box. Since the surface is frictionless, there is no force of friction, and the net force is 10 Newtons. The force moves the box 10 meters.
F= 10 N d= 10 mSubstitute the values into the formula.
[tex]W= 10 \ N * 10 \ m[/tex]
Multiply.
[tex]W= 100 \ N*m[/tex]
Let's convert the units. 1 Newton meter is equal to 1 Joule, therefore our answer of 100 Newton meters is equal to 100 Joules.
[tex]W= 100 \ J[/tex]
100 Joules of work was done to move the box.
A 771.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 300.0 K. How much heat must the smelter produce to completely melt the copper bar? For solid copper, the specific heat is 386 J/kg • K, the heat of fusion is 205 kJ/kg, and the melting point is 1357 K.
Answer:
4.73 × 10^5
Explanation:
the lamp cord is 85cm long and comprises cupper wire. Calculate the wire‘s resistance?
radius of a wire is 1.8mmm,Use value of resistivity for Cu as 1.75 × 10-8Ωm.
Answer:
R = 0.0015Ω
Explanation:
The formula for calculating the resistivity of a material is expressed as;
ρ = RA/l
R is the resistance
ρ is the resistivity
A is the area of the wire
l is the length of the wire
Given
l = 85cm = 0.85m
A = πr²
A = 3.14*0.0018²
A = 0.0000101736m²
ρ = 1.75 × 10-8Ωm.
Substitute into the formula
1.75 × 10-8 = 0.0000101736R/0.85
1.4875× 10-8 = 0.0000101736R
R = 1.4875× 10-8/0.0000101736
R = 0.0015Ω
Consider two closely spaced and oppositely charged parallel metal plates. The plates are square with sides of length L and carry charges Q and -Q on their facing surfaces. What is the magnitude of the electric field in the region between the plates
Answer:
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
Explanation:
In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector
E_ {total} = E₁ + E₂
E_ {total} = 2 E
where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction
to calculate the field created by a plate let's use Gauss's law
Ф = ∫ E . dA = q_{int} /ε₀
As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.
E 2A = q_{int} / ε₀
where the 2 is due to the surface has two faces
indicate that the surface has a uniform charge for which we can define a surface density
σ = q_{int} / A
q_{int} = σ A
we substitute
E 2A = σ A /ε₀
E = σ / 2ε₀
therefore the total field is
E_ {total} = σ /ε₀
let's substitute the density for the charge of the whole plate
σ= Q / L²
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed
Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:
[tex]V=1.4*10^5m/s[/tex]
Explanation:
From the question we are told that:
Electric field [tex]B=1.5*10N/C[/tex]
Distance [tex]d=2 x 10^{-3}[/tex]
At negative plate
Generally the equation for Velocity is mathematically given by
[tex]V^2=2as[/tex]
Therefore
[tex]V^2=\frac{2*e_0E*d}{m}[/tex]
[tex]V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}[/tex]
[tex]V=\sqrt{19.2*10^9}[/tex]
[tex]V=1.4*10^5m/s[/tex]
Four identical balls are thrown from the top of a cliff, each with the same speed. The
first is thrown straight up, the second is thrown at 30° above the horizontal, the third
at 30° below the horizontal, and the fourth straight down. How do the speeds and
kinetic energies of the balls compare as they strike the ground? Ignore the effects of
air resistance. Explain fully using the concepts from this unit.
The comparison of the speeds and kinetic energy of the identical balls are as follows
The speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
The reason for the above comparison results areas follows;
Known parameters;
First ball is thrown straight up
Second ball is thrown 30° above the horizontal
Third ball it thrown 30° below the horizontal
The fourth ball is thrown straight down
Unknown:
Comparison of the speed and kinetic energy of the four balls
Method:
The kinetic energy, K.E. = (1/2) × m × v²
The velocity of the ball, v = u × sin(θ)
Where;
u = The initial velocity of the ball
θ = The reference angle
For the first ball thrown straight up, we have;
θ = 90°
∴ [tex]v_y[/tex] = u
The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh
Where;
h = The height of the cliff
∴ Kinetic energy of first ball, K.E.₁ = (1/2) × m × (u₁² + 2gh)²
For the second ball thrown 30° to the horizontal, we have;
K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₂ = (1/2) × m × ((0.5·u₂)² + 2·g·h)²
For the third ball thrown at 30° below the horizontal, we have;K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₃ = (1/2) × m × ((0.5·u₃)² + 2·g·h)²
For the fourth ball thrown straight down, we have;Kinetic energy K.E.₄ = (1/2) × m × (u₄² + 2gh)²
Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
u₁ = u₄, K.E₁ = K.E.₄, u₂ = u₃, K.E₂ = K.E.₃
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A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be:_______.
A. in the positive y direction
B. in the negative y direction
C. in the positive x direction
D. in the negative x direction
A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be in the negative y direction. Thus, option B is correct.
To determine the force that could generate a torque in the negative z direction, we need to consider the right-hand rule for cross products. The torque vector, denoted by τ, is given by the cross product of the position vector, r, and the force vector, F:
[tex]τ = r × F[/tex]
In this case, the position vector, r, points along the positive x-axis. The negative z-direction torque indicates that the force vector must be perpendicular to both the position vector and the negative z-axis.
Using the right-hand rule, we can determine that the force vector must be in the negative y-direction, which is option B.
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