Answer:
the angular speed of the person increases, being able to make more turns and faster.
K₂ = K₁ I₁ / I₂
Explanation:
When the divers are turning the system is isolated, so all the forces are internal and therefore also the torque, therefore the angular momentum is conserved
initial, joint when starting to turn
L₀ = I₁ w₁
final. When you shrink your arms and legs
Lf = I₂ w₂
L₀ = Lf
I₁ w₁ = I₂ w₂
when you shrink your arms and legs the distance to the turning point decreases and since the moment of inertia depends on the distance squared, the moment of inertia also decreases
I₂ <I₁
w₂ = I₁ / I₂ w₁
therefore the angular speed of the person increases, being able to make more turns and faster.
When it goes into the water it straightens the arm and leg, so the moment of inertia increases
I₁> I₂
w₁ = I₂ / I₁ w₂
therefore we see that the angular velocity decreases, therefore the person trains the water like a stone and can go deeper faster.
In both cases the kinetic energy is
K = ½ I w²
the initial kinetic energy is
K₁ = ½ I₁ w₁²
the final kinetic energy is
K₂ = ½ I₂ w₂²
we substitute
K₂ = ½ I₂ (I₁ / I₂ w1² 2
K₂ = ½ I₁² / I₂ w₁² = (½ I₁ w₁²) I₁ / I₂
K₂ = K₁ I₁ / I₂
therefore we see that the kinetic energy increases by factor I₁/I₂
The only force acting on a 3.2 kg canister that is moving in an xy plane has a magnitude of 6.7 N. The canister initially has a velocity of 3.3 m/s in the positive x direction, and some time later has a velocity of 6.9 m/s in the positive y direction. How much work is done on the canister by the 6.7 N force during this time
Answer:
The work done by the force is 5.76 J
Explanation:
Given;
mass of canister , m = 3.2 kg
magnitude of force, f = 6.7 N
initial velocity of the canister on x-axis, [tex]v_i[/tex]= 3.3i m/s
final velocity of the canister on y- axis, [tex]v_f[/tex] = 6.9j m/s
The work done on the canister = change in the kinetic energy of the canister
[tex]W = K.E_f - K.E_i[/tex]
where;
K.Ei is the initial kinetic energy
K.Ef is the final kinetic energy
The initial kinetic energy:
[tex]K.E_i = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_i = \frac{1}{2} *3.2\sqrt{3.3^2 +0^2+0^2}\\\\K.E_i = 5.28 \ J[/tex]
The final kinetic energy:
[tex]K.E_f = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_f = \frac{1}{2} *3.2\sqrt{0^2 +6.9^2+0^2}\\\\K.E_f = 11.04 \ J\\[/tex]
W = 11.04 - 5.28
W = 5.76 J
Therefore, work done on the canister by the 6.7 N force during this time is 5.76 J
A 0.150 kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce.
What is the magnitude of the impulse imparted to the clay by the floor during the impact? Assume that the acceleration due to gravity is =9.81 m/s2.
Answer:
J = 0.800 kg m/s
Fmax = 291 N
Explanation:
During the fall, energy is conserved.
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × 9.81 m/s² × 1.45 m)
v = 5.33 m/s
Alternatively, you can use kinematics to find the velocity.
Impulse = change in momentum
J = Δp
J = mΔv
J = (0.150 kg) (5.33 m/s)
J = 0.800 kg m/s
Impulse = area under F vs t graph
J = ∫ F dt
J = ½ Fmax Δt
(0.800 kg m/s) = ½ Fmax (0.0055 ms)
Fmax = 291 N
A book of 500 leaves has a mass of 1kg if its thickness is 5cm what are the mass and thickness of each leaf
Answer:
0.002kg and 0.01cm
Explanation:
500 leaves has a thickness is 5cm
Means I leaf has a thickness of 5/500= 0.01cm
Similarly the mass of one leaf would be 1/500 =0.002kg
How do you convert 1.3*10^6cal into joules
Answer:
5.4×10⁶J
Explanation:
1 cal = 4.184 J
1.3×10⁶ cal × (4.184 J/cal) = 5.4×10⁶J
A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). To the nearest degree, at what angle relative to the normal to the interface of the two liquids is light totally reflected?
a. 78 degree
b. 88 degree
c. 68 degree
d. 49 degree
e. the critical angle isundefined
Answer:
a. 78 degree
Explanation:
According to Snell's Law, we have:
(ni)(Sin θi) = (nr)(Sin θr)
where,
ni = Refractive index of medium on which light is incident
ni = Refractive index of ethyl alcohol = 1.361
nr = Refractive index of medium from which light is refracted
nr = Refractive index of ethyl alcohol = 1.333
θi = Angle of Incidence
θr = Angle of refraction
So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:
θi = θc
when, θr = 90°
Therefore, Snell's Law becomes:
(1.361)(Sin θc) = (1.333)(Sin 90°)
Sin θc = 1.333/1.361
θc = Sin⁻¹ (0.9794)
θc = 78.35° = 78° (Approximately)
Therefore, correct answer will be:
a. 78 degree
The angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
From the information given;
the refractive index of the ethyl alcohol [tex]\mathbf{n_1= 1.361}[/tex]the refractive index of the water [tex]\mathbf{n_2 = 1.333}[/tex] the angle of incidence is the critical angle [tex]\theta_i = \theta_c[/tex] the angle of refraction [tex]\theta _r = 90^0[/tex]According to Snell's Law of refraction;
[tex]\mathbf{n_1 sin \theta _c = n_2 sin \theta_r}[/tex]
[tex]\mathbf{1.361 \times sin \theta _c = 1.333 \times sin 90}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times sin 90}{1.361}}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times 1}{1.361}}[/tex]
[tex]\mathbf{ \theta _c = sin^{-1} (0.9794)}[/tex]
[tex]\mathbf{ \theta _c =78.35^0}[/tex]
[tex]\mathbf{ \theta _c \simeq78^0}[/tex]
Therefore, we can conclude that the angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
Learn more about Snell Law of refraction here:
https://brainly.com/question/14029329?referrer=searchResults
You are at a stop light in your car, stuck behind a red light. Just before the light is supposed to change, a fire engine comes zooming up towards you traveling at a horrendous 85.0 km/h. If the siren has a rated frequency 665 Hz, what frequency of the sound do you hear
Answer:
The frequency of the sound you will hear is 713.85 Hz
Explanation:
Given;
speed of your car, [tex]v_s[/tex] = 85.0 km/h
frequency of the siren, f = 665 Hz
Speed of sound in air, v = 345 m/s
The frequency of the sound you hear, can be calculated as;
[tex]f' = f(\frac{v}{v-v_s})[/tex]
Convert the speed of the car to m/s
[tex]85 \ km/h =\frac{85 \ km}{h} (\frac{1000\ m}{1 \ km})(\frac{1 \ h}{3600 \ s} ) = 23.61 \ m/s[/tex]
[tex]f' = f(\frac{v}{v-v_s} )\\\\f' = 665(\frac{345}{345-23.61} )\\\\f' = 665 (1.07346)\\\\f' = 713.85 \ Hz[/tex]
Therefore, the frequency of the sound you will hear is 713.85 Hz
Suppose a stone is thrown vertically upward from the edge of a cliff on Mars (where the acceleration due to gravity is only about 12 ft/s2 with an initial velocity of 64 ft/s from a height of 192 ft above the ground. The height s of the stone above the ground after t seconds is given by
s=−6t2+64t+192
a. Determine the velocity v of the stone after t seconds. b. When does the stone reach its highest point? c. What is the height of the stone at the highest point? d. When does the stone strike the ground? e. With what velocity does the stone strike the ground?
Answer:
a) v = -12t + 64
b) t = 5.33s
c) s = 362.66ft
d) t = 13.10s
e) v = 93.2ft/s
Explanation:
You have the following equation for the height of a stone thrown in Mars:
[tex]s(t)=-6t^2+64t+192[/tex] (1)
a) The velocity of the stone after t seconds is obtained with the derivative of s in time:
[tex]v=\frac{ds}{st}=-12t+64[/tex] (2)
The equation for the speed of the stone is v = -12t + 64
b) The highest point is obtained when the speed of the stone is zero. Then, from the equation (2) equal to zero, you can obtain the time when the stone is at its maximum height:
[tex]-12t+64=0\\\\t=5.33s[/tex]
The time in which the stone is at the maximum height is 5.33s
c) For this time the stone is at the maximum height. Then, you replace t in the equation (1):
[tex]s(1)=-6(5.33)^2+64(5.33)+192=362.66ft[/tex]
the maximum height is 362.66 ft
d) To find the time when the stone arrive to the ground you equal the equation (1) to zero and you solve for t:
[tex]0=-6t^2+64t+192[/tex]
you use the quadratic formula:
[tex]t_{1,2}=\frac{-64\pm\sqrt{64^2-4(-6)(192)}}{2(-6)}\\\\t_{1,2}=\frac{-64\pm 93.29}{-12}\\\\t_1=13.10s\\\\t_2=-2.44s[/tex]
You use the result with positive values because is the onlyone with physical meaning.
The time for the stone hits the ground is 13.10 s
e) You replace 13.10s in the equation (2) to obtain the velocity of the stone when it strike the ground:
[tex]v=-12t+64=-12(13.10)+64=-93.2\frac{ft}{s}[/tex]
The minus sign is because the stone's direction is downward.
The speed of the stone just when it strikes the ground is 93.2ft/s
A ball is projected upward at time t= 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to:____________A. 9.0 sB. 9.4 sC. 9.7 sD. 8.7 sE. 10 s
Answer:
B. 9.4 s
Explanation:
In order to calculate the total time taken by the ball to hit the ground, we first analyze the upward motion. We will use subscript 1 for upward motion. Now, using 1st equation of motion:
Vf₁ = Vi₁ + gt₁
where,
Vf, = Final Velocity in upward motion = 0 m/s (ball stops at highest point)
Vi = Initial Velocity in upward motion = 36.2 m/s
g = - 9.8 m/s² (negative due to upward motion)
t₁ = Time taken in upward motion = ?
Therefore,
0 m/s = 36.2 m/s + (-9.8 m/s²)(t₁)
t₁ = (36.2 m/s)/(9.8 m/s²)
t₁ = 3.7 s
Now, using 2nd equation of motion:
h₁ = (Vi₁)(t₁) + (0.5)(g)(t₁)²
where,
h₁ = distance from top of building to highest point ball reaches = ?
Therefore,
h₁ = (36.2 m/s)(3.7 s) + (0.5)(-9.8 m/s²)(3.7 s)²
h₁ = 133.58 - 66.86 m
h₁ = 66.72 m
No, considering downward motion and using subscript 2, for it.
Using 2nd equation of motion:
h₂ = (Vi₂)(t₂) + (0.5)(g)(t₂)²
where,
h₂ = height of the highest point from ground = h₁ + height of building
h₂ = 66.72 m + 90 m = 156.72 m
Vi₂ = Initial Speed during downward motion = 0 m/s (ball stops for a moment at highest point)
t₂ = Time Taken in downward motion = ?
g = 9.8 m/s²
Therefore,
156.72 m = (0 m/s)(t₂) + (0.5)(9.8 m/s²)(t₂)²
t₂² = (156.72 m)/(4.9 m/s²)
t₂ = √31.98 s²
t₂ = 5.7 s
Now, the total time taken by ball to reach the ground is"
Total Time = T = t₁ + t₂
T = 3.7 s + 5.7 s
T = 9.4 s
Therefore, the correct answer is:
B. 9.4 s
You rub a balloon on your head and it becomes negatively charged. The balloon will be most attracted to what?
Answer:
To things that are positive charged
A bicycle wheel has an initial angular velocity of 1.10 rad/s . Part A If its angular acceleration is constant and equal to 0.200 rad/s2 , what is its angular velocity at t = 2.50 s ? (Assume the acceleration and velocity have the same direction) Express your answer in radians per second. ω = nothing rads Request Answer Part B Through what angle has the wheel turned between t = 0 and t = 2.50 s ? Express your answer in radians. Δθ = nothing rad Request Answer Provide Feedback
Let [tex]\theta[/tex], [tex]\omega[/tex], and [tex]\alpha[/tex] denote the angular displacement, velocity, and acceleration of the wheel, respectively.
(A) The wheel has angular velocity at time [tex]t[/tex] according to
[tex]\omega=\omega_0+\alpha t[/tex]
so that after 2.50 s, the wheel will have attained an angular velocity of
[tex]\omega=1.10\dfrac{\rm rad}{\rm s}+\left(0.200\dfrac{\rm rad}{\mathrm s^2}\right)(2.50\,\mathrm s)=\boxed{1.60\dfrac{\rm rad}{\rm s}}[/tex]
(B) The angular displacement of the wheel is given by
[tex]\theta=\theta_0+\omega_0t+\dfrac\alpha2t^2\implies\Delta\theta=\omega_0t+\dfrac\alpha2t^2[/tex]
After 2.50 s, the wheel will have turned an angle [tex]\Delta\theta[/tex] equal to
[tex]\Delta\theta=\left(1.10\dfrac{\rm rad}{\rm s}\right)(2.50\,\mathrm s)+\dfrac12\left(0.200\dfrac{\rm ram}{\mathrm s^2}\right)(2.50\,\mathrm s)^2=\boxed{3.38\,\mathrm{rad}}[/tex]
