Answer:
4.437 m/s
Explanation:
Diameter of rotation d is 1.7 m
Radius of rotation = d/2 = 1.7/2 = 0.85 m
If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s
We convert to rad/s
Angular speed = 2 x pi x 0.83
= 2 x 3.142 x 0.83 = 5.22 rad/s
Speed is equal to the angular speed times the radius of rotation
Speed = 5.22 x 0.85 = 4.437 m/s
In the given case, the speed of the discus at release, If the thrower takes 1.2s to complete one revolution, starting from rest would be - 8.90 m/s.
Given:
diameter of the circle = 1.7 mradius f the circle would be = 1.7/2 = 0.85 m
time taken for one revolution t = 1.2 sThis rotation exercise can be treated using the rotation kinematics.
Angular acceleration:
θ = w₀ t + ½ α t²
t = 1.2 s to give a revolution (T = 2π rad) and with part of the rest the initial angular velocity is zero (wo = 0)
=> θ = 0 + ½ α t²
=> α = 2θ / t²
=> α= 2 × 2π / 1.2²
=> α = 4π = 8.7266 rad / s²
Let's calculate the angular velocity:
=> w = wo + α t
=> w = 0 + α t
=> w = 8.7266 × 1.2
=> w = 10.47192 rad / s
The relationship between linear and angular velocity is
=> r = d / 2
=> r = 1.7 / 2 = 0.85 m
=> v = w r
=> v = 10.47192 × 0.85
=> v = 8.90 m / s
Thus, the correct speed would be - 8.90 m/s
Learn more:
https://brainly.com/question/14663644
The self-referencing effect refers to ________.
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
What is a substance?
During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial speed v0 = 38 m/s at an angle θ = 35° above horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Air resistance may be ignored throughout this problem.
Part (a) Express the magnitude of the ball's initial horizontal velocity Or in terms of vo and 20%
Part (b) Express the magnitude of the ball's initial vertical velocity vOy in terms of vo and 0. 20%
Part (c) Find the ball's maximum vertical height Amat in meters above the ground.
Part (d) Create an expression in terms of vo-e, and g for the time-ur İt takes te ball to travel to its maximum vertical height.
Part (e) Calculate the horizontal distance in meters the ball has traveled when it returns to ground level.
Answer:
a) v₀ₓ = v₀ cos θ , b) v_{oy} = v₀ sin θ , c) y = v_{oy}² / 2g, y = 24.25 m
e) R = 138.46 m
Explanation:
This is a projectile launch exercise
a) let's use trigonometry to find the components of the initial velocity
cos θ = v₀ₓ / v₀
v₀ₓ = v₀ cos θ
v₀ₓ = 38 cos 35
v₀ₓ = 31.13 m / s
b) sin θ = [tex]v_{oy}[/tex] / v₀
v_{oy} = v₀ sin θ
v_{oy} = 38 sint 35
v_{oy} = 21 80 m / s
c, d) to find the maximum height, the vertical speed is zero
v_{y}² = v_{oy}² - 2 g y
0 = [tex]v_{oy}[/tex]² - 2 gy
y = v_{oy}² / 2g
let's calculate
y = 21.80 2 / (2 9.8)
y = 24.25 m
e) They ask to find the horizontal distance
for this we can use the expression of reaches
R = v₀² sin 2θ / g
let's calculate
R = 38² sin (2 35) / 9.8
R = 138.46 m
5.Which of the following does not affect rate of evaporation?
O Wind speed
O Surface area
O Temperature
O Insoluble heavy impurities
Answer:
D
Explanation:
Insoluble impurities would not change the constituent of the substance. Soluble would for example salt water takes longer time for the water to become vapour when subjected to the same temperature that normal water.
Wind would affect, the more windy the tendency for particles of the liquid to be moved into the atmosphere.
With an increase in surface area, the evaporation rate increase . Take a clue from water placed on the ground and exposed to the atmosphere and that same quantity of water is placed in a cup. That on the floor would evaporate faster.
Similarly the higher the temperature a substance is subjected to the easier is it's rate of evaporation. Take for instance water in a cup placed in the sun and that same placed in a room with mild temperatures than that of the sun.With time that in the sun decreases in volume faster than that in the room.
the distance between 2 station is 5400 m find the time taken by a train to cover this distance, if the train travels with speed 60m/s
Answer:
I dont know bro
Explanation:
Ask an expert
Answer:
Time=90s
Explanation:
Speed=distance /time
[tex]60 = \frac{5400}{t} where \: t \: is \: time \\60t = 5400 \\ t = \frac{5400}{60} \\ t =90 \\ hope \: this \: helps..good \: luck [/tex]
first law of equilibrium
Answer:
for an object to be in equilibrium, it must be experiencing no acceleration. Both the net force and the net torque must be zero.
Hope I helped
Answer:
An object in static equilibrium has zero net force acting upon it.
The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes. This can be written as
F = F1+ F2+ F3+ F4+. . . = 0
If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the inductance per unit length does not exceed 50 nH per meter? Express your answer using two significant figures.
Answer:
Inner radius = 2 mm
Explanation:
In a coaxial cable, series inductance per unit length is given by the formula;
L' = (µ/(2π))•ln(R/r)
Where R is outer radius and r is inner radius.
We are given;
L' = 50 nH/m = 50 × 10^(-9) H/m
R = 2.6mm = 2.6 × 10^(-3) m
Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m
Plugging in the relevant values, we have;
50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)
Rearranging, we have;
(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)
0.25 = ln((2.6 × 10^(-3))/r)
So,
e^(0.25) = (2.6 × 10^(-3))/r)
1.284 = (2.6 × 10^(-3))/r)
Cross multiply to give;
r = (2.6 × 10^(-3))/1.284)
r = 0.002 m or 2 mm
A ship can float on water as long as it weighs less than water.
O A. True
O B. False
Answer:
It's true
Explanation:
Because the ship is mafe up of aluminium, which is a light metal.
Answer:
False
Explanation:
Took The Quiz
Question 10
Air with a density of 1.20 kg/m3 flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?
Answer:
75.0 cm
Explanation:
becouse i don,t no the right answer
A 328-kg car moving at 19.1 m/s in the x direction hits from behind a second car moving at 13.0 m/s in the same direction. If the second car has a mass of 790 kg and a speed of 15.1 m/s right after the collision, what is the velocity of the first car after this sudden collision
Answer:
14.04 m/s
Explanation:
To find the velocity of the first car after the collision, we can use the equation of conservation of momentum:
m1v1 + m2v2 = m1'v1' + m2'v2'
We have the following data:
m1 = m1' = 328,
m2 = m2' = 790,
v1 = 19.1,
v2 = 13,
v2' = 15.1.
Using this data, we can find v1' (final velocity of the first car):
328 * 19.1 + 790 * 13 = 328 * v1' + 790 * 15.1
16534.8 = 328 * v1' + 11929
328 * v1' = 4605.8
v1' = 14.04 m/s