Physics : asked on ashley2816
 02.06.2020

2. Am 80.0 kg astronaut is training for accelerations that he will experience upon re-entry to earth’s gravity from space. He is placed in a centrifuge (r = 25.0 m) and spun at a constant angular velocity of 10.0 rpm (revolutions per minute). a. Find the linear velocity of the centrifuge in m/s. Show your work

b. Find the magnitude and direction of the centripetal acceleration when he is spinning at this constant velocity.

c. How many g’s is the astronaut experiencing? (at constant velocity)

d. Find the linear deceleration and torque required to bring the centrifuge (5000.0 kg) to a stop over a 5 minute time period.

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Step-by-step answer

24.06.2023, solved by verified expert
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a)   v = 26.2 m / s, b) acceleration is radial, a = 27.4 m / s², c)   a = 2.8 g and

d) a = - 8.73 10⁻² m / s²,  τ = 1.09 10⁴ N m

Explanation:

a) For this exercise we can use the relationships between rotational and linear motion

           v = w r

let's reduce the magnitudes to the SI system

          w = 10 rpm (2pi rad / 1 rev) 1 min / 60s) = 1,047 rad / s

          r = 25.0 m

let's calculate

          v = 1.047 25.0

          v = 26.2 m / s

b) When the body is rotating at constant speed, the relationship must be perpendicular to the speed, therefore the direction of acceleration is radial, that is, towards the center of the circle and its magnitude is

            a = v² / r

            a = 26.2²/25

            a = 27.4 m / s²

c) Let's look for the relationship between the centripetal acceleration and the acceleration due to gravity

           a / g = 27.4 / 9.8

           a / g = 2.8

           a = 2.8 g

d) let's find the deceleration and torque to stop the centripette in 5 min

           t = 5 min (60 s / 1min) = 300 s

           

let's use the rotational kinematics relations

           w = w₀ + α t

initial angular velocity is wo = 1,047 rad / s and the final as is stop do w = 0

           α = - w₀ / t

           α = - 1,047 / 300

           α = -3.49 10⁻³ rad / s²

angular and linear are related

           a = α r

           a = -3.49 10⁻³ 25

           a = - 8.73 10⁻² m / s²

the negative sign indicates that the acceleration is stopping the movement

torque is

           τ = F r

The force can be found with Newton's second law

          F = m a

we substitute

         τ = m a r

         τ = 5000.0   8.73 10⁻²  25

         τ = 1.09 10⁴ N m

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Physics
Step-by-step answer
P Answered by PhD

a)   v = 26.2 m / s, b) acceleration is radial, a = 27.4 m / s², c)   a = 2.8 g and

d) a = - 8.73 10⁻² m / s²,  τ = 1.09 10⁴ N m

Explanation:

a) For this exercise we can use the relationships between rotational and linear motion

           v = w r

let's reduce the magnitudes to the SI system

          w = 10 rpm (2pi rad / 1 rev) 1 min / 60s) = 1,047 rad / s

          r = 25.0 m

let's calculate

          v = 1.047 25.0

          v = 26.2 m / s

b) When the body is rotating at constant speed, the relationship must be perpendicular to the speed, therefore the direction of acceleration is radial, that is, towards the center of the circle and its magnitude is

            a = v² / r

            a = 26.2²/25

            a = 27.4 m / s²

c) Let's look for the relationship between the centripetal acceleration and the acceleration due to gravity

           a / g = 27.4 / 9.8

           a / g = 2.8

           a = 2.8 g

d) let's find the deceleration and torque to stop the centripette in 5 min

           t = 5 min (60 s / 1min) = 300 s

           

let's use the rotational kinematics relations

           w = w₀ + α t

initial angular velocity is wo = 1,047 rad / s and the final as is stop do w = 0

           α = - w₀ / t

           α = - 1,047 / 300

           α = -3.49 10⁻³ rad / s²

angular and linear are related

           a = α r

           a = -3.49 10⁻³ 25

           a = - 8.73 10⁻² m / s²

the negative sign indicates that the acceleration is stopping the movement

torque is

           τ = F r

The force can be found with Newton's second law

          F = m a

we substitute

         τ = m a r

         τ = 5000.0   8.73 10⁻²  25

         τ = 1.09 10⁴ N m

Physics
Step-by-step answer
P Answered by Specialist
Answer: Option B and C are True.

Explanation:
The weight of the two blocks acts downwards.
Let the weight of the two blocks be W. Solving for T₁ and T₂:
w = T₁/cos 60° -----(1);
w = T₂/cos 30° ----(2);
equating (1) and (2)
T₁/cos 60° = T₂/cos 30°;
T₁ cos 30° = T₂ cos 60°;
T₂/T₁ = cos 30°/cos 60°;
T₂/T₁ =1.73.
Therefore, option a is false since T₂ > T₁.
Option B is true since T₁ cos 30° = T₂ cos 60°.
Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.
Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.
Answer: Option B and C are True.

Explanation:  
The weight of the two blocks acts downwards.
Le
Physics
Step-by-step answer
P Answered by Master

Answer:

see below.

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

Physics
Step-by-step answer
P Answered by PhD

Answer:

9.6 meters

Step-by-step explanation:

Time taken by the tomatoes to each the ground

using h = 1/2 g t^2 

t^2 = 2h/g = 2 x 50/ 9.8 = 10.2

t = 3.2 sec 

horizontal ditance = speed x time = 3 x 3.2 = 9.6 meters

Physics
Step-by-step answer
P Answered by PhD

The question specifies the diameter of the screw, therefore the IMA of this screw is 0.812? / 0.318 = 8.02

Physics
Step-by-step answer
P Answered by PhD
First sum applied the Newton's second law motion: F = ma
Force = mass* acceleration
This motion define force as the product of mass times Acceleration (vs.Velocity). Since acceleration is the change in velocity divided by time,
force=(mass*velocity)/time
such that, (mass*velocity)/time=momentum/time
Therefore we get mass*velocity=momentum
Momentum=mass*velocity
Elephant mass=6300 kg; velocity=0.11 m/s
Momentum=6300*0.11
P=693 kg (m/s)
Dolphin mass=50 kg; velocity=10.4 m/s
Momentum=50*10.4
P=520 kg (m/s)
The elephant has more momentum(P) because it is large.
Physics
Step-by-step answer
P Answered by PhD
Weight of barbell (m) = 100 kg
Uplifted to height (h) = 2m
Time taken= 1.5 s
Work done by Jordan = potential energy stored in barbell = mgh
= 100×2×9.8
= 1960J
Power = energy/time
= 1960/1.5
1306.67watts
Physics
Step-by-step answer
P Answered by PhD
Weight of jasmine (m) = 400 N
Height climbed on wall (h) = 5m
Total time taken in climbing = 5 sec
Work done in climbing the wall = rise in potential energy = mgh
= 400×9.8×51
= 19600J
Power generated by Jasmine = potential energy / time
= 19600/5
= 3920Watts
Physics
Step-by-step answer
P Answered by PhD
The horizontal and vertical motions of balloons are independent from each other.
Let vertical component of initial velocity U' horizontal component of initial velocity U"
Time of landing (t) is found with the help of vertical motion.
Since vertical component of initial velocity of balloon is zero(U' = 0)
From equation h = U't + 1/2gt^2
h = 1/2gt^2
t = √(2h/g)
t = √( 2×150/9.8)
t = 5.53 sec
Horizontal velocity = 50m/s
Horizontal range of balloon, R = U"t
= 50× 5.53
= 27.65m
So the balloon will go 27.65 metre away from the bridge

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