Physics : asked on babiibri4771
 27.01.2020

A particle moving in the xy-plane has velocity v⃗ =(2ti^+(3−t2)j^)m/s, where t is in s. Part A

What is the x component of the particle's acceleration vector at t = 6 s?

Express your answer with the appropriate units.

Part B

What is the y component of the particle's acceleration vector at t = 6 s?

Express your answer with the appropriate units.

. 1

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24.06.2023, solved by verified expert
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A)   2 m/s²

B) -12 m/s²

Explanation:

A)

Applying the definition of instantaneous acceleration as the derivative of the velocity with respect to time, the acceleration vector can be expressed as follows:a = 2i^ + (-2t) j^ m/s²So, the x-component of the acceleration vector is constant, and equal to 2m/s², at any time.When t=6 s, aₓ= 2 m/s²

B)

Applying the definition of instantaneous acceleration as the derivative of the velocity with respect to time, the acceleration vector can be expressed as follows:a = 2i^ + (-2t) j^ m/s²So, when t= 6s, the y component of a is as follows:ay = -2(6s) = -12 m/s²
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Physics
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P Answered by PhD

A)   2 m/s²

B) -12 m/s²

Explanation:

A)

Applying the definition of instantaneous acceleration as the derivative of the velocity with respect to time, the acceleration vector can be expressed as follows:a = 2i^ + (-2t) j^ m/s²So, the x-component of the acceleration vector is constant, and equal to 2m/s², at any time.When t=6 s, aₓ= 2 m/s²

B)

Applying the definition of instantaneous acceleration as the derivative of the velocity with respect to time, the acceleration vector can be expressed as follows:a = 2i^ + (-2t) j^ m/s²So, when t= 6s, the y component of a is as follows:ay = -2(6s) = -12 m/s²
Physics
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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.
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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
Answer:
7.25 secs.

Explanation:
First find the distance it takes to stop
s = [v^2-u^2]/2a = 0^2 - 8.7^2/2[-2.4] = 8.7^2/4.8
Next find the time it takes to go that distance , s = ut +[1/2] at^2
8.7^2/4.8 = 8.7t +[1/2] [ -2.4]t^2 , rearrange and
t^2 -[8.7/1.2]+ 8.7^2/[(1.2)(4.8)]=0 complete the square
[t - (8.7/2.4)]^2=0
t = 8.7/2.4 = 3.625 secs
At this stage the deceleration will push the object back in the direction it came from for another 3.625 secs when it will be 8.7 m/s again
Total time , T =2t = 7.25 secs.

Note:
The term differential is used in calculus to refer to an infinitesimal (infinitely small) change in some varying quantity. For example, if x is a variable, then a change in the value of x is often denoted Δx (pronounced delta x). The differential dx represents an infinitely small change in the variable x.
Physics
Step-by-step answer
P Answered by PhD
The change in temperature is 9.52°CExplanation:Since, the heat supplied by the electric kettle is totally used to increase the temperature of the water.Thus, from the law of conservation of energy can be stated as:Heat Supplied by Electric Kettle = Heat Absorbed by WaterHeat Supplied by Electric Kettle = m C ΔTwhere,Heat Supplied by Electric Kettle = 20,000 JMass of water = m = 0.5 kgSpecific Heat Capacity of Water = C = 4200 J/kg.°CChange in Temperature of Water = ΔTTherefore,20,000 J = (0.5 kg)(4200 J/kg.°C) ΔTΔT = 20,000 J/(2100 J/°C)ΔT = 9.52°C

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