Physics : asked on AutumnJoy12
 06.08.2021

Aparticle moves along the x axis. it is initially at the position 0.250 m, moving with velocity 0.050 m/s and acceleration -0.240 m/s2. suppose it moves with constant acceleration for 3.70 s. (a) find the position of the particle after this time. m (b) find its velocity at the end of this time interval. m/s

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30.05.2023, solved by verified expert
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(a) -1.208 m

The position of the particle at time t is given by

Aparticle moves along the x axis. it is initially, №16482774, 06.08.2021 11:38

where:

Aparticle moves along the x axis. it is initially, №16482774, 06.08.2021 11:38 is the initial position

Aparticle moves along the x axis. it is initially, №16482774, 06.08.2021 11:38 is the initial velocity

Aparticle moves along the x axis. it is initially, №16482774, 06.08.2021 11:38 is the acceleration

Substituting into the equation t=3.70 s, we find the position after 3.70 seconds:

Aparticle moves along the x axis. it is initially, №16482774, 06.08.2021 11:38

(b) -0.838 m/s

The velocity of the particle at time t is given by:

Aparticle moves along the x axis. it is initially, №16482774, 06.08.2021 11:38

where

Aparticle moves along the x axis. it is initially, №16482774, 06.08.2021 11:38 is the initial velocity

Aparticle moves along the x axis. it is initially, №16482774, 06.08.2021 11:38 is the acceleration

Substituting t = 3.70 s, we find the velocity after 3.70 seconds:

Aparticle moves along the x axis. it is initially, №16482774, 06.08.2021 11:38

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Physics
Step-by-step answer
P Answered by PhD

(a) -1.208 m

The position of the particle at time t is given by

x(t) = x_0 + v_0 t + \frac{1}{2}at^2

where:

x_0 = 0.250 m is the initial position

v_0 = 0.050 m/s is the initial velocity

a=-0.240 m/s^2 is the acceleration

Substituting into the equation t=3.70 s, we find the position after 3.70 seconds:

x(3.70 s) = 0.250 m + (0.050 m/s)(3.70 s) + \frac{1}{2}(-0.240 m/s^2)(3.70 s)^2=-1.208 m

(b) -0.838 m/s

The velocity of the particle at time t is given by:

v(t) = v_0 + at

where

v_0 = 0.050 m/s is the initial velocity

a=-0.240 m/s^2 is the acceleration

Substituting t = 3.70 s, we find the velocity after 3.70 seconds:

v(3.70 s) = 0.050 m/s + (-0.240 m/s^2)(3.70 s)=-0.838 m/s

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 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
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
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
Gravity acceleration (g) = 9.8m/s^2
Time (t) = 3sec
Acceleration = velocity/time
Velocity = acceleration×time
= 9.8×3
= 29.4m/s
Physics
Step-by-step answer
P Answered by PhD
Initial velocity (u) = 0
Time taken = 4.5 seconds
Gravitational acceleration (g) = 9.8m/s^2
By the second equation of motion under gravity,
The distance that object fell down (h)
h = ut + (1/2)gt^2
h = 0×4.5 + (1/2)×9.8×(4.5)^2
h = 99.225 m

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