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GalinKa [24]
3 years ago
7

A particle moves in a circular path of radius 0.10 m with a constant angular speed of 5 rev/s. The acceleration of the particle

is: A. 0.10p m/s2 B. 0.50 m/s2 C. 500p m/s2 D. 1000p2 m/s2 E. 10p2 m/s2
Physics
1 answer:
Darya [45]3 years ago
7 0

Answer:

2.5 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s²

For circular motion, the expression for acceleration is given as,

a = ω²r ................ Equation 1

Where a = acceleration of the particle, ω = angular speed of the particle, r = radius of the circular path.

Given: ω = 5 rev/s = 31.42 rad/s, r = 0.10 m.

Substitute into equation 1

a = 5²(0.10)

a = 25(0.10)

a = 2.5 m/s²

Hence the acceleration of the particle = 2.5 m/s²

Hence, none of the option  is correct

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you nose out another runner to win 100.000 m dash. if your total time for the race was 13.800 s and you aced out the other runne
sammy [17]

Your average speed was

(100 m) / (13.8 s) = 7.25 m/s .

If you finished 0.001s ahead of him, then at your average speed, that corresponds to

(7.25 m/s) x (0.001 s) = 0.00725 m

That's 7.25 millimeters ... about 0.28 of an inch !

NOTE:. I think this is only valid if your speed was a constant ~7.25 m/s all the way.

3 0
3 years ago
The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

3 0
3 years ago
Read 2 more answers
How do organisms use communication to survive?
just olya [345]

Answer: Im not entirly sure but I think It's D all the above. I think all but B because I never really heard of that but if you look in our history I think that happen im not sure I would wait untill you know that somone knows for sure.

Explanation:

3 0
2 years ago
A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude
miskamm [114]

Answer:

Explanation:

Given

Weight of person W=652\ N

At highest point Magnitude of the normal force F=585\ N

net force at highest point

F_{net}=W+F_c

where F_c= centripetal force

F_c=\dfrac{mv^2}{r}

F_{net}=F_{n}= Normal Force

585=652+F_c

F_c=-67\ N

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

F_n=F_c+W

F_n=652+67

F_n=719\ N  

8 0
3 years ago
How can i calculate distances between objects by using the concepts of echo location
dangina [55]
Send wave from your location to the object and wait until echo is back.
Measure the time taken.

If you know the speed of wave (say sound wave), than just multiply by half time taken wave to return
5 0
2 years ago
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