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prisoha [69]
3 years ago
14

an athlete whirls an 8.71 kg hammer tied to the end of a 1.5 m chain in a simple horizontal circle where you should ignore any v

ertical deviations the hammer moves at the rate of 1.16 rev/s what is the tension in the chain
Physics
1 answer:
-BARSIC- [3]3 years ago
5 0

Answer:

T = 692.42 N

Explanation:

Given that,

Mass of hammer, m = 8.71 kg

Length of the chain to which an athlete whirls the hammer, r = 1.5 m

The angular sped of the hammer, \omega=1.16\ rev/s=7.28\ rad/s

We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :

F=m\omega^2r\\\\=8.71\times (7.28)^2\times 1.5\\\\=692.42\ N

So, the tension in the chain is 692.42 N.

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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha
krek1111 [17]

Answer:

F=18.58\times 10^{-11}\ N

\theta=30.276^{\circ}

Explanation:

Given:

mass of first particle, m_1=6.7\ kg

mass of second particle, m_2=5.1\ kg

mass of third particle, m_3=3.7\ kg

coordinate position of first particle in meters, (x_1,y_1)\equiv(4.2,0)

coordinate position of second particle in meters, (x_2,y_2)\equiv(0,2.8)

coordinate position of third particle in meters, (x_3,y_3)\equiv(0,0)

<u>Now, gravitational force on particle 3 due to particle 1:</u>

F_{31}=G\frac{m_1.m_3}{r_{31}^2}

F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}

F_{31}=9.37\times 10^{-11}\ N

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

F_{32}=G\frac{m_2.m_3}{r_{21}^2}

F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}

F_{32}=16.05\times 10^{-11}\ N

towards positive X axis.

<u>Now the net force</u>

F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}

F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}

F=18.58\times 10^{-11}\ N

<em>For angle in counterclockwise direction from the +x-axis</em>

tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}

\theta=30.276^{\circ}

4 0
3 years ago
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Answer:CCCCCCCCCCCCCCCCC

Explanation:CCCCCCCCCCCCC

4 0
3 years ago
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A vacuum tube can be used to__. A. change alternating current into direct current B. increase the strength of a signal.. C. turn
vesna_86 [32]

The correct answer of this question is :  A) Change alternating current into direct current.

EXPLANATION :

As per the question, we are given vacuum tube. Vacuum tube can be of various types. Normally it contains two electrodes called cathode and anode which are enclosed in an evacuated glass chamber . There are also other types of vacuum tubes which contain extra electrodes like control grid .

The vacuum tube can be used as a rectifier. It means that it can be used as an electronic device which will convert alternating current into direct current. It may be a half wave rectifier or a full wave rectifier. Actually the direct current obtained during the rectification of alternating current is pulsating in nature.

Hence, the correct answer is that a vacuum tube can be used to change alternating current into direct current.


4 0
3 years ago
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An object is thrown upward from the top of a 144​-foot building with an initial velocity of 128 feet per second. The height h of
ankoles [38]

Answer:

After 9 seconds the object reaches ground.

Step-by-step explanation:

We equation of motion given as h = -16t²+128t+144,

We need to find in how many seconds will the object hit the ground,

That is we need to find time when h = 0

                        0 = -16t²+128t+144

                       16t²-128t-144= 0

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Negative time is not possible, hence after 9 seconds the object reaches ground.

8 0
3 years ago
Jupiter has the strongest gravity of any planet in our solar system. Which quantity will be a maximum there?
Mamont248 [21]

weight will be maximum there

3 0
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