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pav-90 [236]
4 years ago
7

When using a crowbar to remove a nail, the person should hold it at which of the following spots to increase the IMA of the leve

r? nearest the end prying out the nail furthest from the end prying out the nail right in the middle the location where the person holds it will not affect the IMA
Physics
1 answer:
Ilia_Sergeevich [38]4 years ago
5 0

Answer: the furthest from the end prying out the nail

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A 12 N force is used to pull a wagon 10 m to the left in 15 seconds. Which of the following represents the magnitude of the forc
Lana71 [14]

1. - 12N

2.- 2 hours

3. - 0.5 m/s/s

7 0
3 years ago
Read 2 more answers
1.
Gwar [14]

<u>Answer:</u>

For 1: The correct option is Option C.

For 3: The final velocity of the opponent is 1m/s

<u>Explanation: </u>

During collision, the energy and momentum remains conserved. The equation for the conservation of momentum follows:

m_1u_1+m_2u_2=m_1v_1+m_2v_2      ...(1)

where,

m_1,u_1\text{ and }v_1 are the mass, initial velocity and final velocity of first object

m_2,u_2\text{ and }v_2 are the mass, initial velocity and final velocity of second object

<u>For 1:</u>

We are Given:

m_1=150g=0.15kg\\u_1=?m/s\\v_1=0.85m/s\\m_2=3500g=3.5kg\\u_2=0m/s\\v_2=0.85m/s

Putting values in equation 1, we get:

(0.15\times u_1)+(3.5\times 0)=(3.5+0.15)\times 0.85\\\\u_1=20.683\approx 21m/s

Hence, the correct answer is Option C.

  • <u>For 2: </u>

Impulse is defined as the product of force applied on an object and time taken by the object.

Mathematically,

J=F\times t

where,

F = force applied on the object

t = time taken

J = impulse on that object

Impulse depends only on the force and time taken by the object and not dependent on the surface which is stopping the object.

Hence, the impulse remains the same.

  • <u>For 3:</u>

Let the speed in right direction be positive and left direction be negative.

We are Given:

m_1=240kg\\u_1=0m/s\\v_1=-1m/s\\m_2=80kg\\u_2=-2m/s\\v_2=?m/s

Putting values in equation 1, we get:

(240\times 0)+(80\times (-2))=(240\times (-1))+(80\times v_2)\\\\v_2=1m/s

Hence, the final velocity of the opponent is 1m/s and has moved backwards to its direction of the initial velocity.

4 0
3 years ago
????????????????!?!?!??
azamat

The amount of cytoplasm I think.


7 0
3 years ago
Read 2 more answers
Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz​
Ronch [10]

Answer:

This is because these metals are used for minting (making) coins.

8 0
3 years ago
The y-position of a damped oscillator as a function of time is shown in the figure.
NISA [10]

(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

<h3>at time, t = 0, y = 3.5</h3>

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

<h3>at time, t = 1 cm, y = - 3cm</h3>

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) =  e^(-bx)

ln[-0.857 / cos(ω)] = -bx  

ln[-0.857 / cos(ω)] / b = - x  ---- (1)

<h3>at time, t = 2 cm, y = - 2cm</h3>

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx  

ln[-0.57 / cos(2ω)] /2b = - x  ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) =  cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω  - cosω - 6 = 0

let cosω  = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω  = arc cos(-0.92)

ω  = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b  ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

Learn more about damping coefficient here: brainly.com/question/14058210

#SPJ1

4 0
2 years ago
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