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emmasim [6.3K]
3 years ago
12

When a downward force is applied at a point 0.60 m to the left of a fulcrum, equilibrium is obtained by placing a mass of 10–7 k

g at a point 0.40 m to the right of the fulcrum. What is the magnitude of the downward force
Physics
1 answer:
Sliva [168]3 years ago
4 0

Answer:

force = 6.53×  10^{-7} N

Explanation:

Given data

downward force = 0.60 m

mass m  = 10^{-7} kg

distance h = 0.40 m

to find out

magnitude of the downward force

solution

we know here mg is apply 0.4 m away from support and

so applied force is d =  0.6 m from support

so

by balancing torque we can calculate force

that is

force = mass  × g ×  h / d

put here all these value

force =  mass  × g ×  h / d

force =  ( 10^{-7}  × 9.81 ×  0.4 )  / 0.6

force = 6.53×  10^{-7} N

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A standard inverting op-amp circuit has an R1 of 10 kΩ and an Rf of 220 kΩ. If the offset current is 100 nA the output offset vo
kiruha [24]

Answer:

The value is  V_{os} = 0.001 \  V

Explanation:

From the question we are told that

     The circuit resistance is  R_1 =  10 \ k \Omega

     The feedback resistance  is  R_f =  220 \ k \Omega

      The offset current is  I_{os } = 100 \  nA  =  100 * 1)^{-9} \ A

Generally the offset voltage is mathematically reparented as

           V_{os} =  R_f * I_{os}

=>        V_{os} = 10 *10^{3}*  100 *10^{-9}

=>        V_{os} = 0.001 \  V

6 0
3 years ago
Using the
xxMikexx [17]

Answer:

The principle of momentum conservation states that if there no external force the total momentum of the system before and after the collision is conserved.

Since momentum is a vector, we should investigate the directions and magnitudes of initial and final momentum.

\vec{P}_{initial} = \vec{P}_{final}\\\vec{P}_{initial} = m_1\vec{v}_1 + 0\\\vec{P}_{final} = m_1\vec{v}_1' + m_2 \vec{v}_2'

If the first ball hits the second ball with an angle, we should separate the x- and y-components of the momentum (or velocity), and apply conservation of momentum separately on x- and y-directions.

6 0
3 years ago
A student's life was saved in an automobile accident because an airbag expanded in front of his head. If the car had not been eq
dexar [7]

Answer:    

Compared to windshield the airbag exerts much lesser force  

Explanation:

Impulse is defined as change in momentum of the object when it is acted upon by a force during interval of time

<em>Impulse = Impulsive force *time</em>

I = F*Δt

If the object should be bought to rest from certain velocity there should be change in momentum. If the duration in which the momentum is increased then there would be less force applied and hence less damage.

Airbags are used to reduce the force experience by the people when they are met with accident by extending the time required to stop the momentum.

During the collision, the passenger is carried towards the<em> windshield</em> and if they are stopped by collision with wind shield the force will be larger and more damage.But if they are hit with airbag then the force will be less due to increased time.

The change is momentum will be the same with or without momentum but its the time that decides the impact of force.By making it longer the force become less.

<em>Thus compared to the windshield the airbag exerts much lesser force.</em>

<em>   </em>

8 0
3 years ago
A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa
Jlenok [28]

Answer:

The answer is 18 N.

Explanation:

A force can be divided into components x and y components. The component along the x-axis is called the horizontal component and along the y-axis is called the vertical component. In this case, as the force is in a horizontal direction and is also known as x-component of force. The x- component of force is  

Fx = Fcosθ

Fx = 22(cos 35°)

Fx = 22 x 0.819

Fx =  18 N

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5 0
3 years ago
A cyclist rides at 6.20 m/s through a intersection. A stationary car begins to
Xelga [282]

Answer:

The width of the intersection is 20 meters

Explanation:

The speed with which the cyclist is riding, v₁ = 6.20 m/s

The rate at which the car starts to accelerate, a = 3.844 m/s²

The initial velocity of the car = The car is stationary at the start = 0 m/s

The time at which the cyclist and the car reach the other side of the intersection = The same time;

Let 't' represent the time at which the cyclist and the car both reach the other side of the intersection, we have;

The distance travelled by the cyclist = The distance traveled by the car

∴ v₁ × t = 1/2 × a × t²

Plugging in the values for 'v₁', and 'a' in the above equation, we get;

6.20 × t = 1/2 × 3.844 × t²

∴ 1.922·t² - 6.20·t = 0

∴ t·(1.922·t - 6.20) = 0

t = 0, or t = 6.20/1.922 = 100/31

The time at which the cyclist and the car both reach the other side of the intersection, t = 100/31 seconds

The with of the intersection, w = v₁ × t

∴ w = 6.20 × 100/31 = 100/5 = 20

The width of the intersection, w = 20 meters.

8 0
3 years ago
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