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IRINA_888 [86]
3 years ago
14

A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of

the see saw in order to perfectly balance the child and rock?
Physics
1 answer:
ivann1987 [24]3 years ago
6 0

Answer:

a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw

Explanation:

τchild=τrock  

Use the equation for torque in this equation.

(F)child(d)child)=(F)rock(d)rock)

The force of each object will be equal to the force of gravity.

(m)childg(d)child)=(m)rockg(d)rock)

Gravity can be canceled from each side of the equation. for simplicity.

 (m)child(d)child)=(m)rock(d)rock)  

Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.

(25kg)(1m)=(50kg)drock

Solve for the distance between the rock and the center of the seesaw.

drock=25kg⋅m50kg

drock=0.5m

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While water skiing behind her father’s boat, Letty is pulled at constant speed by a force of 164 N from the tow rope that makes
kap26 [50]

Answer:

0.265

Explanation:

Draw a free body diagram.  There are four forces:

Normal force Fn pushing up.

Weight force mg pulling down.

Tension force T at an angle θ.

Friction force Fn μ pushing left.

Sum the forces in the y direction:

∑F = ma

Fn + T sin θ − mg = 0

Fn = mg − T sin θ

Sum the forces in the x direction:

∑F = ma

T cos θ − Fn μ = 0

Fn μ = T cos θ

μ = T cos θ / Fn

μ = T cos θ / (mg − T sin θ)

Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:

μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)

μ = 0.265

7 0
3 years ago
Light of wavelength 623.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 76.5 cm from the slit.
Dvinal [7]

Answer:

Explanation:

wave length of light λ = 623 x 10⁻⁹ m .

Distance of screen D = 76.5 x 10⁻² m

width of slit        =      d

Distance on the screen between the second order minimum and the central maximum       =  2  λ D / d

1.11 x 10⁻²  = (2 x 623 x 10⁻⁹ x 76.5 x 10⁻² )/ d

d =  ( 2 x 623 x 10⁻⁹ x 76.5 x 10⁻²) / 1.11 x 10⁻²

= 85872.97 x  10⁻⁹

=  85.87297 x  10⁻⁶

= 85.87 μm

width a of the slit is = 85.87 μm

6 0
3 years ago
A student makes the following statement
Aliun [14]

Answer:

The current is not used up. The electrons flow through the entire circuit and this travel is the current. They flow until they are not charged anymore. That is also why the circuit must be closed or else electrons would escape not just light it up for a second then go out.

Explanation:

8 0
4 years ago
When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured
zaharov [31]

Answer:

\gamma=6.07*10^5\frac{N}{m^2}

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

\gamma=\frac{F/A}{\Delta L/L_0}

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, \Delta L is the amount by which the length of the object changes and  L_0 is the original length of the object. In this case the force is the weight of the mass:

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Replacing the given values in Young's modulus formula:

\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}

6 0
3 years ago
What's the difference between meteoroids, meteorites, and meteors?
SVEN [57.7K]

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-- While it's drifting in space, it's a <em>meteoroid</em>.

-- While it's falling through the atmosphere burning up and making a bright streak of light, it's a <em>meteor</em>.

-- If it doesn't completely burn up and there's some of it left to fall on the ground, then the leftover piece on the ground is a <em>meteorite</em>.

4 0
4 years ago
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