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IRISSAK [1]
3 years ago
7

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

istance from the left end of the rod should a 0.6kg mass be hung to balance the rod?
a.48 cm
b. 50 cm
c. 45 cm
d. the rod can not be balanced with this mass. e.42 cm
NO LINKS. ​
Physics
1 answer:
Aleks04 [339]3 years ago
8 0
I’m pretty sure it’s b
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A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands
pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}

v=\infty

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}

v=34

The image distance is 34 cm right to the converging lens.

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5 0
4 years ago
What is the rising of rock layers called
Alex777 [14]

The answer of this question is called uplift i.e the rising of rock layers in geology is called uplift.

Uplift is the process in which there will be vertical elevation of earth surface due to various natural processes occurring inside the surface of earth like earth quakes. Sometimes uplift is also caused by the mountain building which hampers the balance of a particular region.

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6 0
3 years ago
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galina1969 [7]

Answer:

Force on the proton will be .73\times 10^{-14}N

Explanation:

We have given speed of proton is 20.3% of speed of light

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Magnetic field B = 0.00629 T

Charge on proton q=1.6\times 10^{-16}C

Angle between velocity and magnetic field \Theta =137^{\circ}

Force on the proton is equal to F=qvBsin\Theta =1.6\times 10^{-19}\times 6.9\times 10^7\times 0.00629\times sin(137^{\circ})=4.73\times 10^{-14}N

4 0
3 years ago
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