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saul85 [17]
3 years ago
6

Help plzzz, if a question has more than one answer, pur both letters! Plzz I need 5 letters, this is the third time I’m posting

this for help, someone plzzzz help!!

Physics
1 answer:
Kaylis [27]3 years ago
4 0

Answer:

1) e

2)c

3)b

4)d

i dont know if these are right, but something along these lines if you understand the pattern

Explanation:

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A ball thrown into the air has its greatest kinetic energy as it reaches the highest point in its path. please select the best a
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F - False.

Its greatest kinetic energy is at the point of release.

It has the least kinetic energy, zero, at its highest point in its path.
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The fulcrum is the
olga_2 [115]

Answer:

point of support on which a lever rotates.

Explanation:

The fulcrum is the point of support on which a lever rotates. Fulcrum is a pivotal part of simple machines.

The fulcrum provides the platform for a lever to torque.

  • The force that opposes motion by the applied force is termed the frictional force.
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  • The stored energy of an object is its potential energy.
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3 years ago
With a 12 V battery, what is the voltage drop across each resistor?
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In a series circuit, the voltage divides across the resistors in proportion to
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8 0
3 years ago
Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

Explanation:

Given there are three blocks of masses M_{a}, M_{b} and M_{c} (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *a    ................(equation 1)

Also it is given that M_{a} does not move with respect to M_{c}, which gives tension T  is exerted on pulley  by M_{a} only, Hence tension T is

T = M_{a} *a    ..........(equation 2)

There is also also tension exerted by M_{b}. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = M_{b} \sqrt{a^{2} +g^{2} }   ................(equation 3)

From equation 2 and 3, we get

M_{a} *a  = M_{b} \sqrt{a^{2} +g^{2} }  

Squaring both sides we get

M_{a} ^{2} *a^{2} = M_{b} ^{2} * (a^{2}+g^{2})

M_{a} ^{2} *a^{2} = (M_{b} ^{2} * a^{2})+ (M_{b} ^{2} *g^{2})

(M_{a} ^{2}  -  M_{b} ^{2}) * a^{2} = M_{b} ^{2} *g^{2}

a^{2} = M_{b} ^{2} *g^{2}/(M_{a} ^{2}  -  M_{b} ^{2})

Taking square root on both sides, we get acceleration a

a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

Hence substituting the value of a in equation 1, we get

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

3 0
3 years ago
Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?
Schach [20]
Using lens equation;

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6 0
3 years ago
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