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Serhud [2]
4 years ago
14

PLEASE HELP ASAP. Best answer gets BRAINIEST:

Physics
1 answer:
nikklg [1K]4 years ago
3 0

Answer:

1. Force = -623.43 N

2.Tension = 106.65 N

Explanation:

Let us call the bigger mass M, and the smaller mass m.

Since the two masses are connected to each other, they must experience same acceleration (if they didn't, the unequal acceleration will cause the string to break. )

From the free body diagram, the forces acting on the mass M are

T - Mg-F_p,

and according to Newton Second Law, this causes acceleration a; therefore,

(1).  \boxed{ T-Mg-F_p=Ma}.

Similarly, the forces acting on the mass m are

T -mg,

which causes the acceleration -a (upward); therefore,

T-mg=-ma \\

or

\boxed{ mg-T = ma}

From this equation we solve for T and get:

T = mg-ma \\\\T =m(g-a).

We put this into equation (1) and get:

m(g-a)-Mg-F_p=Ma

F_p = m(g-a)-M(g+a)

putting in M=62.4kg,m=13.5kg, and a=1.9m/s^2, we get:

F_p=13.5(9.8-1.9)-62.4(9.8+1.9)\\\\\boxed{ F_p=-623.43N}

The tension in the wire is

T =m(g-a)\\\\T = 13.5(9.8-1.9)\\\\\boxed{ T= 106.65N}

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Electromagnetic waves are produced when _____.
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A net force of 10.N accelerates an object at 5.0 m/s^2. What net force would be required to accelerate the same object at 1.0 m/
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I believe the answer is 2.0

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A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surf
liberstina [14]

Answer:

The apparent depth of (a) the fish is 5.3 cm and (b) the image of the fish is 24.8 cm.

Explanation:

According to the following equation:

\frac{n_{w} }{s} +\frac{n_{a} }{s'} = \frac{n_{a}- n_{w}}{R_{c} } \\

where <em>nw</em> and <em>na</em> is the refractive indices of water (1.33) and air (1.00) respectively; <em>s</em> is the depth of the fish below the surface of the water; s' is the apparent depth of the fish from normal incidence and Rc is the radius of curvature of the mirror at the bottom of the tank.

Note that the bottom of the tank is assumed to be a flat mirror, therefore the radius of curvature is very large (R⇒∞).

Therefore, the above equation can be expressed as:

\frac{n_{w}}{s} +\frac{n_{a}}{s'}=0

Now we can solve for the apparent depth of the fish.

(a) s'=-(\frac{n_{a}}{n_{w}})x s (Make s' subject of the formula from the above equation)

s'=(\frac{1.00}{1.33} )x7cm

∴ s'=5.3 cm.

(b) The motionless fish floats 13 cm above the mirror, therefore the image of the fish will be situated at 13 + 20 =33 cm away from the real fish.

Therefore, s = 33 cm

s'=-(\frac{n_{a}}{n_{w}})x s

s'=(\frac{1.00}{1.33} )x 33 cm

s'=24.8 cm.

NB: Here, it is assumed that the water is pure, as impurities may alter the refractive index of water.

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