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3 years ago

Answer and I will give you brainiliest Please heeeelp

Physics

1 answer:

Lostsunrise [7]3 years ago

5 0

Answer:

T = 4.905[N]

Explanation:

In order to solve this problem we must perform a sum of forces on the vertical axis.

∑Fy = 0

We have two forces acting only, the weight of the body down and the tension force T up, as the body does not move we can say that it is system is in static equilibrium, therefore the sum of forces is equal to zero.

$T-m*g=0\\T=0.5*9.81\\T=4.905[N]$

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You and a friend are playing with a bowling ball to demonstrate some ideas of Rotational Physics. First, though, you want to cal

RideAnS [48]

Answer:

K_{total} = 19.4 J

Explanation:

The total kinetic energy that is formed by the linear part and the rotational part is requested

$K_{total} = K_{traslation} + K_{rotation}$

let's look for each energy

linear

$K_{traslation}$ = ½ m v²

rotation

$K_{rotation}$ = ½ I w²

the moment of inertia of a solid sphere is

I = 2/5 m r²

we substitute

$K_{total}$ = ½ mv² + ½ I w²

angular and linear velocity are related

v = w r

we substitute

K_{total} = ½ m w² r² + ½ (2/5 m r²) w²

K_{total} = m w² r² (½ + 1/5)

K_{total} = $\frac{7}{10}$ m w² r²

let's calculate

K_{total} = $\frac{7}{10}$ 6.40 16.0² 0.130²

K_{total} = 19.4 J

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2 years ago

What is parallax mapping?

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3 years ago

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V batte

Lelechka [254]

Answer:

$2.25\mu C$

Explanation:

At the beginning, we have:

V = 4.0 V potential difference across the capacitor

$Q=9.0 \mu C=9.0\cdot 10^{-6}C$ charge stored on the capacitor

Therefore, we can calculate the capacitance of the capacitor:

$C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F$

Later, the battery is replaced with another battery whose voltage is

V = 5.0 V

Since the capacitance of the capacitor does not change, we can calculate the new charge stored:

$Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C$

Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from $9.0 \mu C$ to $11.25 \mu C$ . Therefore, the additional charge that moved to the positive plate is