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2 years ago

If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its kinetic energy right before it hits the ground?

Physics

1 answer:

Mariana [72]2 years ago

7 0

The kinetic energy of the phone right before it hits the ground is 9J.

<h3>Kinetic energy of the phone</h3>

The kinetic energy of the phone right before it hits the ground is calculated as follows;

K.E = ¹/₂mv²

where;

m is mass of the phone
v is velocity of the phone

K.E = ¹/₂(0.08)(15)²

K.E = 9 J

Thus, the kinetic energy of the phone right before it hits the ground is 9J.

Learn more about kinetic energy here: brainly.com/question/25959744

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A 3.0kg mass tied to a string

dem82 [27]

Answer:

$\boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN}$

Given:

Mass (m) = 3.0 kg

Uniform speed (v) = 20 m/s

Length of string (r) = 40 cm = 0.4 m

To Find:

Tension in the string (T)

Explanation:

Tension (T) is the string will be equal to centripetal force ( $\sf F_c$ ).

$\boxed{ \bold{ T = F_c = \frac{m {v}^{2} }{r} }}$

Substituting value of m, v & r in the equation:

$\sf \implies T = \frac{3 \times {20}^{2} }{0.4} \\ \\ \sf \implies T = \frac{3 \times 400}{0.4} \\ \\ \sf \implies T =3 \times 1000 \\ \\ \sf \implies T =3000 \: N \\ \\ \sf \implies T =3 \: kN$

$\therefore$

Tension in the string (T) = 3 kN

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3 years ago

Determine

djyliett [7]

(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

<h3>C1 and C2 are in series </h3>

$\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F$

<h3>C1 and C2 are parallel to C3</h3>

$C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F$

<h3>C(123) is series to C5 and C6</h3>

$\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F$