Can someone complete this assignment will give brainliest!<br> The assignment is attached

A thin rod of length 1.4 m and mass 180 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

prisoha [69]

Answer:

$K.E = 0.1905 J$

Explanation:

From the question we are told that:

Length $L=1.4m$

Mass $m=180g$

Angular Velocity $\omega=1.80rads/s$

Generally the equation for Kinetic energy K.E is mathematically given by

$K.E =0.5 (1/3 ML^2 )w^2$

$K.E =0.5 ( 1/3 * 0.18 * 1.4^2 ) 1.8^2$

$K.E = 0.1905 J$

5 0

3 years ago

A 36N force applied vertically upwards on an object causes it to accelerate at 2.0m/s2. What is the object's mass?

Natalija [7]

Force = mass * acceleration

mass = Force/acceleration

= 36/2

mass = 18 kg

7 0

3 years ago

Estimate the kinetic energy (in GJ) of a 93,000 metric ton aircraft carrier moving at a speed of at 32 knots. You will need to l

Wewaii [24]

Answer:

kinetic Energy = 12.58 GJ

Explanation:

1 metric ton is equal to 1000 kg

then,

93000 metric ton

Mass is m = 93000 x 1000 kg

speed is v = 32 knots

1 knot is 0.514 m/s

then ,

v = 32 x 0.514 = 16.448 m/s

To solve for the Kinetic Energy (KE), we have;

KE = 0.5 x m x v²

KE = 0.5*93000*1000*(16.448)²

= 12.58 x $10^{9}$ J

= 12.58 GJ

6 0

3 years ago

A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2

Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S

2) t = 1 mu S

3) t = 50 mu S

1) $22.4 \mu C$

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

$C=\frac{Q_0}{V_0}$

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

$C=4.4\mu F = 4.4\cdot 10^{-6}F$

$V_0 = 5.1 V$

Solving for $Q_0$ ,

$Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C$

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) $20.0\mu C$

To find the charge on the capacitor at any other time t, we use the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

where

$Q_0 = 22.4 \mu C$

t is the time

$R=2.0 \Omega$ is the resistance

$C=4.4\mu F$ is the capacitance

Therefore, at time $t=1 \mu s$ , we have:

$Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C$

3) $0.08 \mu C$

As before, we use again the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

However, here the time to consider is

$t=50 \mu C$

Substituting into the formula,

$Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C$

4 0

3 years ago

A pendulum has a mass of 0.060 kg swinging at a small angle from a light string, with a period of 1.4 s. What is the length of t

jok3333 [9.3K]

Answer:

0.5 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 0.060 kg

Period (T) = 1.4 s

Lenght (L) =?

NOTE:

1. Acceleration due to gravity (g) = 10 m/s²

2. Pi (π) = 3.14

The length of the pendulum can be obtained as follow:

T = 2π√(L/g)

1.4 = 2 × 3.14 × √(L/10)

1.4 = 6.28 × √(L/10)

Divide both side by 6.28

1.4 / 6.28 = √(L/10)

Take the square of both side

(1.4 / 6.28)² = L/10

Cross multiply

L = 10 × (1.4 / 6.28)²

L = 0.5 m

Therefore, the length of the pendulum is 0.5 m

4 0

3 years ago

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Can someone complete this assignment will give brainliest! The assignment is attached