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3 years ago

A speedboat initially at rest accelerates uniformly at 4.0 m/s (squared) for 7.0 s. How fast is the boat moving after 7.0 s?

1 answer:

8 0

Well if the boat initially at rest accelerates at uniformly at 4.0 m/s (squared) then it would be best to muitlply it so 4.0 squared equals 2 by multiplying that by 7.0 your answer would be 14 s

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1.Calculate the energy transferred by a 12V hairdryer, running on a current of 0.50A, that is left on for 8.0 minutes.

CaHeK987 [17]

Answer:

1. Energy = 2880 Joules.

2. Energy = 60 Joules.

3. Quantity of charge = 120 Coulombs.

Explanation:

Given the following data;

1. Voltage = 12 Volts

Current = 0.5 Amps

Time, t = 8 mins to seconds = 8 * 60 = 480 seconds

To find the energy;

Power = current * voltage

Power = 12 * 0.5

Power = 6 Watts

Next, we find the energy transferred;

Energy = power * time

Energy = 6 * 480

Energy = 2880 Joules

2. Charge, Q = 4 coulombs

Potential difference, p.d = 15V

To find the total energy transferred;

Energy = Q * p.d

Energy = 4 * 15

Energy = 60 Joules

3. Voltage = 6 Volts

Current = 1 Amps

Time = 2 minutes to seconds = 2 * 60 = 120 seconds

To find the quantity of charge;

Quantity of charge = current * time

Quantity of charge = 1 * 120

Quantity of charge = 120 Coulombs

8 0

3 years ago

Karen has a mass of 51.9 kg as she rides the up escalator at Woodley Park Station of the Washington D.C. Metro. Karen rode a dis

kifflom [539]

Answer:

28852 J

Explanation:

When a force applied in a body produces a displacement in it, the force realized a work. The force that moves Karen is contrary to her weight and must be equal to it.

The work (W) is:

W = F.d.cos(θ), where F is the force, d is the displacement, and θ is the angle.

Knowing that cos(26°) = 0.899, and F = m*g

W = 51.9*9.8*63.1*0.899

W = 28852 J

4 0

3 years ago

A box of weight 280 N is being pulled to the right by a pulling force vec F of magnitude 50 N. The box is moving at a constant s

boyakko [2]

Answer:

what diagram. I can't see any

5 0

3 years ago

two circular loops of wire, each containing a single turn, have the same radius of 4.0 cm and a common center. the planes of the

REY [17]

The magnetic field at center of circular loops of wire is 3.78 x 10¯⁵ T.

We need to know about the magnetic field at the center of circular loops of wire to solve this problem. The magnetic field at the center can be determined as

B = μ₀ . I / 2r

where B is magnetic field, μ₀ is vacuum permeability (4π×10¯⁷ H/m), I is the current and r is radius.

From the question above, we know that:

r = 4 cm = 0.04 m

I = 1.7 A

By substituting the parameter, we get

B = μ₀ . I / 2r

B = 4π×10¯⁷ . 1.7 / (2.0.04)

B = 2.67 x 10¯⁵ T

Due to the perpendicular plane of loops, the total magnetic field at center will be

Btotal = √(2(B²))

Btotal = √(2(2.67 x 10¯⁵²))

Btotal = 3.78 x 10¯⁵ T

Find more on magnetic field at: brainly.com/question/7802337

#SPJ4

3 0

1 year ago

The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.

S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

$\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$m$ - Mass of the physical pendulum, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$d$ - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

$I_{O}$ - Moment of inertia with respect to pivot point, measured in $kg\cdot m^{2}$ .

In addition, frequency and angular frequency are both related by the following formula:

$\omega =2\pi\cdot f$

Where:

$f$ - Frequency, measured in hertz.

If $f = 0.658\,hz$ , then angular frequency of the physical pendulum is:

$\omega = 2\pi \cdot (0.658\,hz)$

$\omega = 4.134\,\frac{rad}{s}$

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

$\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}$

$I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}$

Given that $m = 1.15\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $d = 0.425\,m$ and $\omega = 4.134\,\frac{rad}{s}$ , the moment of inertia associated with the physical pendulum is:

$I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}$

$I_{o} = 0.280\,kg\cdot m^{2}$

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

8 0

3 years ago