The term wavelength refers to which of the following?

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 15.0 ∘ west of north,

Darina [25.2K]

Answer:

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

Explanation:

The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.

The angle between the 2 forces and displacement is ∅ = 15°.

First we have to calculate the work done by the individual force and then we can calculate the total work.

The work done on a particle by a constant force F during a straight line displacement s is given by following formula:

W = F*s

W = F*s*cos∅

With ∅ = the angles between F and s

The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N

The total work done can be calculated as followed:

Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2

Wtotal = 2Fs*cos∅

Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°

Wtotal = 3.44*10^9 Nm = <u>3.44 *10^9 J</u>

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The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

5 0

2 years ago

When the starter motor on a car is engaged, there is a 290 A current in the wires between the battery and the motor. Suppose the

Tema [17]

Answer:

$33.6\times10^{-4}}\text{ m}$ = 3.36 mm

Explanation:

From Ohm's law,

$V = IR$ (Voltage = Current * Resistance)

$R = \dfrac{V}{I}$

The geometric definition of resistance is

$R = \rho\dfrac{l}{A}$

where $\rho$ is the resistivity of the material, $l$ and $A$ are the length and cross-sectional area, respectively.

$A = \rho\dfrac{l}{R}$

$A = \rho\dfrac{l\timesI}{V}$

Since the wire is assumed to have a circular cross-section, its area is given by

$A = \pi\dfrac{d^2}{4}$ where $d$ is the diameter.

$\pi\dfrac{d^2}{4} = \rho\dfrac{l\timesI}{V}$

$d = \sqrt{\dfrac{4\rho l I}{\pi\times V}}$

Resistivity of copper = $1.68\times10^{-8}$ . With these and other given values,

$d = \sqrt{\dfrac{4\times1.68\times10^{-8}\times1.5\times290}{3.14\times 0.55}}$

$d = \sqrt{1128.43\times10^{-8}}$

$d = 33.6\times10^{-4} \text{ m}$

5 0

3 years ago

Can you respond this two questions, please? :

Andrews [41]

Where's the diagram for question 1?

6 0

3 years ago

IM SOO CONFUSED PLS HELP!! The mass of the nucleus is approximately EQUAL to the mass number multiplied by ____ Atomic Mass unit

nevsk [136]

Answer:

option a.

Explanation:

We can think of an atom as a nucleus (where the protons and neutrons are) and some electrons orbiting it.

We also know that the mass of an electron is a lot smaller than the mass of a proton or the mass of an electron.

So, if all the protons and electrons of an atom are in the nucleus, we know that most of the mass of an atom is in the nucleus of that atom.

Then we define the mass number, which is the total number of protons and neutrons in an atom. Such that the mass of a proton (or a neutron) is almost equal to 1u

Then if we define A as the total number of protons and neutrons, and each one of these weights about 1u

(where u = atomic mass unit)

Then the weight of the nucleus is about A times 1u, or:

A*1u = A atomic mass units.

Then the correct option is:

The mass of the nucleus is approximately EQUAL to the mass number multiplied by __1__ Atomic Mass unit.

option a.

5 0

3 years ago

(c) Due to up thrust

garri49 [273]

Answer:

B1. Pascal's law is a principal in fluid mechanics given by Blaise Pascal that states that a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid that the same change occur everywhere. 2 applications of Pascal's law are hydraulic lifts, hydraulic jacks, hydraulic hydraulic brakes ,hydraulic pumps. mark me as a braintalist list plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

5 0

3 years ago