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musickatia [10]

3 years ago

11

if a force of 50N is used to pull a box along the ground, at a distance of 8m. The box moves in the same direction as the force.

Calculate the work done.

1 answer:

ruslelena [56]3 years ago

5 0

Answer:

<h2>400 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 50 × 8

We have the final answer as

<h3>400 J</h3>

Hope this helps you

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A wire 6.90 m long with diameter of 2.15 mm has a resistance of 0.0320 Ω. Find the resistivity of the material of the wire. rho

spayn [35]

<h2>Answer:</h2>

1.68 x 10⁻⁸Ωm

<h2>Explanation:</h2>

The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;

R = ρL/A ------------------------(i)

Where;

A = πd² / 4 [where d = diameter of the wire]

From the question;

L = 6.90m

d = 2.15mm = 0.00215m

R = 0.0320Ω

First calculate the crossectional area (A) of the wire as follows;

A = πd² / 4

[Take π = 3.142]

d = 0.00215m

∴ A = 3.142 x (0.00215)² / 4

∴ A = 0.000003631m²

Now, substitute the values of A, L, and R into equation (i) as follows;

R = ρL/A

0.0320 = ρ x 6.90 / 0.000003631

0.0320 = 1900302.95 x ρ

Solve for ρ;

=> ρ = 0.0320 / 1900302.95

=> ρ = 1.68 x 10⁻⁸Ωm

Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm

4 0

2 years ago

At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6

Elis [28]

The particle has constant acceleration according to

$\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k$

Its velocity at time $t$ is

$\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du$

$\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t$

$\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k$

Then the particle has position at time $t$ according to

$\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du$

$\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k$

At at the point (3, 6, 9), i.e. when $t=0$ , it has speed 8, so that

$\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64$

We know that at some time $t=T$ , the particle is at the point (5, 2, 7), which tells us

$\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}$

and in particular we see that

$v_{0y}=-2v_{0x}$

and

$v_{0z}=-v_{0x}$

Then

${v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3$

$\implies v_{0y}=\mp\dfrac{8\sqrt6}3$

$\implies v_{0z}=\mp\dfrac{4\sqrt6}3$

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

$\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k$

4 0

3 years ago

A balloon contains 0.04m3 of air at a pressure of 1.20 x 105Pa. Calculate the pressure required to reduce its volume to 0.025m3

sdas [7]

Answer:

you can simply answer p1v1=p2v2 p2 = p1v1÷ p2 then inter what given in the formula you get it pl=1.20x 105pa vl= 0.40m3 v2=0.025m3

4 0

3 years ago

Which identifies a machine that converts mechanical energy into electrical energy?

yKpoI14uk [10]

The answer is the option c. generator. Generator is a machine that converts mechanical energy into electricity. The way it is made is because the mechanical movement may be transformed in rotation of magnets placed inside a coiled wire, and this induces the movement of electrons through the coiled wire. As you know electricity is the flow of electrons.

8 0

2 years ago

You throw a rock straight up into the air with a speed of 14.2 m/s. how long does it take the rock to reach its highest point?

slega [8]

The acceleration of gravity on or near the Earth's surface is 9.8 m/s² downward.
Is that right ? I don't hear any objection, so I'll assume that it is.

That means that during every second that gravity is the only force on an object,
the object either gains 9.8m/s of downward speed, or it loses 9.8m/s of upward
speed. (The same thing.)

If the rock starts out going up at 14.2 m/s, and loses 9.8 m/s of upward speed
every second, it runs out of upward gas in (14.2/9.8) = <em>1.449 seconds</em> (rounded)

At that point, since it has no more upward speed, it can't go any higher. Right ?

(crickets . . .)

4 0

3 years ago