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3 years ago

Calculate the work done if a boy lifts a bag of cement 500N to the floor of a lorry 2.5m above the ground

Physics

1 answer:

eduard3 years ago

4 0

Answer:

W = 1250 J = 1.25 KJ

Explanation:

The work done by the boy is due to the change in the position of the cement vertically. Hence, the work done in this case will be equal to the potential energy of the cement:

$W = P.E\\W = mgh$

where,

W = Work done = ?

mg = W = weight of cement = 500 N

h = height covered = 2.5 m

Therefore,

$W = (500\ N)(2.5\ m)$

W = 1250 J = 1.25 KJ

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The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

Circular Motion

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

3 years ago

A 1050 W carbon-dioxide laser emits light with a wavelength of 10μm into a 3.0-mm-diameter laser beam. What force does the laser

avanturin [10]

The force exerted by the laser beam on a completely absorbing target is $3.5 \times 10^{-6} \ N$ .

The given parameters;

power of the laser light, P = 1050 W
wavelength of the emitted light, λ = 10 μm

The speed of the emitted laser light is given as;

v = 3 x 10⁸ m/s

The force exerted by the laser beam on a completely absorbing target is calculated as follows;

P = Fv

$F = \frac{P}{v} \\\\F = \frac{1050}{3\times 10^8} \\\\F = 3.5 \times 10^{-6} \ N$

Thus, the force exerted by the laser beam on a completely absorbing target is $3.5 \times 10^{-6} \ N$ .

Learn more here:brainly.com/question/17328266

3 0

2 years ago

A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an

AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

3 0

3 years ago

Sound waves cannot carry energy through. A water B air C a mirror D a vacuum

Ber [7]

I looked up the question and got D- a vacuum

3 0

3 years ago

If rho(x,y) is the density of a wire (mass per unit length), then

lidiya [134]

Answer:

See description

Explanation:

With the given information we have:

$x(t) = 1 + cos(t)\\ y(t)=sin(t)\\ \rho(x,y) = 3x$

the interval is $[0,\pi ]$

now the mass $m$ has the given expression:

$m = \int \rho(x,y) dS$

we will use the formula for a line integral and let:

$dS=\sqrt{x'(t)^2 + y'(t)^2}=\sqrt{cos(t)^2 + sin(t)^2}dt=dt$

therefore we have:

$m=\int \rho(x,y)dS=\int\limits^\pi_0 {3*x}dS=\int\limits^\pi _0{3*(1+cos(t))dS\\=\int\limits^\pi _0{3*(1+cos(t))dt$

we solve the integral:

$m=3*\int\limits^\pi _0{(1+cos(t))dt= 3*(t+sin(t))\limits^\pi _0=3*\pi=9.42$

7 0

3 years ago