Ask question

Ask question

All categories

2 years ago

15

A ledge on a building is 18 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up ove

r a pulley and straight down to a 3.0-kg can of nails on the ground.If the can of paint is accidently knocked off the ledge, what time interval does a carpenter have to catch the can before it smashes on the floor?

1 answer:

True [87]2 years ago

5 0

Answer:t=5.07 s

Explanation:

Given

height of Building h=18 m

mass of Paint can $m_1=4 kg$

mass of second can $m_2=3 kg$

let T be the Tension in the rope

For 4 kg can

$4g-T=4a$

$T=4(g-a)$ ----1

For 3 kg can

$T-3g=3a$

$T=3(g+a)$ ----2

From 1 and 2

$4(g-a)=3(g+a)$

$g=7a$

$a=\frac{g}{7}$

So time taken to cover 18 m is

$h=ut+\frac{at^2}{2}$

$18=0+\frac{g\cdot t^2}{7\times 2}$

$t^2=\frac{18\times 2\times 7}{g}$

$t=5.07 s$

You might be interested in

A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad

maksim [4K]

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

Where,

F = Force/Weight

A = Area

$\Delta L$ = Compression

$L_0$ = Original Length

According to the values given we have to

$\Upsilon_{steel} = 200*10^9Pa$

$\Delta L = 5.6*10^{-7}m$

$L_0 = 3.2m$

$r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2$

Replacing this values at our previous equation we have,

$\Upsilon = \frac{F/A}{\Delta L/L_0}$

$200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}$

$F = 38272.5N$

Therefore the Weight of the object is 3.82kN

4 0

1 year ago

A device used to increase or decrease the emf in the second of two unconnected coils is a

grigory [225]

<span>The answer is transformer. They utilize electromagnetic induction to generate current. This is only possible in alternating current due to the differential increase and decrease of electrical current that induces changes in magnetic flux in the coil. This varies the magnetic flux of the primary coil that generates current in the secondary coil.</span>

4 0

2 years ago

A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.

Anna007 [38]

Answer:

$\mu_k = \frac{2(vt - L)}{(g + a) t^2}$

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

So here as per force equation in vertical direction we know that

$N - mg = ma$

so normal force on the block is given as

$N = mg + ma$

now the magnitude of kinetic friction on the block is given as

$F_f = \mu N$

$F_f = \mu (mg + ma)$

now when bag is sliding on the floor then net deceleration of the block due to friction is given as

$a = - \frac{F_f}{m}$

$a = -\mu_k(g + a)$

now we know that bag hits the opposite wall at L distance away in time t

so we have

$d = v t + \frac{1}{2}at^2$

$L = vt - \frac{1}{2}(\mu_k)(g + a) t^2$

$\mu_k = \frac{2(vt - L)}{(g + a) t^2}$

8 0

1 year ago

Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a

Phantasy [73]

The question is incomplete. It comes with a set of answer choices.

These are the answer choices:

Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.

Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.

Both Alex and his friend observe it as 10 m/s.

Both Alex and his friend observe it as less than 10 m/s.

Answer: Both Alex and his friend observe it as 10 m/s.

Justification:

1) The speed is relative to the frame of reference.

2) It is said that the both Alex and his friend are standing still.

3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).

Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

7 0

2 years ago

Read 2 more answers

Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

matrenka [14]

<span> (a) does an electric field exert a force on a stationary charged object?
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span> $F=qE$
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with charge q and speed v is
</span> $F=qvB \sin \theta$
where $\theta$ is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object?
Yes, The intensity of the electric force is still
</span> $F=qE$
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
$F=qvB \sin \theta$
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so?
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span> $F=qE$
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span> $F=qvB \sin \theta$
<span>is different from zero because v is different from zero.</span>

6 0

2 years ago