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Vsevolod [243]
3 years ago
6

Two particles collide, one of which was initially moving and the other initially at rest. Is it possible for both particles to b

e at rest after the collision?
Physics
1 answer:
solmaris [256]3 years ago
4 0

Answer:

Not possible

Explanation:

Unless there's some extra external force to keep both particles at rest after the collision, the momentum must be conserved before and after the collision.

So before the collision, 1 particle is at rest, 1 not ->  total momentum is non-zero

After the collision, both particles are at rest -> total momentum is zero which is different from before.

Therefore this is not possible.

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Consider a turnbuckle that has been tightened until the tension in wire AD is 350 N. Draw the FBD that is required to determine
Mars2501 [29]

Answer:

yes

Explanation:

yes

6 0
2 years ago
A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential
castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge q=+10.5\times10^{-6}\ C

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

\Delta U=-W

\Delta U=-F\cdot d

\Delta U=-q(E\cdot d)

Put the value into the formula

\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}

\Delta U=-3.32\times10^{-3}\ J

The change in electric potential energy  is -3.32\times10^{-3}\ J

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

\Delta U=-W

\Delta U=-F\cdot (-d)

\Delta U=-q(E\cdot (-d))

Put the value into the formula

\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})

\Delta U=3.32\times10^{-3}\ J

The change in electric potential energy  is 3.32\times10^{-3}\ J

Hence, This is the required solution.

3 0
3 years ago
If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.
AleksAgata [21]

Answer:

a) f_1=5.587Hz

b) f_{n+1}-f_n=5.587Hz

Explanation:

The frequency of the n^{th} harmonic of a vibrating string of length <em>L, </em>linear density \mu under a tension <em>T</em> is given by the formula:

f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}

a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:

f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz

3 0
2 years ago
15.One girl was involved in a bus crash. She told the police that she was seated in the last line of seats and that when the bus
Alona [7]

Answer: It is not likely.

Explanation:

When the bus is moving forward, all the objects inside of it also are moving forward.

Now, as the objects inside the buss are not fixed to the bus, if the bus suddenly stops the objects inside of it will keep moving forward, because of the conservation of the momentum, defined as the quantity of motion (Similar to when you are in a car and it suddenly stops, you can feel the forward impulse).

Then is not likely that, in a case where the bus stops suddenly, an object inside the bus flies backward in opposite direction to the previous movement of the bus.

5 0
2 years ago
a block with length 1.5m width 1m height 0.5m and mass 300kg lays on the table.what is the pressure at the bottom surface of the
Nesterboy [21]

Answer:

your answer will be 320kg that would be the pressure at the bottom surface of the block

6 0
2 years ago
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