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3 years ago

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A particle has velocity v⃗1 v → 1 as it accelerates from 1 to 2. What is its velocity vector v⃗2 v → 2 as it moves away from poi

nt 2 on its way to point 3?

1 answer:

Jobisdone [24]3 years ago

8 0

Answer:

The velocity of the particle will be downward.

Explanation:

Given that,

The velocity of a particle is v₁. It is accelerated from 1 to 2.

If it moves away from point 2 on its way then

We need to find the velocity of particle

According to figure,

A particle moves downward from 1 to 2 with the velocity v₁ and after that the particle moves downward from 2 to 3 with the velocity v₂.

Hence, The velocity of the particle will be downward.

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The greater the mass of an object, the greater the _____. (fill in question)

g100num [7]

Answer:

momentum

Explanation:

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3 years ago

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Calculate the magnitude of the total impulse applied to the car to bring it to rest

n200080 [17]

Answer:

Total impulse = $mv$ = Initial momentum of the car

Explanation:

Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.

The final velocity of the car is 0 m/s as it is brought to rest.

Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.

The magnitude of impulse is the absolute value of the change in momentum.

$|J|=|p_f-p_i|$

Momentum of an object is equal to the product of its mass and velocity.

So, the initial momentum of the car is given as:

$p_i=mv$

The final momentum of the car is given as:

$p_f=m(0)=0$

Therefore, the impulse is given as:

$|J|=|p_f-p_i|=|0-mv|=|-mv|=mv$

Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.

5 0

3 years ago

A vehicle moving at 5m/s, i . what should be the constant declaration in order to stop it within 15m? ii. How long it takes to s

Flura [38]

Answer:

Refer to the attachment for solution (1).

<h3><u>Calculating time taken by it to stop (t) :</u></h3>

By using the second equation of motion,

→ v = u + at

v denotes final velocity
u denotes initial velocity
t denotes time
a denotes acceleration

→ 0 = 5 + (-5/6)t

→ 0 = 5 - (5/6)t

→ 0 + (5/6)t = 5

→ (5/6)t = 5

→ t = 5 ÷ (5/6)

→ t = 5 × (6/5)

→ t = 6 seconds

→ Time taken to stop = 6 seconds

7 0

3 years ago

Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien

jeka94

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis $\theat =45^{\circ}$

$S_1=S_0\cos ^2\theta$

here $S_0=\frac{S}{2}$

$S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}$

$S_1=\frac{S}{4}$

When it is passed through third Polarizer with its axis $90^{\circ}$ to first but $\theta =45^{\circ}$ to second thus $S_2$

$S_2=S_0\cos ^2\theta$

$S_2=\frac{S}{4}\times \frac{1}{2}$

$S_2=\frac{S}{8}$

When middle sheet is absent then Final Intensity will be zero

3 0

3 years ago

g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie

polet [3.4K]

Answer: $71.16\ Hz$

Explanation:

Given

Capacitance $C=100\ \mu F$

Resistance $R=500\ \Omega$

Inductance $L=50\ mH$

In LCR circuit, current is maximum at resonance frequency i.e.

$X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}$

Insert the values

$\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}$

Also, frequency is given by

$\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}$

$\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz$

8 0

3 years ago