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2 years ago

A scooter is accelerated from rest at the rate of 8m/s . How long will it take to cover a distance of 32m?

Physics

1 answer:

TiliK225 [7]2 years ago

6 0

Speed = 8 m/s
Distance = 32 m

We need to find time

$\begin{gathered} {\underline{\boxed{ \rm {\red{Speed \: = \: \frac{Distance}{Time} }}}}}\end{gathered}$

$\begin{gathered} {\underline{\boxed{ \rm {\orange{Time \: = \: \frac{Distance}{Speed} }}}}}\end{gathered}$

<h3>Substuting the values</h3>

$\begin{cases} \large\rm{\purple{ {\leadsto} }} \: \: Time \: = \: \frac{32}{8} \\ \\ \large\rm{\purple{ {\leadsto} }} \: \: Time \: = \: \cancel\frac{32}{8} \\ \\ \large\rm{\purple{ {\leadsto} }} \: \: Time \: = \: 4 \: seconds \end{cases}$

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A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i

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Answer:

Explanation:

a )

The stored elastic energy of compressed spring

= 1 / 2 k X²

= .5 x 19.6 x (.20)²

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b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .

c ) Let h be the distance along the incline which the block ascends.

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3 years ago

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3 years ago

You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown

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Answer:

The mass is $m_2 =21.75*10^{-27} \ kg$

The velocity is $v_2 = 3.0*10^{6} m/s$

Explanation:

From the question we are told that

The speed of the protons is $u_1 = 2.10*10^{7} m/s$

The mass of the protons is $m$

The speed of the rebounding protons are $v_1 = -1.80 * 10^{7} \ m/s$

The negative sign shows that it is moving in the opposite direction

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

$m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1$

Where $m_1$ is the mass of a single proton

So substituting values

$m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1$

$m_2 =13 m_1$

The mass of on proton is $m_1 = 1.673 * 10^{-27} \ kg$

So $m_2 =13 ( 1.673 * 10^{-27} )$

$m_2 =21.75*10^{-27} \ kg$

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

$m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2$

Now $u_2$ because before collision the the nucleus was at rest

$m_1 u_1 = m_1 v_1 + m_2v_2$

=> $v_2 = \frac{m_1(u_1 -v_1)}{m_2}$

Recall that $m_2 =13 m_1$

$v_2 = \frac{m_1(u_1 -v_1)}{13m_1}$

=> $v_2 = \frac{(u_1 -v_1)}{13}$

substituting values

$v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}$

$v_2 = 3.0*10^{6} m/s$

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3 years ago

A certain substance has a heat of vaporization of 37.51 kJ / mol. At what Kelvin temperature will the vapor pressure be 3.50 tim

stellarik [79]

Answer:

T2=336K

Explanation:

Clausius-Clapeyron equation is used to determine the vapour pressure at different temperatures:

where:

In(P2/P1) = ΔvapH/R(1/T1 - 1/T2)

p1 and p2 are the vapour pressures at temperatures

T1 and T2

ΔvapH = the enthalpy of vaporization of the liquid

R = the Universal Gas Constant

p1=p1, T1=307K

p2=3.50p1; T2=?

ΔvapH=37.51kJ/mol=37510J/mol

R=8.314J.K^-1moL^-1

In(3.50P1/P1)= (37510J/mol)/(8.314J.K^-1)*(1/307 - 1/T2)

P1 and P1 cancelled out:

In(3.50)=4511.667(T2 - 307/307T2)

1.253=14.696(T2 - 307/T2)

1.253=(14.696T2) - (14.696*307)/T2

1.253T2=14.696T2 - 4511.672

Therefore,

4511.672=14.696T2 - 1.253T2

4511.672=13.443T2

So therefore, T2=4511.672/13.443=335.61

Approximately, T2=336K

6 0

3 years ago

A scooter is accelerated from rest at the rate of 8m/s . How long will it take to cover a distance of 32m?​

A scooter is accelerated from rest at the rate of 8m/s . How long will it take to cover a distance of 32m?