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3 years ago

Show that the expression vf = vi + at is dimensionally correct

1 answer:

4 0

To show that the expression given is dimensionally correct, we can do a dimensional analysis where we write the units of each term in the expression.

vf [=] vi + at
m/s [=] m/s+ (m/s²) (s)
<span>m/s [=] m/s+ m/s
</span><span>m/s [=] m/s
</span>
Therefore, the expression is dimensionally correct.

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Two carts have a compressed spring between them and are initially at rest. One of the carts has total mass, including its conten

ladessa [460]

Answer:

A) - 1.8 m/s

Explanation:

As we know that whole system is initially at rest and there is no external force on this system

So total momentum of the system must be conserved

so we will have

$m_1v_1 + m_2v_2 = 0$

now plug in all data into above equation

$5(v) + 3(3)$

$5v = -9$

$v = -1.8 m/s$

so correct answer is

A) - 1.8 m/s

3 0

3 years ago

The kinetic energy of the molecules inside the balloon _______ which

11111nata11111 [884]

Answer:

Increase,.faster

Explanation:

The kinetic energy of the molecules inside the balloon

increases

which means they are moving

faster

I hope this helps you :)

6 0

3 years ago

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

6 0

3 years ago

What would the person’s mass be on the earth? Part B pls

Dafna1 [17]

Answer:

I hopes it helps

thank you

4 0

3 years ago

Just this last one!!

Pepsi [2]

Answer:

$47 \ \frac{m}{s}$

Explanation:

s = displacement (m)

u = initial velocity $(\frac{m}{s})$

v = final velocity $(\frac{m}{s})$

a = acceleration $(\frac{m}{s^{2} })$

t = time (s)

s = 235

a = -4.7

v = 0

v² = u² + 2as

(0)² = u² + 2(-4.7)(235)

u² - 2209 = 0

u² = 2209

u = 47

4 0

2 years ago

Read 2 more answers