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2 years ago

Find the magnitude of the vector 34 i^+ 14 j^m.

Physics

1 answer:

Tanya [424]2 years ago

6 0

I believe you square both numbers and take the square root. √34^2+14^2= so it would be 36.77

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Please help!

nignag [31]

Since there is no air resistance, gravity is the only force acting on the two objects - the bowling ball and the arrow. Therefore, they will hit the ground at the same time. Thus the answer is D. Both hit the ground at the same time. Hope this helped and have a great day!

5 0

2 years ago

Read 2 more answers

Suppose that the average speed (vrms) of carbon dioxide molecules (molar mass 44.0 g/mol) in a flame is found to be 2.67 105 m/s

34kurt

Answer:

$T = 1.26 \times 10^8 K$

Explanation:

As we know that rms speed of ideal gas is given by the formula

$v_{rms} = \sqrt{\frac{3RT}{M}}$

here we know that

$v_{rms} = 2.67 \times 10^5 m/s$

molecular mass of gas is given as

$M = 44 g/mol = 0.044 kg/mol$

now from above formula we have

$2.67\times 10^5 = \sqrt{\frac{3(8.31)T}{0.044}}$

now we have

$T = 1.26 \times 10^8 K$

7 0

3 years ago

According to Newton’s first law of motion, if a person is swinging a rock st the end of a string and the string breaks, what wil

balu736 [363]

Travel in a straight path with constant velocity as the rock doesn't want to break its inertial frame.

3 0

3 years ago

A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R−x a

Colt1911 [192]

Answer:

The intensity of sound wave at the surface of the sphere $I = \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }$

Explanation:

B = Bulk modulus

Intensity, $I = \frac{P_{max} ^{2} }{2\sqrt{\rho B} }$

The amplitude of oscillation of the sphere is given by:

$P_{max} = BkA\\k = \frac{2\pi }{\lambda} \\$

$A = \triangle R\\$

Substitute v and A into Pmax

$P_{max} = (2\pi f)\sqrt{\rho B} \triangle R\\ P_{max} ^{2} = 4\pi^{2} *f^{2} \rho B (\triangle R)^{2}$

$I = \frac{ 4\pi^{2} f^{2} \rho B (\triangle R)^{2}}{2\sqrt{\rho B} }$

$P_{total} = 4\pi R^{2} I$

$P_{total} =4\pi R^{2} \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} }$

The intensity of the sound wave at a distance is given by:

$I = \frac{P_{total} }{4\pi d^{2} }$

$I = 4\pi R^{2} \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} } * \frac{1}{4\pi d^{2} } \\I = \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }$

5 0

3 years ago

When an electron falls from a higher to a lower energy level in an atom, the photon released has a wavelength of 121.6 nm. What

yarga [219]

Answer:

$\Delta E=1.64*10^{-18}J$

Explanation:

The energy difference between the energy levels involved in the transition of the electron is directly proportional to the frequency of the emitted photon:

$\Delta E=h\nu(1)$

Where h is the Planck constant. The photon's frequency is inversely proportional to its wavelegth:

$\nu=\frac{c}{\lambda}(2)$

Here c is the speed of light. Replacing (2) in (1):

$\Delta E=\frac{hc}{\lambda}\\\Delta E=\frac{(6.63*10^{-34}J\cdot s)(3*10^8\frac{m}{s})}{121.6*10^{-9}m}\\\Delta E=1.64*10^{-18}J$

6 0

2 years ago