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3 years ago

What is the net force on a bathroom scale when a 150-pound person stands on it

Physics

1 answer:

andreev551 [17]3 years ago

7 0

Answer:

Zero, the support force and gravitational force together equal a zero net force

Explanation:

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For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency

Angelina_Jolie [31]

Answer:

2.7 J

Explanation:

The energy of one photon is given by

$E=hf$

where

h is the Planck constant

f is the frequency

For the photons in this problem,

$f=6.8\cdot 10^9 Hz$

So the energy of one photon is

$E_1=(6.63\cdot 10^{-34})(6.8\cdot 10^9 )=4.5\cdot 10^{-24} J$

The number of photons contained in 1.0 mol is

$N_A = 6.022\cdot 10^{23} mol^{-1}$ (Avogadro number)

So the total energy of $N_A$ photons contained in 1.0 mol is

$E=N_A E_1 =(6.022\cdot 10^{23})(4.5\cdot 10^{-24})=2.7 J$

3 0

3 years ago

You ride on an elevator that is moving with constant upward acceleration while standing on a bathroom scale. the reading on the

Blababa [14]

The reading on the scale is greater than your actual weight.

4 0

3 years ago

You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close toget

aleksklad [387]

Answer:

35.7 m

Explanation:

Let

$\mid A\mid=18.5 m$

$\mid B\mid=41 m$

We have to find the distance between Joe's and Karl'e tent.

$A_x=Acos\theta$

$A_y=Asin\theta$

Substitute the values then we get

$A_x=18.5cos23^{\circ}=17 m$

$A_y=18.5sin 23^{\circ}=7.2 m$

$B_x=41cos37.5^{\circ}=32.5 m$

$B_y=41sin37.5^{\circ}=-24.96 m$

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

$B=A+C$

$C=B-A$

$C_x=B_x-A_x=32.5-17=15.5 m$

$C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m$

$\mid C\mid=\sqrt{C^2_x+C^2_y}$

$\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m$

Hence, the distance between Joe's and Karl's tent=35.7 m

6 0

3 years ago

Salmon often jump waterfalls to reach their

natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

$v_x = u cos \theta$

where u is the initial speed and $\theta=37.7^{\circ}$ . The horizontal distance travelled by the salmon is

$d=v_x t = (ucos \theta)t$

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

$t=\frac{d}{v_x}=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$y=h+u_y t -\frac{1}{2}gt^2$

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

$u_y = u sin \theta$ is the vertical component of the initial velocity of the salmon

$g=9.81 m/s^2$ is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

$y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}$

and we can solve this formula for u, the initial speed of the salmon:

$y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s$

5 0

3 years ago

A car has wheels which spin forwards. As the wheels spin forwards, they grip the road and push the road backwards. Since forces

alexandr402 [8]

Answer:

the law of motion

Explanation:

because the wheels are moving it means motion i am not sure which number law it is but I believe that it is 2nd but u should look it up to be safe

6 0

3 years ago

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