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3 years ago

How far from the base of the platform does she land?

Physics

1 answer:

seropon [69]3 years ago

3 0

When Janet leaves the platform, she's moving horizontally at 1.92 m/s. We assume that there's no air resistance, and gravity has no effect on horizontal motion. There's no horizontal force acting on Janet to make her move horizontally any faster or slower than 1.92 m/s.

She's in the air for 1.1 second before she hits the water.

Moving horizontally at 1.92 m/s for 1.1 second, she sails out away from the platform

(1.92 m/s) x (1.1 sec) = <em>2.112 meters</em>

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An elevator supported by a single cable descends a shaft at a constant speed. The only forces acting on the elevator are the ten

VikaD [51]

Answer:

C. The net work done by the two forces is zero joules.

Explanation:

Work (W) is defined as the product of force F) by the distance (d)the body travels due to this force.

W= F*d Formula ( 1)

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

The component of the force that performs work must be parallel to the displacement.

Calculation of the forces

We apply Newton's second law:

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

∑F = m*a

a = 0 because the elevator moves at constant speed.

∑F=0

T-W = 0

T = W

Calculation of work done by the forces

W₁ : work done by the gravity= (+ W*d ) J

W₂ : work done by the tension force=( -T*d) J

Net work done by the two forces(W₁-W₂):

W₁= W₂ .because T = W

W₁-W₂ =0

3 0

2 years ago

A wave has a period of 2 seconds and a wavelength of 4 meters.Calculate its frequency and speed.

Lana71 [14]

F=l/t
f=4/2
f=2m/s

Hope this helps

3 0

3 years ago

Read 2 more answers

Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t

Nataliya [291]

Answer:

Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

Q = cos^-1 ( sqrt (0.464))

Q = cos^-1 (0.6811754546)

Q = 47.06 degrees

8 0

3 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

2 years ago

Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.

dezoksy [38]

The De Broglie wavelength of the electron is
$\lambda=34.0 nm=34 \cdot 10^{-9} m$
And we can use De Broglie's relationship to find its momentum:
$p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s$

Given $p=mv$ , with m being the electron mass and v its velocity, we can find the electron's velocity:
$v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}= 2.13 \cdot 10^4 m/s$

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
$K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J$

3 0

3 years ago