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3 years ago

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A student says that a speed of 50 m/s is faster than a speed of 140 km/h because the number is bigger. What would you say to the

m?
Pls can someone help me pls!
It’s so important!!!!

1 answer:

ladessa [460]3 years ago

4 0

The units are not consistent - 1 m/s is not the same as 1 km/h.

First thing to do would be to convert from one unit of speed to the other, say km/h to m/s. There are 1000 meters (m) for every kilometer (km) and 3600 seconds (s) for every hour (h), so

$1\,\dfrac{\mathrm{km}}{\mathrm h}\cdot\dfrac{10^3\,\mathrm m}{\mathrm{km}}\cdot\dfrac{\mathrm h}{3600\,\mathrm s}\approx0.278\,\dfrac{\mathrm m}{\mathrm s}$

So in fact 1 km/h is about 4 times slower than 1 m/s.

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Standing still, Bruce, the quarterback, gets tackled by Biff, the 90.0-kg tackle, who is traveling at 7.0 m/s. Upon collision, B

tatyana61 [14]

$\\ \sf\longmapsto \Delta P=P$

$\\ \sf\longmapsto m1v1=m2v2$

$\\ \sf\longmapsto 90(7)=m2(10)$

$\\ \sf\longmapsto 10m2=630$

$\\ \sf\longmapsto m2=\dfrac{630}{10}$

$\\ \sf\longmapsto m2=63kg$

Bruces mass is 63kg

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3 years ago

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A double slit that is illuminated with coherent light of wavelength 644 nm produces a pattern of bright and dark fringes on a sc

shutvik [7]

Answer:

$2.77$ cm

Explanation:

$d$ = separation between the slits = 2783 x 10⁻⁹ m

$\lambda$ = wavelength of coherent light = 644 nm = 644 x 10⁻⁹ m

$D$ = Distance of the screen = 6 cm = 0.06 m

$y_{n}$ = Position of nth bright fringe

Position of nth bright fringe is given as

$y_{n} = \frac{nD\lambda }{d}$

for n = 2

$y_{2} = \frac{nD\lambda }{d}$

$y_{2} = \frac{(2)(0.06)(644\times10^{-9}))}{2783\times10^{-9}}$

$y_{2} = 0.0278$ m

for n = 4

$y_{4} = \frac{nD\lambda }{d}$

$y_{4} = \frac{(4)(0.06)(644\times10^{-9}))}{2783\times10^{-9}}$

$y_{4} = 0.0555$ m

Distance between 4th and 2nd bright fringes is given as

$w = y_{4} - y_{2} = 0.0555 - 0.0278 = 0.0277 m$

$w = 2.77$ cm

8 0

3 years ago

A device for training astronauts and jet fighter pilots is designed to rotate the trainee in a horizontal circle of radius 11.0

kvv77 [185]

The velocity of the trainee is 29 m/s or 0.42 rev/s

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration (m / s²)v = final velocity (m / s)</em>

<em>u = initial velocity (m / s)</em>

<em>t = time taken (s)</em>

<em>d = distance (m)</em>

Centripetal Acceleration of circular motion could be calculated using following formula:

$\large {\boxed {a_s = v^2 / R} }$

<em>a = centripetal acceleration ( m/s² )</em>

<em>v = velocity ( m/s )</em>

<em>R = radius of circle ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

<u>Unknown:</u>

Velocity of Rotation = v = ?

<u>Solution:</u>

$F = ma$

$F = m\frac{v^2}{R}$

$7.80w = m\frac{v^2}{R}$

$7.80mg = m\frac{v^2}{R}$

$7.80g = \frac{v^2}{R}$

$7.80 \times 9.8 = \frac{v^2}{11.0}$

$v^2 = 840.84$

$v \approx 29 ~m/s$

$\omega = \frac{v}{R}$ → in rad/s

$\omega = \frac{v}{2 \pi R}$ → in rev/s

$\omega = \frac{29}{2 \pi \times 11.0}$

$\omega \approx 0.42 ~ rev/s$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Uniform Circular Motion : brainly.com/question/2562955
Trajectory Motion : brainly.com/question/8656387

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

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3 years ago

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When you go out into the sun, the UV light can give you a tan. UV light has a frequency of 7.89 x 1014 Hz. What is the wavelengt

Olin [163]

Answer:

Explanation:

Formula and givens

λ = c / f
λ is the wavelength
c = the speed of light
f = the frequency
c = 3*10^8
f = 7.89 * 10^14

λ = ?

Solution

λ = 3*10^8 / 7.89*10^14

λ = 3*10^8/7.89*10^14

λ = 2.36 * 10^7

λ = 236 nanometers. What you use as your solution depends on what what you have been taught.

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3 years ago

A 4.75 kg block is sent up a ramp inclined at an angle ????=31.5° from the horizontal. It is given an initial velocity ????0=15.

Colt1911 [192]

Answer:

d = 13.7 m

Explanation:

When block is moving upwards along the inclined plane

then the block is decelerated due to gravity as well as due to friction and speed of the block by which it is projected upwards is given

$v_i = 15 m/s$

deceleration caused to the block due to net force opposite to the motion is given as

$F = - mg sin\theta - \mu mgcos\theta$

$a = \frac{F}{m}$

$a = -g(sin\theta + \mu cos\theta)$

since block is sliding on the inclined plane

so here we can say that the coefficient of the friction must be kinetic friction here

$a = -9.81(sin31.5 + 0.368 cos31.5)$

$a = - 8.2 m/s^2$

now for finding the distance upto which it will stop is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 15^2 = 2(-8.2) d$

$d = 13.7 m$

8 0

3 years ago