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3 years ago

7

A toy car accelerates at a constant rate from rest to a speed of 4 m/s in a time of 0.55 s. What was the magnitude of the accele

ration, in units of m/s2?

1 answer:

Korvikt [17]3 years ago

3 0

initial velocity=u=0m/s
Final velocity=v=4m/s
Time=t=0.5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{4-0}{0.5}$

$\\ \sf\longmapsto Acceleration=\dfrac{4}{0.5}$

$\\ \sf\longmapsto Acceleration=8m/s^2$

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In order to rendezvous with an asteroid passing close to the earth, a spacecraft must be moving at 8.50×103m/s relative to the e

Mariulka [41]

Answer:

v₀ = 13.9 10³ m / s

Explanation:

Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.

F = m a

G m M / x² = m dv / dt = m dv/dx dx/dt

G M / x² = dv/dx v

GM dx / x² = v dv

We integrate

v² / 2 = GM (-1 / x)

We evaluate between the lower limits where x = Re = 6.37 10⁶m and the velocity v = vo and the upper limit x = 2.50 10⁸m with a velocity of v = 8.50 10³ m/s

½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))

72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)

72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3 10⁻⁸)

72.25 10⁶ - v₀² = -1.213 10⁸

v₀² = 72.25 10⁶ + 1,213 10⁸

v₀² = 193.6 10⁶

v₀ = 13.9 10³ m / s

6 0

4 years ago

A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli

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Answer:

Part a)

$A = 1.78 m$

Part b)

$f = 2 rev/s$

Explanation:

Part A)

As we know that time period of the motion is given as

$T = 2.68 s$

so we have

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.68}$

$\omega = 2.34 rad/s$

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

$mg = m\omega^2 A$

so we have

$A = \frac{g}{\omega^2}$

$A = \frac{9.81}{2.34^2}$

$A = 1.78 m$

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

$mg = m\omega^2 A$

$g = \omega^2 (0.0623)$

$\omega = 12.5 rad/s$

so now we have

$2\pi f = 12.5$

$f = 2 rev/s$

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3 years ago

Why are earbud wires for an mp3 player coated with plastic

saw5 [17]

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3 years ago

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NeTakaya

Answer:

D. 22 Hz and 42 Hz

Explanation:

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<em><u>Therefore; in this case since the frequency of the beat is 20 Hz, that is from 20 beats per second.</u></em>
We need to find a pair from the choices whose frequency difference is 20 Hz.
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daser333 [38]

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Projectile Motion.

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so we get as,

T = 50/9.8

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thus the time taken to reach max height is 5.10 seconds.

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3 years ago