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3 years ago

A car accelerates from 6.00 m/s to 20.0 m/s, over a distance of 50.0 m. what is its rate of acceleration?

1 answer:

6 0

v² = u² + 2as 
2as = v² - u²
a = (v² - u²) / 2s 

a = (20.0² m²/s² - 6.00² m²/s²) / (2 * 50.0 m) 
a = (400 m²/s² - 36 m²/s²) / (100 m) 
a = (364 m²/s²) / (100 m) 
a = 3.64 m/s²

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3 years ago

Read 2 more answers

I drop an egg from a certain distance and it takes the egg 3.74 seconds to reach the ground. How high up was the egg?

Ulleksa [173]

Answer:

B. 68.6m

Explanation:

Free Fall Motion

When a body is left to move in the air with no friction, the motion is ruled only by the force of gravity. The vertical distance a body travels in the air after a time t is .

$\displaystyle y=\frac{gt^2}{2}$

We know the egg takes 3.74 seconds to reach the ground. The height it was launched from is

$\displaystyle y=\frac{(9.8)(3.474)^2}{2}$

$\displaystyle y=68.54\ m$

The closest correct option is

B. 68.6m

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3 years ago

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3 years ago

A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

Mila [183]

Answer:

$v=6.65m/sec$

Explanation:

From the Question we are told that:

Mass $m=97.6$

Coefficient of kinetic friction $\mu k=0.555$

Generally the equation for Frictional force is mathematically given by

$F=\mu mg$

$F=0.555*97.6*9.8$

$F=531.388N$

Generally the Newton's equation for Acceleration due to Friction force is mathematically given by

$a_f=-\mu g$

$a_f=-0.555 *9.81$

$a_f=-54455m/sec^2$

Therefore

$v=u-at$

$v=0+5.45*1.22$

$v=6.65m/sec$

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3 years ago

In a certain two slit diffraction experiment, two slits 0.02mm wide are spaced0.2mm between centers.(a) How many fringes appear

Kamila [148]

Answer:

a) m = 10 and b) λ = 3.119 10⁻⁷ m

Explanation:

In the diffraction experiments the maximums appear due to the interference phenomenon modulated by the envelope of the diffraction phenomenon, for which to find the number of lines within the maximum diffraction center we must relate the equations of the two phenomena.

Interference equation d sin θ = m λ

Diffraction equation a sin θ = n λ

Where d is the width between slits (d = 0.2 mm), a is the width of each slit (a = 0.02 mm). θ is the angle, λ the wavelength, m and n are an integer.

Let's find the relationship of these two equations

d sin θ / a sin θ = m Lam / n Lam

The first maximum diffraction (envelope) occurs for n = 1, let's simplify

d / a = m

Let's calculate

m = 0.2 / 0.02

m = 10

This means that 10 interference lines appear within the first maximum diffraction.

b) let's use the interference equation, remember that the angles must be given in radians

θ = 0.17 ° (π rad / 180 °) = 2.97 10⁻³ rad

d sin θ = m λ

λ = d sin θ / m

λ = 0.2 10⁻³ sin (2.97 10⁻³) / 2

λ = 3.119 10⁻⁷ m

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3 years ago