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3 years ago

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A ball is moving with a velocity of 0.5 m/s its velocity is decreasing at the rate of 0.05 m/s² what is its velocity in 5 second

s? how much time will it take from start to stop?what is the distance to travelled by it before stopping?

1 answer:

Law Incorporation [45]3 years ago

4 0

Answer:

v after 5s = 0.25 m/s, it took 10s to stop, it has traveled 2.5m before stopping

Explanation

We can use the equation of motion with constant acceleration

Given: v0= 0.5 m/s a= -0.05 m/s²

v(5s) = v0 + a×t = 0.25 m/s

Stop => v=0 => v0 + a×t = 0 => t=10s

Distance at t=10s ⇒ x(10) = 0.5×10 + 0.5x(-0.05)x10² = 2.5m

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A cyclist overcomes a resistive force of 30 N in order to cycle 30 m. It takes her 6 seconds to cycle this distance. Calculate t

zmey [24]

Answer:

Power = 15[W]

Explanation:

This problem can be solved using the work definition.

Work is equal to the product of the force by the distance, for this problem we have:

F = force = 30 [N]

d = distance = 30 [m]

w = work = F * d = 30*30 = 900 [J], "units in joules"

The power is defined as the work done in an interval of time.

P = w / t

where:

t = time [s]

therefore

P = w / t

P = 900 / 6

P = 150 [W] "units in watts"

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3 years ago

A water balloon is dropped out of a fifth floor window and hits an unsuspecting person below 4.0 seconds later. The window is ap

andrew11 [14]

Answer:

80 meters high

Explanation:

The velocity of the balloon would be g*t (I won't calculate, but will us this later)

We know that the kinetic energy at the bottom equals the potential at the top.

KE = PE

1/2 * m * v^2 = m * g * h

1/2 * m * (g * t)^2 = m * g * h (substitution)

1/2 * m * g^2 * t^2 = m * g * h

1/2 * g * t^2 = h (simplification by dividing the commons between both sides)

h = 1/2 * 9.81 * 4^2

h = 78.48 m (roughly 80 m)

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3 years ago

Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i

koban [17]

Answer:

A

$N = 1340.86 \ slits / cm$

B

$\theta = 15.7^o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 650 \ nm = 650 *10^{-9} \ m$

The angle of first bright fringe is $\theta = 5^o$

The order of the fringe considered is n =1

Generally the condition for constructive interference is

$dsin (\theta ) = n * \lambda$

=> $d = \frac{1 * 650 *10^{-9 }}{ sin(5)}$

=> $d = 7.458 *10^{-6} \ m$

Converting to cm

$d = 7.458 *10^{-6} \ m = 7.458 *10^{-6} * 100 = 0.0007458 \ cm$

Generally the number of grating pre centimeter is mathematically represented as

$N = \frac{1}{d}$

=> $N = \frac{1}{0.0007458}$

=> $N = 1340.86 \ slits / cm$

Considering question B

From the question we are told that

The first wavelength is $\lambda_1 = 650 \ nm = 650 *10^{-9} \ m$

The second wavelength is $\lambda_2 = 429 \ m = 420 *10^{-9 } \ m$

The order of the fringe is $n = 2$

The grating is $N = 5000 \ slits / cm$

Generally the slit width is mathematically represented as

$d = \frac{1}{N }$

=> $d = \frac{1}{ 5000 }$

=> $d = 0.0002 \ c m = 2.0 *10^{-6} \ m$

Generally the condition for constructive interference for the first ray is mathematically represented as

$d sin(\theta_1) = n * \lambda_1$

=> $\theta_1 = sin^{-1} [\frac{ 2 * \lambda }{d}]$

=> $\theta_1 = sin^{-1} [\frac{ 2 * 650 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_1 = 40.5 ^o$

Generally the condition for constructive interference for the second ray is mathematically represented as

$d sin(\theta_2) = n * \lambda_2$

=> $\theta_2 = sin^{-1} [\frac{ 2 * \lambda_1 }{d}]$

=> $\theta_2 = sin^{-1} [\frac{ 2 * 420 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_2 = 24.8 ^o$

Generally the angular separation is mathematically represented as

$\theta = \theta_1 - \theta_1$

=> $\theta = 42.5^o - 24.8^o$

=> $\theta = 15.7^o$

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3 years ago

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Ronch [10]

ANSWER:
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3 years ago

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Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a small moon,

Nady [450]

Answer:

The orbital speed of Dactyl is $5.55m/s$

Explanation:

The orbital speed can be determined by the combination of the universal law of gravity and Newton's second law:

$F = G\frac{M \cdot m}{r^{2}}$ (1)

Where G is gravitational constant, M is the mass of the asteroid, m is the mass of the moon and r is the distance between them

In the other hand, Newton's second law can be defined as:

$F = ma$ (2)

Where m is the mass and a is the acceleration

Then, equation 2 can be replaced in equation 1

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (2)

However, a will be the centripetal acceleration since the moon Dactyl describe a circular motion around the asteroid

$a = \frac{v^{2}}{r}$ (3)

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$ (4)

Therefore, v can be isolated from equation 4:

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (5)

Finally, the orbital speed can be found from equation 5:

Notice, that it is necessary to express r in units of meters.

$r = 95km \cdot \frac{1000m}{1km}$ ⇒ $95000m$

$v = \sqrt{\frac{(6.672x10^{-11}N.m^{2}/kg^{2})(4.4x10^{16}kg)}{95000m}}$

$v = 5.55m/s$

Hence, the orbital speed of Dactyl is $5.55m/s$

3 0

3 years ago