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2 years ago

What is the velocity of a jet that travels 785 m [W] in 2.39 s?

Physics

1 answer:

Cerrena [4.2K]2 years ago

7 0

Answer:

v = 328.4

Explanation:

Data:

Distance (d) = 785 m
Time (t) = 2.39 s
Velocity (v) = ?

Use formula:

$\boxed{v = \frac{d}{t}}$

Replace:

$\boxed{v = \frac{785m}{2.39s}}$

It divides:

$\boxed{v = 328.4\frac{m}{s}}}$

What is the velocity?

The velocity is <u>328.4 meters per second.</u>

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An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

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2 years ago

A spring of spring constant 25 N/m is hung vertically and a 0.300 kg mass is attached to one end, causing a displacement of the

Mamont248 [21]

Answer:k=1175

Explanation:thank you for asking

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2 years ago

f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t

inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

$h(t)=35t-0.83t^2$

We know that the rate of change of displacement is equal to the velocity of an object.

$v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t$

Velocity of the arrow after 3 seconds will be :

$v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s$

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

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2 years ago

What is the acceleration of a 19 kg bike pushed with a force of 340 N? F = ma

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Answer:

Explanation:

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2 years ago

How fast, in rpm, would a 5.8 kg, 22-cm-diameter bowling ball have to spin to have an angular momentum of 0.21 kgm2/s

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Answer:

91 rpm

Explanation:

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2 years ago