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cluponka [151]
3 years ago
12

A car tire rotates with an average angular speed of 32 rad/s. In what time interval will the tire rotate 3.5 times? Answer in un

its of s.
Physics
1 answer:
stiks02 [169]3 years ago
3 0

Answer:

\Delta t = 0.687\ s

Explanation:

given,

Angular speed of the tire = 32 rad/s

Displacement of the wheel = 3.5 rev

Δ θ = 3.5 x 2 π

        = 7 π rad

now,

Time interval of the car to rotate 7π rad

using equation

\omega_{avg} =\dfrac{\Delta \theta}{\Delta t}

\Delta t=\dfrac{7\pi}{32}

\Delta t = 0.687\ s

Time taken to rotate 3.5 times is equal to 0.687 s.

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Answer:

Mg and HCl

Explanation:

Here, we want to get the combination that could be used in the production of the needed hydrogen

An important chemical property of inorganic acids is that when they react with metals, they give off hydrogen gas in conjunction with the formation of a salt

HCl is a mineral acid while Mg is a metallic substance

So the reaction between this metal and the mineral acids will give the needed hydrogen gas to be produced

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3 years ago
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An object falling straight down without air resistance is said to be exhibiting
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Explanation:

If the object is falling straight down it is in free fall.  The difference between that and two-dimensional motion is that 2D motion is parabolic (projectile)

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3 years ago
What is a point value for an arrow that is on two different colors?
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It depends on what it is closest to but I would say for instance black is 5 points and red is 6 if u land on the line 5.5
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3 years ago
Two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 9.8 m above the ground a
xenn [34]

Answer:

Time = 0.55 s

Height = 8.3 m

Explanation:

The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, h_b.

The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, h_d. Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be h_d and the distance travelled by the ball measured from the top be h_b.

It follows that h_d+h_b=9.8.

Applying the equation of motion to each body (h = v_0t + 0.5at^2),

Ball:

h_b=0\times t + 0.5\times 9.8t^2 (since v_{b0} =0.)

h_b=4.9t^2

Dart:

h_d=17.8\times t - 0.5\times9.8t^2 (the acceleration is opposite to the displacement, hence the negative sign)

h_d=17.8\times t - 4.9t^2

But

h_b+h_d =9.8

17.8\times t - 4.9t^2+4.9t^2 =9.8

17.8\times t = 9.8

t = 0.55

The height of the collision is the height of the dart above the ground, h_d.

h_d=17.8\times t - 4.9t^2

h_d=17.8\times 0.55 - 4.9\times(0.55)^2

h_d=9.79 - 1.48225

h_d=8.3

8 0
4 years ago
In a nuclear reactor, each atom of uranium (of atomic mass 235 u) releases about 200 MeV when it fissions. What is the change in
Ket [755]

Answer:

0.002372187708 kg

Explanation:

Each atom of Uranium 235 releases 200 MeV = 200×10⁶×1.60218×10⁻¹⁹

= 200×1.60218×10⁻¹³ Joule

Number of atoms in a 2.6 kg sample mass = (2.6/0.235)×6.02214076×10²³

⇒Number of atoms in a 2.6 kg sample mass = 66.627×10²³ atoms

Change in energy = Change in mass / (speed of light)²

ΔE = Δmc²

⇒200×1.60218×10⁻¹³×66.627×10²³ = Δm×(3×10⁸)²

⇒Δm = 200×1.60218×10⁻¹³×66.627×10²³/(3×10⁸)²

⇒Δm = 2372.187708×10⁻⁶ kg

∴Change in mass = 0.002372187708 kg

7 0
3 years ago
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