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mr Goodwill [35]
2 years ago
6

Your 64-cm-diameter car tire is rotating at 3.5 rev/s when suddenly you press down hard on the accelerator. After traveling 200

m, the tire’s rotation has increased to 6.0 rev/s. What was the tire’s angular acceleration? Give your answer in rad/s2.
Physics
1 answer:
andreyandreev [35.5K]2 years ago
4 0

Answer: angular acceleration = 0.748rad/s²

Explanation: according to the question, our answer needs to be in rad/s², thus all units in rev/s will be converted to rad/s

Assuming the motion of the object is of a constant angular acceleration, then newton's laws of motion is applicable.

The formulae below is used

v² = u² + 2αθ

v = final angular speed =6rev/s = 6*2π = 12π rad/s

u =initial angular speed =3.5rev/s = 3.5 *2π = 7π rad/s

Note 1 rev = 2π rad.

α = angular acceleration.

θ = angular displacement.

Diameter = 64cm = 0.64m, radius = 64/2 = 32cm = 0.32m

The angular displacement can be gotten using the formulae below

S = rθ, where s= linear distance covered = 200m, r = radius = 0.32m

θ = S/r = 200/0.32=625 rad.

By substituting the parameter we have that

(12π)² = (7π)² + 2α(625)

1421.22 = 486.31 + 1250α

1421.22 - 486.31 = 1250α

934.91 = 1250α

α = 934.91/1250

α= 0.748 rad/s²

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A kite is 100m above the ground. If there are 200m of string out, what is the angle between the string and the horizontal? (Assu
belka [17]

Answer:

the answer is 30°

Explanation:

due to:

sin law of sines

\frac{sin 90}{200} =\frac{sin\beta }{100}\\arcsin(100\frac{sin90}{200} )= 30°

3 0
2 years ago
From the edge of a roof you throw a snowball downward that strikes the ground with 100J of kinetic energy. then you throw a seco
Vanyuwa [196]

Answer:

The second snowball hits the ground with a kinetic energy of 100 Joules

Explanation:

Given that,

From the edge of a roof you throw a snowball downward that strikes the ground with 100 J of kinetic energy. It is a case of conservation of energy.

At the highest point,

mgh+\dfrac{1}{2}mu^2=mgh'+0          

100=mgh'

At lowest point,

mgh'=K

From above two equation, we get :

Kinetic energy, K = 100 J

So, the second snowball hits the ground with a kinetic energy of 100 Joules. So, the correct option is (A).                                                                        

7 0
2 years ago
Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th
Oksana_A [137]
<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>

Explanation:

When the charge of 2e is placed in between the plates .

The force applied on this charge by plates is = q E

here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C

and E is the magnitude of electric field intensity

The work done = Force x displacement

Thus W = q E x S

here S is displacement

Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8

= 1.02 x 10⁻¹⁷ J

This work will be converted into the kinetic energy of charge .

Thus K.E = 1.02 x 10⁻¹⁷ J

3 0
2 years ago
When jumping, a flea accelerates at an astounding 1000 m/s2 but over the very short distance of 0.50 mm. If a flea jumps straigh
Nadusha1986 [10]

Answer:

The flea reaches a height of 51 mm.

Explanation:

Hi there!

The equations of height and velocity of the flea are the following:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

While in free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the flea at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration of the flea due to the jump.

v = velocity of the flea at time t.

g = acceleration due to gravity.

First, let's calculate how much time it takes the flea to reach a height of 0.0005 m. With that time, we can calculate the speed reached by the flea during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference on the ground, then, h0 = 0. Since the flea is initially at rest, v0 = 0. Then:

h = 1/2 · a · t²

We have to find the value of t for which h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Now, let's find the velocity reached in that time:

v = v0 + a · t   (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

When the flea is at a height of 0.50 mm, its velocity is 1.00 m/s. This initial velocity will start to decrease due to the downward acceleration of gravity. When the velocity is zero, the flea will be at the maximum height. Using the equation of velocity, let's find the time at which the flea is at the maximum height (v = 0):

v = v0 + g · t

At the maximum height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, let's find the height reached by the flea in that time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

h = 0.051 m

The flea reaches a height of 51 mm.

5 0
2 years ago
A circuit node has current ia entering and currents ib and ic exiting, where it is known that ia=5 ma and ib=9 ma. what is curre
Rudik [331]
The current is the flow of electrons. It is expressed as Coulombs per second, or Amperes. Since it is a flow, all that comes in must go out. The basis here is the node. Since ia is the full flow, it must be greater than ib or ic. So, I think the given information is wrong. It should be ia = 9 mA and ib=5 mA.

Current entering = Current leaving
ia = ib + ic
9 mA = 5 mA + ic
ic = 9 mA - 5 mA = 4 mA 
8 0
2 years ago
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