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Anna71 [15]
3 years ago
15

Describe a ball's motion as it rolls up a slanted

Physics
1 answer:
emmasim [6.3K]3 years ago
4 0

The ball will decelerate as it moves upwards.

The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

The given parameters;

  • initial velocity of the ball, u = 1.25 m/s
  • time of motion of the ball, t = 4.22 s

As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.

Thus, the ball decelerate as it moves upwards.

The acceleration of the ball is calculate as;

a = \frac{v_f -v_0}{t} \\\\

<em>at the highest point on the incline plane, the final velocity </em>v_f<em> is zero</em>

a = \frac{0-1.25}{4.22} \\\\a = -0.3 \ m/s^2

Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

Learn more here:brainly.com/question/23860763

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S.I unit for distance =______
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Answer:

opinion a

Explanation:

the si units of distance is metre (m)

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A police car moving at 36.0 m/s is chasing a speeding motorist traveling at 30.0 m/s. The police car has a siren that emits soun
maxonik [38]

Answer:

The frequency heard by the motorist is 4313.2 Hz.

Explanation:

let f1 be the frequency emited by the police car and f2 be the frequency heard by the motorist, let v1 be the speed of the police car and v2 be the speed of the motorist and v = 343 m/s be the speed of sound.

because the police car is moving towards the motorist at a higher speed, then the motorist will hear a increasing frequency and according to Dopper effect, that frequency is given by:

f1 =  [(v + v2/(v - v1))]×(f2)

   = [( 343 + 30)/(343 - 36)]×(3550)

   = 4313.2 Hz

Therefore, the frequency heard by the motorist is 4313.2 Hz.

5 0
2 years ago
Where do mosquito lay eggs
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4 0
3 years ago
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A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the
kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

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Brrunno [24]

Answer:

The frequency is the same

Explanation:

When a wave is created by a source which is vibrating at a certain frequency, the frequency of the wave itself is equal to the frequency of the source.

This occurs with every kind of wave. For instance, if we consider the radio waves produced by an antenna, the frequency of the radio waves is equal to the frequency of the antenna.

In this case, the waves are created by the vibrating bug. The bug is vibrating with a certain frequency f: as a consequence, the frequency f' of the waves produced by the bug will be equal to the frequency of vibration of the bug:

f'=f.

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