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Zarrin [17]
3 years ago
7

A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r

esistance) the ball will land on the ground with a speed of
Physics
2 answers:
marusya05 [52]3 years ago
6 0

As we know that here no air resistance while ball is moving in air

So here we will say that

initial total energy = final total energy

KE_i + U_i = KE_f + U_f

here we know that

Ui = U_f = 0 (as it will be on ground at initial and final position)

so we will say

KE_i = KE_f

since mass is always conserved

so we will say that final speed of the ball must be equal to the initial speed of the ball

so we have

v_f = v_i = 30 m/s

andreev551 [17]3 years ago
5 0

Answer:

When ball hits the ground its velocity will be 30 m /sec

Explanation:

We have given that there is no air resistance

So from energy conservation initial energy will be equal to final energy

So U_I+KE_I=U_F+KE_F

As the ball is initially at the ground and finally also on the ground

So U_I=U_F=0

As we know that mass is also conserved

So initial velocity will be equal to final velocity

So when ball hit the ground its velocity will be 30 m /sec

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Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1
miv72 [106K]

Answer:

w = 4,786 rad / s ,  f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

       L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

      L₀ = 0 + I₂ w₂

Final after coupling

      L_{f} = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

      I = ½ M R²

How the moment is preserved

      L₀ = L_{f}

      I₂ w₂ = (I₁ + I₂) w

      w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

      d₁ = 60 cm = 0.60 m

      d₂ = 40 cm = 0.40 m

      f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

      w₂ = 2 π f₂

      w₂ = 2π 3.33

      w₂ = 20.94 rad / s

Let's replace

       w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

       w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

      w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

      w = 20.94 1.28 / 5.6

      w = 4,786 rad / s

Angular velocity and frequency are related.

      w = 2π f

      f = w / 2π

      f = 4.786 / 2π

      f = 0.76176 Hz

7 0
3 years ago
Light in the air is incident at an angle to the surface of (12.0 A) degrees on a piece of glass with an index of refraction of (
Orlov [11]

The question is incomplete. You dis not provide values for A and B. Here is the complete question

Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.

A = 12

B = 18

Answer:

18.5⁰

Explanation:

Angle of incidence i = 12.0 + A

A = 12

= 12.0 + 12

= 14

Refractive index u = 1.10 + B/100

= 1.10 + 18/100

= 1.10 + 0.18

= 1.28

We then find the angle of refraction index u

u = sine i / sin r

u = sine24/sinr

1.28 = sine 24 / sine r

1.28Sine r = sin24

1.28 sine r = 0.4067

Sine r = 0.4067/1.28

r = sine^-1(0.317)

r = 18.481

= 18.5⁰

4 0
2 years ago
The melting point of a substance is the same as its _____ point.
kirza4 [7]

Answer:

Boiling point

Explanation:

5 0
2 years ago
An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on i
Sergeu [11.5K]

Answer:

a = 2.77~{\rm m/s^2}

Explanation:

Since the pulley has a mass concentrated on its rim, the pulley can be considered as a ring.

The moment of inertia of a ring is

I = mr^2 = (2.3)(23.5\times 10^{-2})^2 = 0.127

The mass on the left is heavier, that is the pulley is rotating counterclockwise.

By Newton's Second Law, the net torque is equal to moment of inertia times angular acceleration.

\tau = I \alpha

Here, the net torque is the sum of the weight on the left and the weight on the right.

\tau = m_1gR - m_2gR = (1.65)(9.8)(23.5\times 10^{-2}) - (1)(9.8)(23.5\times 10^{-2}) = 1.497~{\rm Nm}

Applying Newton's Second Law gives the angular acceleration

\tau = I\alpha\\1.497 = 0.127\alpha\\\alpha = 11.78~{\rm rad/s^2}

The relation between angular acceleration and linear acceleration is

a = \alpha R

Then, the linear acceleration of the masses is

a = 11.78 \times 23.5\times 10^{-2} = 2.77~{\rm m/s^2}

5 0
3 years ago
Calculate the efficiency of given pulley System if 40N
jonny [76]

Answer:

20n

Explanation:

60n - 40n = efficiency of pulley system

20n = efficiency of pulley system

4 0
2 years ago
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