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Harrizon [31]
3 years ago
13

As the pendulum swings through the origin, it has a speed of v0 = 1.1 m/s. If the mass on the end of the pendulum is 15 g, and t

he pendulum is 30 cm long, how high does the pendulum swing before coming back down? What is the angle corresponding to that height?

Physics
1 answer:
serg [7]3 years ago
4 0

Answer:

1: 6.18 cm

2: 52.5609 degrees

Explanation:

We have the pendulum speed at the origin, and in that moment, all energy is kinetic, so we can calculate the pendulum energy by:

Ec = 0.5*m*v^2 = 0.5*0.015*1.1^2 = 0.0091 J

Now with that energy, we can calculate the height the pendulum will reach, as in that moment, the kinetic energy is totally converted to gravitational potencial energy:

Eg = m*g*h = 0.0091

0.015 * 9.81 * h = 0.0091

h = 0.0091 / (0.015 * 9.81 ) = 0.0618 m = 6.18 cm

Looking at the image attached, we can see that the pendulum will form a triangle, and one of the cathetus will be the length of the pendulum minus the height it went up, and the hypotenusa will be the pendulum length.

So, we know that the sine of the angle will be the division between the opposite cathetus and the hypotenusa:

sin(angle) = (30-6.18)/30 = 23.82/30 = 0.794 -> angle = 52.5609 degrees

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Norma-Jean [14]

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same  provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\

assume the bullet goes to right side and the gravitational acceleration =10 ms^{-2}

so now the weight of the rifle=\frac{25}{10}  

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}

this is a negative velocity to the right side. that means the rifle recoils to the left side

3 0
2 years ago
1.
ale4655 [162]

Answer:

5.0 atm

Explanation:

P₁V₁=P₂V₂

P₁V₁/V₂=P₂

(1)(2.5)/(0•50)=P₂

P₂=5

Pressure is now 5.0 atm

8 0
3 years ago
A speeding car traveling at 24.8 m/s passes a police car that is at rest. The police car begins its pursuit at the instant the s
marusya05 [52]

Answer:

the acceleration required is 1.37m/s^2

Explanation:

The car is having a constant velocity movement, so if we calculate the time to reach 897m, we can use it to find the acceleration the policeman need to apply to reach the car.

x=v*t\\t=\frac{x}{v}\\t=\frac{897m}{24.8m/s}\\t=36.17s

the policeman is traveling with a constant acceleration starting from rest so:

x=\frac{1}{2}*a*t^2\\\\a=\frac{2*x}{t^2}\\\\a=1.37 m/s2

7 0
2 years ago
A train travels at a speed of 30 m/s. The train starts at an initial position of 1000 meters and travels for 30 seconds. What is
pychu [463]
1000 + 30x30 = 1900. Hope that helps
6 0
3 years ago
An aircraft has a liftoff speed of 33 m/s. What is the minimum constant acceleration an
JulsSmile [24]

Answer: The minimum acceleration for the air plane is 2.269m/s2.

Explanation: To solve such problem the equation of motion are applicable.

The initial velocity is 0 since the airplane was initially standing. We are going to use this equation

V^2=U^2+2as

33^2=0+2a (240)

a= 2.269m/s2

5 0
3 years ago
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