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Evgen [1.6K]
1 year ago
13

What is the period if the block’s mass is doubled? note that you do not know the value of either m or k , so do not assume any p

articular values for them. The required analysis involves thinking about ratios.
Physics
1 answer:
Oksanka [162]1 year ago
5 0

The period of the block's mass is changed by a factor of √2 when the mass of the block was doubled.

The time period T of the block with mass M attached to a spring of spring constant K is given by,

T = 2π(√M/K).

Let us say that, when we increased the mass to 2M, the time periods of the block became T', the spring constant is not changed, so, we can write,

T' = 2π(√2M/K)

Putting T = 2π(√M/K) above,

T' =√2T

So, here we can see, if the mass is doubled from it's initial value. The time period of the mass will be changed by a factor of √2.

To know more about time period of mass, visit,

brainly.com/question/20629494

#SPJ4

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Two capacitors have the same size of plates and the same distance 4 mm between the plates The potentials of the two plates in ca
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we reduce the magnitudes to the SI system

          s = 4 mm (1 m / 1000 mm) = 0.004 m

we calculate

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3 years ago
A giraffe is slowly walking across a field. Is it balanced or unbalanced?
vodka [1.7K]
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3 0
3 years ago
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

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3 0
3 years ago
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