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3 years ago

6

If the mass of an ideal pendulum is doubled how will the oscillation period change?

1 answer:

Elanso [62]3 years ago

5 0

Answer: Doubling the mass of the bob will half the period. b) If the length is doubled, the period will increase by a factor of √2 . The mass of the bob will not effect the period

Explanation:

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The viscosity of magma is directly related to the silica content. That being said, the viscosity of magma is higher when the silica content is also high and it is a low viscosity magma when the silica content is also low. The silica content of low viscosity magma is approximately 45 - 55 wt%.

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PERSONAL QUESTION <br> if a bird can fly why can't a fly just bird ?

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4 years ago

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assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves th

finlep [7]

Answer:

A. xmax = 131.49 m

B. t = 8.74 s

C. ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

$y=y_o+v_osin\theta-\frac{1}{2}gt^2$ (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

$0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2$ (2)

You use the quadratic formula to obtain the value of t:

$t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s$

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

$x_{max}=v_ocos\theta t$

$x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m$

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

$y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}$ (3)

By replacing in the equation (3) the values of all parameters you obtain:

$y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m$

The maximum height reached by the cannon ball is 220.33m

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3 years ago

A Fleas can jump up a distance as large as 33 cm horizontally in a single leap (they are only 0.3 mm in size!). What is their la

Alex Ar [27]

Answer:

Launch velocity of the fleas = 1.80 m/s

Explanation:

Optimal launch angle is the angle that guarantees the maximum horizontal distance covered.

Optimum launch angle = 45°

The horizontal distance covered by a projectile is given as Range

R = (u² sin 2θ)/g

u = initial velocity of the projectile (flea) = ?

R = range = 33 cm = 0.33 m

θ = 45°

2θ = 2 × 45° = 90°

Sin 90° = 1

g = acceleration due to gravity = 9.8 m/s²

R = (u²/g)

u² = Rg = 0.33 × 9.8 = 3.234

u = 1.80 m/s

Hope this helps!!!

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3 years ago

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C the thermal equilibrium

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4 years ago

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