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3 years ago

A ball is thrown straight upward and rises to a maximum

Physics

1 answer:

Leviafan [203]3 years ago

3 0

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

$v_f^2 - v_i^2 = 2 a d$

here we know that

$v_f = 0$

also we have

$a = -9.81 m/s^2$

so we have

$0 - v_i^2 = 2(-9.81)(16)$

$v_i = 17.72 m/s$

Now we need to find the height where its speed becomes half of initial value

so we have

$v_f = 0.5 v_i$

now we have

$v_f^2 - v_i^2 = 2 a d$

$(0.5v_i)^2 - v_i^2 = 2(-9.81)h$

$-0.75v_i^2 = -19.62 h$

$0.75(17.72)^2 = 19.62 h$

$h = 12 m$

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A ball on the end of a string is whirled around in a horizontal circle of radius 0.300 m. The plane of the circle is 1.00 m abov

Mamont248 [21]

Answer:

radial acceleration is 41.8 m / s²

Explanation:

The acceleration for circular motion is

a = v² / r

They also give us the X and Y position where the body falls when the rope breaks, let's write the projectile launch equations

x = vox t

y = v₀ₓ t - ½ g t2

Since the circle is horizontally the v₀ₓ is zero (v₀ₓ = 0)

x = v₀ₓ t

t = x / v₀ₓ

y = - ½ g t²

Let's replace and calculate the initial velocity on the X axis

y = - ½ g (x / vox)²

v₀ₓ = √ (g x² / 2 y)

v₀ₓ = √ [- (-9.8) 1.6² / (2 1.00)]

v₀ₓ = 3.54 m / s

This is the horizontal velocity, but since it circle is in horizontal position it is also the velocity of the body at the point of rupture.

Now we can calculate the radial acceleration

a = v² / r

a = 3.54² / 0.300

a = 41.8 m / s²

5 0

3 years ago

A rod of length 35.50 cm has linear density (mass per length) given by λ = 50.0 + 23.0x where x is the distance from one end, an

hram777 [196]

Answer:

(a)20.65g

(b)0.19m

Explanation:

(a) The total mass would be it's mass per length multiplied by the total lenght

0.355(50 + 23*0.355) = 20.65 g

(b) The center of mass would be at point c where the mass on the left and on the right of c is the same

Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g

$c(50 + 23c) = 10.32$

$23c^2 + 50c - 10.32 = 0$

$c \approx 0.19m$

7 0

3 years ago

Neon has 10 protons and 10 electrons. How many additional electrons would it take to fill the outer (second) electron shell?

Yakvenalex [24]

None because Neon is already stable

8 0

3 years ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>

Let's draw the free body diagram of the system using the given data.
From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

$N_x=86.62N$

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

$N_y=F_V=mg-Tsin59\\$