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4 years ago

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After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it rea

ches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.619 s to get directly over Julie's head. What is the speed of the ball when it leaves Sarah's hand?

1 answer:

blsea [12.9K]4 years ago

4 0

Answer:

Vo=20.2m/s

Explanation:

We can calculate both components of initial velocity and the use them to calculate the speed as $\sqrt{V_{ox}^{2}+V_{oy}^{2}}$

From the vertical movement to the maximum height:

$V_{fy}=V_{oy}-g*t$ since $V_{fy}=0$

$V_{oy}=g*t=16.19m/s$

Now, from the horizontal movement, also to the point of maximum height:

$Y_{f}=Y{o}+V_{oy}*t-\frac{g*t^{2}}{2}$ Solving for $V_{oy}$

$V_{oy}=12.1m/s$

Finally we calculate the magnitude of the new vector:

$V_{o}=\sqrt{V_{ox}^{2}+V_{oy}^{2}}=20.2m/s$

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You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauc

pickupchik [31]

a) $15.4^{\circ}$

b) 5.2 m/s

c) 151.2 N

Explanation:

a)

When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by

$mg sin \theta$

where

m =16 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

$\theta$ is the angle of the ramp

According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:

$F=ma\\mgsin \theta = ma$

where a is the acceleration.

From the equation above we get

$a=g sin \theta$

And we are told that the acceleration must not exceed

$a=2.6 m/s^2$

Substituting this value and solving for $\theta$ , we find the maximum angle of the ramp:

$\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}$

b)

Here we are told that the vertical distance of the ramp is

$h=1.4 m$

Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.

So we have:

$GPE=KE\\mgh=\frac{1}{2}mv^2$

where:

m = 16 kg is the mass of the box

$g=9.8 m/s^2$

h = 1.4 m height of the ramp

v = final speed of the box at the bottom of the ramp

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s$

c)

There are two forces acting on the box in the direction perpendicular to the ramp:

- The normal force, N, upward

- The component of the weight perpendicular to the ramp, downward, of magnitude

$mg cos \theta$

Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:

$N-mg cos \theta$

and by substituting:

m = 16 kg

$g=9.8 m/s^2$

$\theta=15.4^{\circ}$

We can find the normal force:

$N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N$

8 0

3 years ago

How long is the image of a metrestick formed by a plane mirror?

Zinaida [17]

A plane mirror doesn't change the apparent size or shape of things.

-- If the meter stick is held parallel to the mirror, or is lying on the mirror,
then its image in the mirror is 1 meter long.

-- If the meter stick is held perpendicular to the mirror, or is standing on
the mirror, or pointing directly at it, then its image in the mirror has no length.

7 0

3 years ago

A laser used for many applications of hard surface dental work emits 2780-nm wavelength pulses of variable energy (0-300 mJ) abo

Bess [88]

Answer:

a

$n = 1.119 *10^{18} \ photons$

b

$P = 1.6 \ W$

Explanation:

From the question we are told that

The wavelength is $\lambda = 2780 nm = 2780 *10^{-9} \ m$

The energy is $E = 80 mJ = 80 *10^{-3} \ J$

This energy is mathematically represented as

$E = \frac{n * h * c }{\lambda }$

Where c is the speed of light with a value $c = 3.0 *10^{8} \ m/s$

h is the Planck's constant with the value $h = 6.626 *10^{-34} \ J \cdot s$

n is the number of pulses

So

$n = \frac{E * \lambda }{h * c }$

substituting values

$n = \frac{80 *10^{-3} * 2780 *10^{-9}}{6.626 *10^{-34} * 3.0 *10^{8} }$

$n = 1.119 *10^{18} \ photons$

Given that the pulses where emitted 20 times in one second then the period of the pulse is

$T = \frac{1}{20}$

$T = 0.05 \ s$

Hence the average power of photons in one 80-mJ pulse during 1 s is mathematically represented as

$P = \frac{E}{T}$

substituting values

$P = \frac{ 80 *10^{-3}}{0.05}$

$P = 1.6 \ W$

6 0

4 years ago

PLEASE HELP!! Newtons first law states that if an objects velocity is changing a __________ must be acting on it. Question 1 opt

Anika [276]

Newtons first law states that if an objects velocity is changing a <u>force</u> must be acting on it.

Explanation:

Newton's first law of motion states that:

"An object at rest (or in uniform motion) will remain at rest (or will continue moving with the same velocity) unless acted upon an unbalanced force"

We can apply this law to a daily life example:

Take a book at rest on a table: the forces acting on the book are balanced. If we do not apply any other force, we know that the book will remain at rest: this is exactly what is summarized in Newton's first law.
Take a space probe moving in the interstellar space, very far from any planet or source of gravitational force. Since there are no forces acting on the proble, the probe will continue moving at the same velocity (same speed and same direction) forever, unless stopped by a new force acting on it.

This means that in order to put an object at rest in motion, or to stop an object already in motion, or to change its velocity, an unbalanced force needs to be applied: otherwise, the object will continue having the same velocity (which can be either zero or non-zero), so it will continue having same speed and same direction.

Learn more about Newton laws of motion:

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5 0

3 years ago

3 A picture is supported by two vertical strings; if the weighi

andrezito [222]

Answer:

<em>The force exerted by each string is 25 N</em>

Explanation:

<u>Net Force</u>

The net force is the vector sum of forces acting on a body. The net force is a single force that represents the effect of the original forces on the body's motion. It gives the particle the same acceleration as all those actual forces together as described by Newton's second law of motion.

The picture described in the problem is hanging at rest supported by two vertical strings. This means that the net force acting on it is zero.

Assume the magnitude of each of these equal forces is F, and the picture has a weight of W=50 N, thus the net force is:

F + F - W

The positive signs indicate an upwards direction and the negative sign means a downwards direction. Since the net force is zero:

F + F - W = 0

2F = W

F = W/2 = 50 N/2

F = 25 N

The force exerted by each string is 25 N

7 0

3 years ago