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Natalka [10]
3 years ago
12

The diagram below shows the muzzle of a cannon located 50. meters above the ground. When the cannon is fired, a ball leaves the

muzzle with an initial horizontal speed of 250. meters per second. [Neglect air resistance.]Which action would most likely increase the time of flight of a ball fired by the cannon?
Physics
1 answer:
Nastasia [14]3 years ago
5 0

Answer:positioning the cannon higher above the ground

Explanation:

❤️

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Which of the following is an example in which you are traveling at constant speed but not at constant velocity?
grin007 [14]

Answer:

c) Driving around in a circle at exactly 100 km/hr.

This examples represents constant speed but not constant velocity.

Explanation:

To answer this question it is importan to know the difference between the concept of speed and velocity.

speed: Is the ratio of change in the displacement per unit of time. JUST  A MAGNITUDE

velocity: Is the ratio of change in the displacement per unit of time in a given direction. It is a composed value by magnitude and direction, this is know as a VECTOR.

So, in conclusion the speed is the magnitude or the scalar value, for the velocity vector.

In the example,

in a) Rolling freely down a hill in a cart, traveling in a straight line. <em>There is a component of acceleration, from earth gravity, so the speed is changing.</em>

<em />

in b) Driving backward at exactly 50 km/hr. <em> The direction and speed are the same, so speed and velocity are constant</em>.

In d) Jumping up and down, with a period of exactly 60 hops per minute. <em>There are changes in the speed, due to the acceleration and decceleration between the changes in the direction. Actually you have to stop to change direction between the ups and downs in the jumps. Also there is the gravitational component, always changing the speed.</em>

In c) Driving around in a circle at exactly 100 km/hr. <em>In this case the speed remains constant, while the direction is changing all the time.</em>

3 0
4 years ago
Problem 18.9 an underwater diver sees the sun 60 ∘ above horizontal. part a how high is the sun above the horizon to a fisherman
julsineya [31]

The angles in the equation are not the angles relative to the horizon but are relative to the "normal" which means that the line that is perpendicular to the surface.

The angle under the water is 90 – 60 = 30. 

n1 for water is 1.33, n2 for air is 1 Which you seem to understand. 

(1.33)(sin (30)) = (1.00)(sin (x2)) 

Rearranging the equation above, will give us: x^2 = sin^-1((1.33 sin 30)/1) = 41.68

But remember that that is the angle relative to the normal so you have to deduct it from 90 to get the angle relative to the horizon and you get (90 – 41.68) = 48.32 degrees.

3 0
3 years ago
You walk 53 m to the north, then you turn 60° to your right and walk another 45 m. Determine the direction of your displacement
Anton [14]

Answer:

NE 60*

Explanation:

4 0
3 years ago
A hockey puck is sliding across a frozen pond with an initial speed of 9.5 m/s. It comes to rest after sliding a distance of 37.
Bond [772]

Answer:

<h3>The coefficient of kinetic friction between the puck and the ice is \mu _{k} = 0.12</h3>

Explanation:

Given :

Initial speed  v_{o} = 9.5 \frac{m}{s}

Displacement x = 37.4 m

From the kinematics equation,

  v^{2} - v^{2} _{o}  = 2ax

Where v^{2}   = final velocity, in our example it is zero (v =0), a = acceleration.

   a =- \frac{90.25}{ 2 \times 37.4}

   a =- 1.21 \frac{m}{s^{2} }

From the formula of friction,

  F =- \mu _{k } N

Minus sign represent friction is oppose the motion

Where N = mg ( normal reaction force )

 ma = -\mu _{k}  m g                                                  ( ∵ g = 9.8 \frac{m}{s^{2} } )

So coefficient of friction,

 \mu_{k} = \frac{1.21}{9.8}

 \mu_{k} = 0.12

Therefore, the coefficient of kinetic friction between the puck and the ice is  \mu _{k} = 0.12 .

4 0
3 years ago
The function of a microscope is most similar to the function of a.............​
Nataly [62]

Answer:

Magnifying glass and a Petri dish.

Explanation:

4 0
3 years ago
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