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3 years ago

Difference between uniform circular motion and non uniform circular motion

1 answer:

6 0

Uniform circular motion is the motion in which an object covers equal distance in equal interval of time in a circular path and in non uniform motion object covers equal distance in unequal interval of time in a circular path

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A suitcase has a width of 55 cm. Which unit conversation fraction should you use to find the width in inches?

Crank

Answer:

2,54 cm are equal to 1 inch

Explanation:

Doing the conversion:

$55[cm]*\frac{1[inch]}{2,54[cm]} =21,65[inch]$

6 0

3 years ago

What can lead to groundwater shortages

Alex

Answer:

most commonly occurs because of the frequent pumping of water from the ground.

Explanation:

8 0

3 years ago

Compare and contrast speed and velocity.

vlabodo [156]

Speed is the time rate of an object moving from one place to another, while velocity is the rate and direction of the object's movement. They are very similar but they don't mean the same thing.

8 0

3 years ago

Consider a river flowing toward a lake at an average velocity of 3 m/s. the river height is 90m above the lake. what is the tota

strojnjashka [21]

Kinetic energy per unit of mass is

$K=\frac{v^{2} }{2}$

Given, $v=3m/s^{2}$

Therefore,

$K=\frac{(3^{2} m/s^{2} )^2}{2}$

$K=4.5 J/kg$

Now potential energy per unit mass is

$p=g\times h$

Given, $h=90 m$

Therefore,

$p= 9.8m/s^2 \times 90$

$p=882.9 J/kg$

Thus, total mechanical energy of the river water per unit mass is

$T=K+p=(4.5+882.9)J/kg$

$T=887.9 J/kg$

$T=0.887 kJ/kg$

6 0

3 years ago

A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s

Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± $\frac{1}{2} at^2$ ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± $\frac{1}{2} at^2$ ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - $\frac{1}{2} at^2$

h = 26.54(1.7) - $\frac{1}{2} (10)(1.7)^2$

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m

7 0

3 years ago