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3 years ago

Mike drove 70 km due west and then 100 km due east.

Physics

1 answer:

Deffense [45]3 years ago

4 0

Distance = 170 km Displacement = 30 km East Speed = 56-2/3 km/hr Velocity = 10 km/hr East

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The pitch of the sound produced by a metal wire fixed between two points depends on A) how fast the wire vibrates. B) what type

mrs_skeptik [129]

It depends on A) how fast the wire vibrates because sound is produced by vibration

7 0

3 years ago

O D. Both objects won't move at all. They will just stay where they were released.

astraxan [27]

Answer:

Do not see a picture or graph but suspect it would show the golf ball falling faster and striking the ground slightly before the soccer ball.

Probably D: Soccer ball was affected by air resistance more than the golf ball.

Explanation:

Even though heavier, friction loss of the greater surface area soccer ball will counter pull of gravity more than the compact golf ball.

In a vacuum, (no friction) both objects fall at the same rate regardless of mass.

6 0

3 years ago

A kite is hovering over the ground at the end of a straight 38-m line. the tension in the line has a magnitude of 17 n. wind blo

postnew [5]

Let the tension in the thread will be at angle theta

now by force balance in x and y directions we can say

$17 cos\theta = 18 cos60$

$17 cos\theta = 9$

$\theta = 58.03 degree$

now for the height of the kite we can use

$h = L sin\theta$

$h = 38 sin58.03$

$h = 32.23 m$

<em>so it will be at height 32.23 m from ground</em>

3 0

3 years ago

Equation for fluid density

MrMuchimi

<span> (ρ = m / V) hope this helps...:)</span>

7 0

4 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

3 years ago