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3 years ago

A storm is moving at 15 km/hr what to do to determine its velocity

1 answer:

6 0

Check the current weather map for 2 different times, and see where the center of the storm is. That tells you what direction it's moving. With its speed and direction, you have its velocity.

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When you take pictures with a camera, the distance between the lens and the film or chip has to be adjusted, depending on the di

Shalnov [3]

Answer:

Explanation:

When a camera shifts focus from a faraway object to a nearby object, the lens-to-film distance must increase. Likewise, when it shifts focus from a nearby object to a distant object, there must be an increase in the lens to film distance (that is, the image distance).

Therefore, if the picture of an object that is far away, the lens must move towards the film.

The focal length cannot be changed because it is fixed for a lens. Nevertheless, in order to focus on an object, the image distance can be changed.

8 0

3 years ago

A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is

lys-0071 [83]

Answer:

$\omega_f=571.42\ rpm$

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, $a=1.5\ m/s^2$

Displacement, d = 1.3 m

The angular acceleration is given by :

$\alpha =\dfrac{a}{r}$

$\alpha =\dfrac{1.5}{0.033}$

$\alpha =45.46\ rad/s^2$

The angular displacement is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{1.3}{0.033}$

$\theta=39.39\ rad$

Using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

Here, $\omega_i=0$

$\omega_f=\sqrt{2\alpha \theta}$

$\omega_f=\sqrt{2\times 45.46\times 39.39}$

$\omega_f=59.84\ rad/s$

Since, 1 rad/s = 9.54 rpm

So,

$\omega_f=571.42\ rpm$

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0

3 years ago

A professional cyclist rides a bicycle that is 92 percent efficient. For every 100 joules of energy he exerts as input work on t

emmainna [20.7K]

Efficiency = Work Output / Work Input

92% = Work Output / 100

0.92 = Work Output / 100

Work Output = 0.92 * 100

Work Output = 92 joules.

8 0

3 years ago

6. If two objects experience the same net force, but they have different masses, which object will accelerate at

Digiron [165]

Answer:

i would say the heavier object

4 0

3 years ago

Read 2 more answers

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

3 years ago