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3 years ago

What is the Hairdryer energy transfers

Physics

1 answer:

ra1l [238]3 years ago

4 0

Answer:

it transfers heat to the hair to make it dry with the hot air which would be thermal energy

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What is the difference between a distance time graph and a speed time graph?

Feliz [49]

Answer:

speed-time graph is the acceleration of the particle or straight lines acceleration is constant

a distance-time graph represents that a particle has constant speed.

Explanation:

3 0

3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

3 years ago

Based on what you have learned from this unit, construct a tri-fold brochure instructing a new freshman as to the best way to le

Rzqust [24]

<span>A tri-fold brochure has two parallel folds, splitting the brochure into three sections. Even when printed on low-weight paper, tri-folds can stand up easily, which makes them a great choice for displaying at conventions. You can fold both folds inwards so that the left and right sections of the brochure sit on top of one another, or you can have one fold inwards and the other outwards, to create an accordion effect, which looks very attractive.</span>

3 0

3 years ago

A man paddles a canoe at 6 km per hour. If he paddles on a river with a current of 6 km per hour, what is the speed of the canoe

Romashka-Z-Leto [24]

Answer:

If the canoe heads upstream the speed is zero. And directly across the river is 8.48 [km/h] towards southeast

Explanation:

When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:

Vr = velocity of the river = 6[km/h}

Vc = velocity of the canoe = -6 [km/h]

We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.

Vt = Vr + Vc = 6 - 6 = 0 [km/h]

For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.

So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).

Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:

$Vt = \sqrt{(6)^{2} +(-6)^{2} } \\Vt=8.48[km/h]$