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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics

2 answers:

Schach [20]3 years ago

5 0

True. Kinetic means to have motion. So yes, bicycling up a tall hill has kinetic energy.

PtichkaEL [24]3 years ago

4 0

This is True

Kinetic energy is the energy of motion. The bicyclist is in motion as he pedals up the tall hill. Therefore, the bicyclist contains kinetic energy.

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A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a spe

Aleksandr-060686 [28]

The direction of motion of the person relative to the water is 16.7° north of east.

Why?

We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)

We can calculate the angle that give the direction using the following formula:

$Tan(\alpha)=\frac{NorthSpeed}{EastSpeed}\\\\Tan(\alpha)^{-1}=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}\\\\\alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$

Now, substituting the given information we have:

$alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}$

$\alpha =Tan(\frac{3.6\frac{m}{s} }{12\frac{m}{s} })^{-1}\\\\\alpha =Tan(0.3)^{-1}=16.69\°(North-East)=16.7\°(North-East)$

Hence, we have that the direction of motion of the person relative to the water is 16.7° north of east.

Have a nice day!

7 0

3 years ago

Select the correct answer. Which electromagnetic waves have the highest frequencies in the electromagnetic spectrum?

Law Incorporation [45]

Answer:

Your answer would be C, Radio waves.

Explanation:

4 0

3 years ago

44.2 cm + 0.123 cm = cm

trapecia [35]

The answer is 44.323 cm

3 0

3 years ago

Read 2 more answers

A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h

timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

$F_n = \frac{mv^2}{r}$

$N+mgcos\theta = \frac{mv^2}{r}$

Here,

Normal reaction of the ring is N and velocity of the ring is v

$N+mgcos\theta = \frac{mv^2}{r}$

$N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})$

$N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}$

$N = 1.374lb$

PART B) Acceleration

$F_t = ma_t$

$-mgsin\theta = ma_t$

$-W sin\theta = \frac{W}{g} a_t$

$-2Sin30\° = (\frac{2}{32.2})a_t$

$a_T = -16.10ft/s^2$

Negative symbol indicates deceleration.

NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.

8 0

3 years ago

The focal length of the lens of a simple digital camera is 40 mm, and it is originally focused on a person 25 m away. In what di

ss7ja [257]

Answer:

Explanation:

Here image distance is fixed .

In the first case if v be image distance

1 / v - 1 / -25 = 1 / .05

1 / v = 1 / .05 - 1 / 25

= 20 - .04 = 19.96

v = .0501 m = 5.01 cm

In the second case

u = 4 ,

1 / v - 1 / - 4 = 1 / .05

1 / v = 20 - 1 / 4 = 19.75

v = .0506 = 5.06 cm

So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )

3 0

3 years ago