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3 years ago

10

Which is an example of potential energy? Question 9 options: an electric current lighting a light bulb releasing the strings on

a guitar a roller coaster speeding down a hill the pulled back string on an archer's bow

1 answer:

Feliz [49]3 years ago

4 0

The pulled back string on an archer’s bow.

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You are driving your motorcycle in a circle of radius 75 m on wet pavement. what is the fastest you can go before you lose tract

stira [4]

On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s

Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.

Maximum velocity of a road with friction is given by the formula,

v = μRg

where, v is the maximum velocity

μ is the coefficient of static friction

R is the radius of the circle road

g is the acceleration due to gravity

Given,

μ = 0.20

R = 75m

g = 9.8m/s²

On substituting the given values in the above formula,

v = 0.20× 75 ×9.8

v = 147m/s

So, the Maximum velocity of the wet road is 147m/s.

Learn more about Velocity here, brainly.com/question/18084516

#SPJ4

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2 years ago

Human-Powered Flight Human-powered aircraft require a pilot to pedal, as in a bicycle, and produce a sustained power output of a

alukav5142 [94]

Answer:

# of Snickers bars 2

Explanation:

Power output= 0.30 HP

=0.3*746

= 0.30 HP (746 W=1.00 HP)

= 224 W

time required 2 h 49 m = 10140 seconds

Since power is work divided by time, then work is:

Work done by the jet = P*t

= 224 *(10140)

= 2.3 MJ (2.3 x $10^{6}$ J)

Converting MJ to Cal

2.3 MJ=549 Cal

# of Snickers bars = 549 Cal / 280 Cal

= 2.0 bars (rounded from 1.96)

8 0

3 years ago

A terain tarvels 120 I in 2 hours and 30 mins .what is its average speed

Zarrin [17]

Answer:

It's speed is 48 km/hr.

Explanation:

6 0

3 years ago

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Control rods in a nuclear reactor are used to:

kiruha [24]

B.absorb neutrons to prevent chain reactions which become uncontrollable

7 0

3 years ago

Read 2 more answers

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ <em> ...Newtons second law Fnet = ma</em>

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ <em>...f is the friction which is zero here.</em>

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0

3 years ago