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2 years ago

Which of the following is NOT a kind of energy store?

Physics

1 answer:

snow_tiger [21]2 years ago

6 0

Answer:

Option D is your answer ☺️☺️. If I'm right so,

Please mark me as brainliest. thanks!!!

You might be interested in

Which is not a simple harmonic motion (S.H.M.) (a) Simple Pendulum (b) Projectile motion (c) None (d) Spring motion

Scrat [10]

Answer:

b) Projectile MOTION

Explanation:

SHM is periodic motion or to and fro motion of a particle about its mean position in a straight line

In this type of motion particle must be in straight line motion

So here we can say

a) Simple Pendulum : it is a straight line to and fro motion about mean position so it is a SHM

b) Projectile motion : it is a parabolic path in which object do not move to and fro about its mean position So it is not SHM

d) Spring Motion : it is a straight line to and fro motion so it is also a SHM

So correct answer will be

b) Projectile MOTION

7 0

2 years ago

The hottest recorded temperature in the history of the United States is 134 °F, which is 57 °C. What would the speed of sound at

guapka [62]

Answer:

365 m/s

Explanation:

Mrs Finley

4 0

2 years ago

An object is moving with the speed of light around the Earth. How much time will it take to complete one round trip along the eq

dimulka [17.4K]

Answer:

0.14 seconds

Explanation:

The speed of light in vacuum is approximately 3.0*10^8. The distance that would be covered by the object would be equivalent to the circumference of the cross-section of the earth on the equator.

Circumference = 2 $\pi$ *6400000 =4.02*10^7

Time = distance/speed = 4.2*10^7 / 3.0*10^8 =0.14s

8 0

2 years ago

A private aviation helicopter's main rotor blades rotate at approximately

Arisa [49]

Answer: 7.5 rev/s

Explanation:

We are given the angular velocity $\omega$ a helicopter's main rotor blades:

$\omega=450 rpm=450 \frac{rev}{min}$

However, we are asked to express this $\omega$ in the International Systrm (SI) units. In this sense, the SI unit for time is second ( $s$ ):

$\omega=450 \frac{rev}{min} \frac{1 min}{60 s}$

$\omega=7.5 \frac{rev}{s}$

4 0

2 years ago

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