Why is it realistic to say all living organisms are solar powered

An airplane flies with a constant speed of 560 miles per hour. How far can it travel in 1 1/2 hours?

Alik [6]

You would multiply the speed by the time. So the answer would be 840 miles.

5 0

3 years ago

A ball moving with an initial velocity of 5 m/s comes to rest after 2s. What was the ball's acceleration?

Inga [223]

Answer:

-2.5m/s²

Explanation:

The acceleration of a body is giving by the rate of change of the body's velocity. It is given by

a = Δv / t ----------------(i)

Where;

a = acceleration (measured in m/s²)

Δv = change in velocity = final velocity - initial velocity (measure in m/s)

t = time taken for the change (measured in seconds(s))

From the question;

i. initial velocity = 5m/s

final velocity = 0 [since the body (ball) comes to rest]

Δv = 0 - 5 = -5m/s

ii. time taken = t = 2s

<em>Substitute these values into equation (i) as follows;</em>

a = (-5m/s) / (2s)

a = -2.5m/s²

Therefore, the acceleration of the ball is -2.5m/s²

NB: The negative sign shows that the ball was actually decelerating.

6 0

3 years ago

2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the

frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum = 2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = $\frac{1}{2}$ mv²

v² = 2gh₁

v = √(2gh₁)

from the second equation; s = ut + $\frac{1}{2}$ at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{5}$ mR²) ω²

v = √( $\frac{10}{7}$ gh₁ )

so we substitute √( $\frac{10}{7}$ gh₁ ) for v and t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$ gh₁ ) × √(2h₂/g)

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{3}$ mR²) ω²

v = √( $\frac{6}{5}$ gh₁ )

so we substitute √( $\frac{6}{5}$ gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( $\frac{6}{5}$ gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ $I_{}$ ω²

mgh₁ = $\frac{1}{2}$ mv² + $\frac{1}{2}$ ( $\frac{2}{5}$ mR²) ω²

v = √( $\frac{10}{7}$ gh₁ )

so we substitute √( $\frac{10}{7}$ gh₁ ) for v and t = √(2h₂/g) in equation 1;

d = vt

d = √( $\frac{10}{7}$ gh₁ ) × √(2h₂/g)

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0

3 years ago

An archer wants to hit a target that is dropped from a tower. At the sound of a horn, the archer is to shoot an arrow; at the sa

bija089 [108]

It has to be D because the arrow will drop as it moves, if it were a gun, you'd lead the target so fire below it, but due to it being an arrow, you aim high not low. Also, they didnt specify how fast anything is, so you'd probably miss if you actually did it.

6 0

3 years ago

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How are energy, force, and the motion of objects related?

OLEGan [10]

Answer:

When two objects interact, each one exerts a force on the other that can cause energy to be transferred to or from the object. For example, when energy is transferred to an Earth-object system as an object is raised, the gravitational field energy of the system increases. This energy is released as the object falls; the mechanism of this release is the gravitational force. Likewise, two magnetic and electrically charged objects interacting at a distance exert forces on each other that can transfer energy between the interacting objects.

Explanation:

Even when an object is sitting still, it has energy stored inside that can be turned into kinetic energy (motion). ... A force is a push or pull that causes an object to move, change direction, change speed, or stop. Without a force, an object that is moving will continue to move and an object at rest will remain at rest.

5 0

3 years ago

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