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3 years ago

10

A slingshot shoots a marble. It travels at an average speed of 15 m/s for 12 seconds before hitting it's target. How far does it

travel?

1 answer:

salantis [7]3 years ago

6 0

Multiply m/s by 12 seconds to get meters by itself. 15(12)=180 m

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Your family decided to go to sea World in San Antonio this weekend. If it takes

Ne4ueva [31]

Answer:

63 mph

Explanation:

252/4 is 63mi

4 0

2 years ago

Read 2 more answers

A spring 1.50 m long with force constant 448 N/m is hung from the ceiling of an elevator, and a block of mass 10.9 kg is attache

GuDViN [60]

Answer:

Explanation:

Let the extension in the spring be x .

restoring force = weight of block

kx = mg

x = $\frac{10.9\times9.8}{448}$

= 23.84 cm

b )

When the elevator is going upwards

Restoring force = mg + ma

k x₁ = 10.9 ( 9.8 + 1.89 )

x₁ = 28.44 cm

( y coordinate will be - ( 28.44 - 23.84 ) = - 4.6 cm )

c ) When the cable snaps , both elevator and block undergo free fall . In this case apparent g = 0

Since the spring is stretched by 28.44 cm , a restoring force continues to act on the block which is equal to

.2844 x 448

= 127.41 N

So a net acceleration a will act on the block

a = 127.41 / 10.9

= 11.68 m / s²

The block will undergo SHM with amplitude equal to 28.44 cm .

3 0

3 years ago

When the acceleration of a mass on a spring is zero, the velocity is at a

Sergeu [11.5K]

1) Maximum

2) Maximum

Explanation:

The force acting on a mass on a spring is given by Hooke's law; in magnitude:

$F=kx$

where

F is the force

k is the spring constant

x is the displacement

Also we know from Newton's second law that we can write

$F=ma$

where

m is the mass

a is the acceleration

So we can write the equation as

$ma=kx$ (1)

From this relationship, we see that the acceleration is directly proportional to the displacement.

On the other hand, we know that the total mechanical energy of the system mass-spring is constant, and it is given by

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const.$ (2)

where the first term is the elastic potential energy while the second term is the kinetic energy, and where

v is the velocity of the mass

From eq. (2), it is clear that when displacement increases, velocity decreases, and vice-versa; however, from eq.(1) we also know that acceleration is proportional to the displacement.

Therefore this means that:

- When acceleration increases, velocity decreases

- When acceleration decreases, velocity increases

Therefore, the two answers here are:

- When the acceleration of a mass on a spring is zero, the velocity is at a maximum

When the velocity of a mass on a spring is zero, the acceleration is at a maximum

6 0

3 years ago

Normally, jet engines push air out the back of the engine, resulting in forward thrust, but commercial aircraft often have thrus

ANEK [815]

Answer:

When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.

Explanation:

Newton’s third law motion states that for every action there will be an equal and opposite reaction.

Thrust reversal is also known as reverse thrust. It acts opposite to the motion of the aircraft by providing the deceleration.

Commercial aircraft moves the ejected air in the forward direction means that the thrust will acts opposite to the motion of the aircraft that is backward direction due to thrust reversal. This thrust force might be used to decelerate the craft.

Uses of thrust reversal in practice:

When the ejected air is moving forward direction then the thrust force moving backward direction due to reversal thrust the speed of the craft slows down.

When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.

6 0

3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago