Ask question

Ask question

All categories

liberstina [14]

3 years ago

6

if a truck falls for 33 seconds after it drives off a bridge into canyon below. how far does it fall before it hits the canyon f

loor? give your answer in meters.

1 answer:

Svetlanka [38]3 years ago

8 0

Here's a formula that's so useful, you should memorize it:

Distance = (1/2) · (acceleration) · (time squared)

On Earth, acceleration of gravity is 9.8 m/s² .

Height of the bridge over the canyon =

Distance the truck falls = (1/2) · (9.8 m/s²) · (33 sec)²

Distance = (4.9 m/s²) · (1,089 sec²)

Distance = 5,336 meters

You might be interested in

How much energy is needed to melt 5 g of ice? The specific latent heat of melting for water is 334000 J/kg.

Katen [24]

Answer:

The needed energy to melt of ice is 1670 J.

Explanation:

Given that,

Mass of ice = 5 g

Specific latent heat = 334000 J/kg

We need to calculate the energy

Using formula of energy

$Q=mL$

Where, m = mass

L = latent heat

Put the value into the formula

$Q=5\times10^{-3}\times334000$

$Q=1670\ J$

Hence, The needed energy to melt of ice is 1670 J.

5 0

3 years ago

In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If t

Vitek1552 [10]

Answer:

So the ratio will be $\frac{T_L}{T_H}=-0.171$

Explanation:

We have given heat engine absorbs 450 joule from high temperature reservoir

So $Q=450j$

As the heat engine expels 290 j

So work done W = 290 J

We know that efficiency $\eta =\frac{W}{Q}=\frac{290}{450}=0.6444$

It is given that efficiency of the engine only 55 % of Carnot engine

So efficiency of Carnot engine $=\frac{0.6444}{0.55}=1.171$

Efficiency of Carnot engine is $\eta =1-\frac{T_L}{T_H}$

$1.171 =1-\frac{T_L}{T_H}$

$\frac{T_L}{T_H}=-0.171$

3 0

3 years ago

Read 2 more answers

Why wil an ice cube melt if it is compressed even if it remains at the same temp

Gnoma [55]

No, it will only melt if the temperature is lowered. If you compress it, it will change the shape, but it will not change the state it is in (i.e. solid).

8 0

3 years ago

Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning

vladimir1956 [14]

Answer:

Explanation:

The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .

When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .

The greatest acceleration is attained at maximum displacement or at one of the two extreme end .

Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .

Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .

Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .

8 0

3 years ago

Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th

Evgen [1.6K]

Answer:

$\omega=0.24\ rad.s^{-1}$

Explanation:

Given:

mass of person, $m=36\ kg$

mass of merry go-round, $M=300\ kg$

radius of merry go-round, $R=2\ m$

velocity of the person running, $v=4\ m.s^{-1}$

<u>We consider merry go-round as a ring:</u>

Now the moment of inertial of the ring is given as,

$I=M.R^2$

$I=300\times 2^2$

$I=1200\ kg.m^{-2}$

<u>Moment of inertia of the person considering as a point mass:</u>

$I_p=m.R^2$

$I_p=36\times 2^2$

$I_p=144\ kg.m^2$

<u>Now according to the conservation of angular momentum:</u>

$I.\omega=I_p.\omega_p$

where:

$\omega =$ angular velocity of the merry-go-round

$\omega_p=$ angular velocity of the person running

$1200\times \omega=144\times \frac{v}{R}$

$\omega=\frac{144}{1200} \times \frac{4}{2}$

$\omega=0.24\ rad.s^{-1}$

4 0

3 years ago

Read 2 more answers