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3 years ago

An object is falling from a height of 7.5 meters. At what height will its velocity be 7 meters/second?

Physics

1 answer:

Ivanshal [37]3 years ago

8 0

Answer:

5.0 m

Explanation:

The velocity of the object at time t is given by

$v=at$

where a=9.8 m/s^2 is the acceleration of gravity and t is the time. If we substitute v=7 m/s, we can find the time t at which the velocity became 7 m/s:

$t=\frac{v}{a}=\frac{7 m/s}{9.8 m/s^2}=0.71 s$

The height of the object at time t, h(t), is given by the equation

$h(t)=h_0 -\frac{1}{2}at^2$

where $h_0 = 7.5 m$ is the initial height. By substituting t=0.71 s inside the equation, we find the height h(t) at which the velocity is 7 m/s:

$h(0.71 s)=7.5 m- \frac{1}{2}(9.8 m/s^2)(0.71 s)^2=5.0 m$

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Rzqust [24]

Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.

At the top of the first hill, the car's potential energy is

PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

   KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:

   KE = 0.9 PE

(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)

Divide each side by (mass):

   (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)

Divide each side by (0.9):

   (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

   Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

      = (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

      = (24,500 / 88.2) (m²/s²) / (m/s²)

      = 277-7/9 meters
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3 0

2 years ago

The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pre

IrinaVladis [17]

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

W = 160(5 - 7)

W = -320 KJ

6 0

3 years ago

On Mars, the acceleration due to gravity is 3.77 m / s 2 . 3.77 m/s2. How far would a 15 g 15 g rock fall from rest in 2.5 s 2.5

Zinaida [17]

Answer:

s = 11.78 m

Explanation:

given,

acceleration due to gravity, g = 3.77 m/s²

mass of the rock = 15 g

time = 2.5 s

distance traveled = ?

using equation of motion

$s = ut +\dfrac{1}{2}at^2$

initial speed = 0 m/s

$s = \dfrac{1}{2}at^2$

$s = \dfrac{1}{2}\times 3.77 \times 2.5^2$

s = 11.78 m

distance traveled by the rock is equal to 11.78 m.

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