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Eduardwww [97]
3 years ago
13

A car travels at a constant velocity of 40 m/s for 5.0 s, determine its displacement during this 5.0 s.

Physics
1 answer:
zimovet [89]3 years ago
3 0

Answer:200 m

Explanation:

V=40m/s

T=5 s

Displacement=v*t

40*5=200

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A new planet has been discovered and given the name Planet X . The mass of Planet X is estimated to be one-half that of Earth, a
harina [27]

Answer:

    vₐ = v_c  ( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

          Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

         Em_f = U = - G Mm / R₂

energy is conserved

           Em₀ = Em_f

           ½ m v_c² - G Mm / R = - G Mm / R₂

           v_c² = 2 G M (1 /R -  1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

           v_c = \sqrt{\frac{2GM}{R} }

now indicates that the mass and radius of the planet changes slightly

            M ’= M + ΔM = M ( 1+ \frac{\Delta M}{M} )

            R ’= R + ΔR = R ( 1 + \frac{\Delta R}{R} )

we substitute

           vₐ = \sqrt{\frac{2GM}{R} } \  \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }

         

let's use a serial expansion

           √(1 ±x) = 1 ± ½ x +…

we substitute

         vₐ = v_ c ( (1 + \frac{1}{2}  \frac{\Delta M}{M} )  \ ( 1 - \frac{1}{2}  \frac{\Delta R}{R} ))

we make the product and keep the terms linear

        vₐ = v_c  ( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )

5 0
2 years ago
A car with a mass of 725 kg is moving at 32 m/s toward the east. What is the momentum of the car? A second car with a mass of 2,
Elden [556K]

Answer:

A compact car, with mass 725 kg, ... at 115 km/h toward the east. ... b. A second car, with a mass of 2175 kg, has the same momentum. What is its ... Glisens : m = 2175 kg;. D 21,32 xrolka. anknown. r = ? 110.6 mis east

Explanation:

7 0
3 years ago
an auditorium measures 30.0 m ✕ 15.0 m ✕ 5.0 m. the density of air is 1.20 kg/m3. (a) what is the volume of the room in cubic fe
Fynjy0 [20]

dimension = 30.0 m ✕ 15.0 m ✕ 5.0 m.

density = 1.20 kg/m3

(a)volume = lenght * breadth * height

      = 30 * 15 * 5

      = 2250 metre cube = 2.25 cubic meter

(b)   mass of air = density * volume

        mass of air = 1.2 * 2250

mass of air = 2700kg

weight  = mass * 9.8

             = 2700 * 9.8

             = 26,460 N

  • The definition of Density is the amount of matter in a given space, or volume
  • Density = mass/volume
  • units for density kg/m^3
  • Density of water 1g/ml
  • Salt water is denser that is why  don't sink as easily.

To know more about density  visit : brainly.com/question/15164682

#SPJ4

5 0
1 year ago
If a spaceship has a momentum of 30,000 kg-m/s to the right and a mass of
m_a_m_a [10]

Answer:

75m/s

Explanation:

...................

8 0
3 years ago
You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi
Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0
3 years ago
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