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3 years ago

How long would it take someone jogging at 2.4 m/s to travel 2300 m?

Physics

1 answer:

Ber [7]3 years ago

7 0

Answer:

958 Second or 15 minute and 58 second approximately 16 minutes

Explanation:

m / m/s

2300/2.4 = 958 second

958/60 = 15 minutes and 58 seconds approximately 16 minutes

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An astronaut has a mass of 75 kg and is floating in space 500 m from his 125,000 kg spacecraft. What will be the force of gravit

nikdorinn [45]

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

5 0

3 years ago

A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800

Arada [10]

Explanation:

The given data is as follows.

m = 5000 kg, h = 800 km = $0.8 \times 10^{6} m$

$R_{e} = 6.37 \times 10^{6} m$ , r = R + h = $7.17 \times 10^{6} m$

$M_{e} = 5.98 \times 10^{24}$ kg, G = $6.67 \times 10^{-11} Nm^{2}/kg^{2}$

As we know that,

$\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}$

v = $\sqrt{\frac{GM_{e}}{r^{2}}}$

And, it is known that formula to calculate angular velocity is as follows.

$\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}$

v = $\sqrt{\frac{GM_{e}}{r^{3}}}$

= $\sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}$

= $1.0402 \times 10^{-3} rad/s$

Thus, we can conclude that speed of the satellite is $1.0402 \times 10^{-3} rad/s$ .

6 0

3 years ago

A monatomic gas is adiabatically compressed to 0.250 of its initial volume. Do each of the following quantities change?

Len [333]

Answer:

Given that

V2/V1= 0.25

And we know that in adiabatic process

TV^န-1= constant

T1/T2=( V1 /V2)^ န-1

So = ( 1/0.25)^ 0.66= 2.5

Also PV^န= constant

So P1/P2= (V2/V1)^န

= (1/0.25)^1.66 = 9.98

A. RMS speed is

Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

Lambda=V/4π√2πr²N

Lambda 2/lambda 1= V2/V1 = 0.25

So the mean free path can be inferred to be 0.25 times the first mean free path

C. Using

Eth= 3/2KT

So Eth2/Eth1= T2/T1

Eth2= 2.5Eth1

Using CV= 3/2R

Cvf= Cvi

So molar specific heat constant does not change

5 0

3 years ago

Read 2 more answers

Russell bradley carried 207 kg of bricks 3.65 m up a ladder. If the amount of work required to perform that task is used to comp

MrRa [10]

Answer:

Explanation:

Work done in carrying bricks

mgh

= 207 x 9.8 x 3.65

-= 7404.4 J

Work done in compressing gas

PΔV

Pressure x change in volume

1.8 x 10⁶ ΔV = 7404.4

ΔV = 7404.4 / 1.8 x 10⁶m³

= 4113.33 x 10⁻⁶ m³

= 4113.33 cc

8 0

3 years ago

Read 2 more answers

The zone along the southern margins of the Sahara is called the

Jobisdone [24]

We have to choose the correct name for the zone along the southern margins of the Sahara. Laterite is a soil in hot and wet tropical areas. Savanna is the tropical grassland. It has tropical savanna climate. Veldt is name for the areas in the South Africa. Finally, the Sahel is the zone along the south margins of the Sahara. It has a semi-arid climate. The Arabic word "sahel" means "coast". Answer: C. Sahel. <span> </span>

7 0

4 years ago