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3 years ago

11

A 95-kg astronaut is stranded from his space shuttle. He throws a 2-kg hammer away from the shuttle with a velocity of 19 m/s .

How fast will he be propelled toward the shuttle, in m/s ? (Round your answer to one decimal place if necessary )

1 answer:

geniusboy [140]3 years ago

5 0

Answer:

The astronaut will be propelled towards the shuttle at the rate of 0.4 m/s.

Explanation:

Given;

mass of the astronaut, m₁ = 95 kg

mass of the hammer thrown, m₂ = 2 kg

velocity of the hammer, v₂ = 19 m/s

let the recoil velocity of the shuttle = v₁

Apply the principle of conservation of linear momentum;

m₁v₁ = m₂v₂

v₁ = m₂v₂/m₁

v₁ = (2 x 19) / 95

v₁ = 0.4 m/s

Therefore, the astronaut will be propelled towards the shuttle at the rate of 0.4 m/s.

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3. What is the mass of a paratrooper who experiences an air resistance of 400 N and an acceleration of 4.5 m/s2

goblinko [34]

Answer:

88.89kg

Explanation:

The formula for mass is m=F/a. If we plug in the values, we get m=400N/4.5m/s^2. The mass is 88.89kg. We know that the unit is in kg because one newton (N) is 1kg*m/s^2. The m/s^2 is cancelled out by the acceleration, and we are left with kg.

4 0

3 years ago

Air at 27oC and 1 atm flows over a flat plate 40 cm in length and 1 cm in width at a speed of 2 m/s. The plate is heated over it

irga5000 [103]

Answer:

Heat transferred = 22.9 watt

Explanation:

Given that:

$T_1$ = 27°C = (273 + 27) K = 300 K

$T_2$ = 600°C = (600 +273) K = 873 K

speed v = 2 m/s

length x = 40 cm = 0.4 cm

width = 1 cm = 0.001 m

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

From the tables of properties of air, the following values where obtained.

$k = 0.02476 \ W/m.k \\ \\ \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg$

To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:

reynolds number = $\dfrac{\rho \times v \times x }{\mu}$

reynolds number = $\dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}$

reynolds number = $\dfrac{0.98}{18.6 \times 10^{-6}}$

reynolds number = 52688.11204

Prandtl number = $\dfrac{c_p \mu}{k}$

Prandtl number = $\dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}$

Prandtl number = $\dfrac{0.018693}{0.02476}$

Prandtl number = 0.754963

The nusselt number for this turbulent flow over the flat plate can be computed as follows:

Nusselt no = $\dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}$

$\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}$

$\dfrac{h \times 0.4}{0.02476} =161.4252008}$

$h =\dfrac{161.4252008 \times 0.02476}{ 0.4}$

h = 9.992 W/m.k

Recall that:

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

Heat transferred = $h\times A \times (T_2-T_1)$

Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)

Heat transferred = 22.9 watt

6 0

3 years ago

A total of 25.6 kJ of heat energy is added to a 5.46 L sample of helium at 0.991 atm. The gas is allowed to expand against a fix

bagirrra123 [75]

Answer:

(a) W = 1329.5 J = 1.33 KJ

(b) ΔU = 24.27 KJ

Explanation:

(a)

Work done by the gas can be found by the following formula:

$W = P\Delta V$

where,

W = Work = ?

P = constant pressure = (0.991 atm)( $\frac{101325\ Pa}{1\ atm}$ ) = 100413 Pa

ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)( $\frac{0.001\ m^3}{1\ L}$ ) = 0.01324 m³

Therefore,

W = (100413 Pa)(0.01324 m³)

<u>W = 1329.5 J = 1.33 KJ</u>

<u></u>

(b)

Using the first law of thermodynamics:

ΔU = ΔQ - W (negative W for the work done by the system)

where,

ΔU = change in internal energy of the gas = ?

ΔQ = heat added to the system = 25.6 KJ

Therefore,

ΔU = 25.6 KJ - 1.33 KJ

<u>ΔU = 24.27 KJ</u>

3 0

3 years ago

Two inductors, L1 and L2, are in parallel. L1 has a value of 25 mH and L2 a value of 50 mH. The parallel combination is in serie

Bond [772]

Answer:

Explanation:

For parallel inductors ,

$\frac{1}{L_R} = \frac{1}{L_1} +\frac{1}{L_2}$

$\frac{1}{L_R} =\frac{1}{25} +\frac{1}{50}$

$L_R=16.67 mH.$

For series combination

Total inductance

= 16.67 + 20

= 36.67 mH .

reactance of total inductance at 300 kHz

= ω $L_{total}$ where ω is angular frequency

= 2πf $L_{total}$

= 2 x 3.14 x 300 x 10³ x 36.67 x 10⁻³

= 69.1 x 10³ ohm

Total rms current = Vrms / reactance

= 60 / 69.1 x 10³ A

= .87 x 10⁻³ A

= .87 mA

7 0

3 years ago

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm

mario62 [17]

Answer:

A) if each astronaut breathes about 500 cm³, the total volume of air breathed in a year is 14716.8m³.

B) The Diameter of this spherical space station should be 30.4m

Explanation:

The breathing frequency (according to Rochester encyclopedia) is about 12-16 breath per minute. if we take the mean value (14 breath per minute), we can estimate the total breaths of a person along a year:

$f_b=14\frac{br}{min} \cdot \frac{60min}{1hr} \frac{24hr}{1day}\frac{365day}{1year}=29433600\frac{br}{year}$

If we multiply this for the number of people in the station and the volume each breath needs, we obtain the volume breathed in a year.

The volume of a sphere is:

$V_{sph}=\frac{4\pi}{3}r^3$

So the diameter is:

$D=2r=2\sqrt[3]{\frac{3V_{sph}}{4\pi}} =30.4m$

6 0

3 years ago