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2 years ago

A(n) 59 kg astronaut becomes separated from

Physics

1 answer:

pantera1 [17]2 years ago

5 0

Answer:

5. 7.452 min

Explanation:

Momentum is conserved, so:

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

0 = (0.695 kg) (-12 m/s) + (59 kg) v

v = 0.1414 m/s

She is 63.2 m away, so the time it takes to reach the shuttle is:

t = 63.2 m / 0.1414 m/s

t = 447.1 s

t = 7.452 min

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An automobile engine delivers 47.4 hp. how much time will it take for the engine to do 6.82 × 105 j of work? one horsepower is e

kogti [31]

1 horsepower is equal to 746 W, so the power of the engine is
$P=47.4 hp \cdot 746 \frac{W}{hp}=35360 W$
The power is also defined as the energy E per unit of time t:
$P= \frac{E}{t}$
Where the energy corresponds to the work done by the engine, which is $E=6.82 \cdot 10^5 J$ . Re-arranging the formula, we can calculate the time t needed to do this amount of work:
$t = \frac{E}{P}= \frac{6.82 \cdot 10^5 J}{35360 W}=19.3 s$

8 0

3 years ago

Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of t

Lelu [443]

Answer:

c) At a distance greater than r

Explanation:

If G= Gravitational constant

M= Mass of earth

r= distance from earth center

then orbital speed is ;

v = $\sqrt{\frac{GM}{r} }$

==> v²=GM/r

If speed of first satellite = V₁

==> V₁² = GM/r

==> r = GM/V₁²

If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.

So our answer is c

5 0

2 years ago

Mary and her younger brother Alex decide to ride the 17-foot-diameter carousel at the State Fair. Mary sits on one of the horses

lorasvet [3.4K]

Answer:

$\omege_A=\omega_M$

$v_M=1.6v_A$

Explanation:

A denotes Alex

M denotes Mary

r = Distance from center

Mary and Alex will have the equal displacements in equal interval of time as they are in uniform circular motion. So,

$\omega_A=\omega_M$

Tangential speed speed is given by

$\dfrac{v_M}{v_A}=\dfrac{r_M\omega_M}{r_A\omega_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{r_M}{r_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{8}{5}\\\Rightarrow \dfrac{v_M}{v_A}=1.6\\\Rightarrow v_M=1.6v_A$