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finlep [7]
2 years ago
15

In Space, an astronaut releases a wrench from his hand. The wrench has a mass of 4 grams and is traveling with a velocity of -15

m/s. The Astronaut’s mass is 70kg. What is his Velocity?
Physics
1 answer:
Cloud [144]2 years ago
8 0

Answer: -8.5 × 10^-4 m/s

Explanation:

The following can be deduced from the question:

m₁ = Mass of a wrench = 4 g = 0.004 kg

v₁ =Speed of wrench = -15 m/s

m₂ = Mass of the Astronaut = 70 kg

The velocity will be calculated as:

m₁v₁ = m₂v₂

v₂ = [0.004 × (-15)] / 70

= -8.5 × 10^-4 m/s

Therefore, the velocity is -8.5 × 10^-4 m/s

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Oil tends to float on water because the density of oil is _____ the density of water.
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Oil is less dense than water so it tends to float on the top of the water. Hope this Helps!
7 0
3 years ago
How much power is required to lift a 60 kg bed to the top of a 3 meter flight of stairs in 10 seconds?
Zigmanuir [339]

Answer:

176.58Watts

Explanation:

Power= work done /time

Where mass(m)=60kg

Height (h) =3m

Time(s)=10s

Force of gravity = 9.81m/s^2

Power=mgh/t

Power= (60kg) * (9.81m/s^2) * (3m)/10s

Power= 176.58Watts

8 0
3 years ago
How do I measure the drag of a paper airplane?
4vir4ik [10]

Answer:

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6 0
2 years ago
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve
Vlad1618 [11]

Answer:

v=0.9833\ c

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

$\text{Density} = \frac{m}{lwh}$

Given :

Side,  b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 kg/m^3

So,

$8100=\frac{3.3}{l \times 0.13 \times 0.13}$

$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$

l = 0.024 m

Then for relativistic length contraction,

$l= l' \sqrt{1-\frac{v^2}{c^2}}$

$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

$0.184=  \sqrt{1-\frac{v^2}{c^2}}$

$0.033=  1-\frac{v^2}{c^2}}$

$\frac{v^2}{c^2}= 0.967$

$\frac{v}{c}=0.9833$

v=0.9833\ c

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

3 0
3 years ago
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What
AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, \alpha=3.3 rad/s^2

Diameter of the wheel, d=21 cm

Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

1m=100 cm

Magnitude of total linear acceleration, a=1.7 m/s^2

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,a_t=\alpha r

a_t=3.3\times \frac{21\times 10^{-2}}{2}

a_t=34.65\times 10^{-2}m/s^2

Radial acceleration,a_r=\frac{v^2}{r}

We know that

a=\sqrt{a^2_t+a^2_r}

Using the formula

1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

Squaring on both sides

we get

2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

v^4=r^2\times 2.7699

v^4=(10.5\times 10^{-2})^2\times 2.7699

v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

v=0.418 m/s

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0
3 years ago
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