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Korvikt [17]
2 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 )
Physics
1 answer:
geniusboy [140]2 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 years ago
A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

4 0
3 years ago
HELP WILL GIVE BRAINLIEST IF CORRECT
anastassius [24]
It’s B i literally jus learned this
6 0
3 years ago
Read 2 more answers
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