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QveST [7]
3 years ago
6

NEED AN ANSWER FAST PLEASE 70 PTS

Physics
1 answer:
lara [203]3 years ago
4 0

Answer:

0.062

m/s

Explanation:

Quantities:

m

a

=

70

kg

m

o

=

26

kg

v

o

f

=

1

6

m/s

v

a

f

=

?

m/s

v

a

i

=

v

o

i

≡

v

i

m/s

p

a

i

+

p

o

i

=

p

a

f

+

p

o

f

m

a

v

a

i

+

m

o

v

o

i

=

m

a

v

a

f

+

m

o

v

o

f

(

m

a

+

m

o

)

v

i

=

m

a

v

a

f

+

m

o

v

o

f

(

70

+

26

)

v

i

=

(

70

)

v

a

f

+

(

26

)

(

1

6

)

96

v

i

=

70

v

a

f

+

13

3

We can assume that the initial speed of the astronaut is zero for the sake of argument. It says "floating in space", so maybe that actually means 0. The actual initial and final speeds are unimportant, because we are just looking for the change (or difference) in the two speeds.

−

13

3

=

70

v

a

f

v

a

f

=

−

0.062

m/s

This means that the astronauts speed will change by  

0.062

m/s and the astronaut will move in the opposite direction of the object (hence the minus sign).

Explanation:

braineist pls

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Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

H_m=1.65m

H_E=1.16307m

Explanation:

From the question we are told that

Mass of ball M=2kg

Length of string L= 2m

Wind force F=13.2N

Generally the equation for \angle \theta is mathematically given as

tan\theta=\frac{F}{mg}

\theta=tan^-^1\frac{F}{mg}

\theta=tan^-^1\frac{13.2}{2*2}

\theta=73.14\textdegree

Max angle =2*\theta= 2*73.14=>146.28\textdegree

Generally the equation for max Height H_m is mathematically given as

H_m=L(1-cos146.28)

H_m=0.9(1+0.8318)

H_m=1.65m

Generally the equation for Equilibrium Height H_E is mathematically given as

H_E=L(1-cos73.14)

H_E=0.9(1+0.2923)

H_E=1.16307m

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2 years ago
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