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belka [17]
3 years ago
14

A man is marooned at rest on level frictionless ice. In desperation, he hurls his shoe to the right at 15 m/s. If the man weighs

720 N and the shoe weighs 4.0 N, the man moves to the left at approximately______.A) 0B) 2.1 x 102 m/s.C) 8.3 x 102 m/s.D) 15 m/s.E) 2.7 x 10-3 m/s 19.
Physics
1 answer:
adell [148]3 years ago
3 0

Answer:

A 0.083 m/s approx 0

Explanation:

mass of the man is 720 N, the mass of the shoe is 4 N, and the man and the shoe were initially at rest. After throwing the shoe, the shoe had a velocity of 15 m/s. Using conservation of momentum:

since the man and the shoe were initially at rest, their initial momentum is zero

0 = M1V1 + M2V2 where M1 is the mass of the shoe ( 4 / 9.81), V1 is the velocity of the shoe 15m/s, M2 is mass of the man ( 720 / 9.81), V2 is the velocity of the man

MAKE V2 subject of the formula

- M1V1 = M2V2

- M1V1 / M2

substitute the values into the equation

- ((4/9.8) × 15) /( 720 / 9.81) = V2

V2 = - 0.0833 m/s approx 0  

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List out some use of atmospheric pressure​
inna [77]

Answer:

<h2>1)Straw : When a man sucks up the fluid from straw the pressure inside is moderately low and accordingly atmospheric pressure outside powers up the fluid into straw. 2)Vacuum cleaner : When a vacuum cleaner is exchanged on, the pressure inside tumble off because of air inside.</h2>
4 0
3 years ago
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At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.855 m/s2 if the acceleration du
natulia [17]
Given: Normal pull of gravity g = 9.8 m/s²; 

 g = 0.855 m/s²  (at a certain distance)

Universal gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²

Mass of the Earth Me = 5.98 x 10²⁴ Kg

Radius r = ?

g = GMe/r²

r = √GMe/g

r = √(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/(0.855 m/s²)

r = 2.16 x 10⁷ m or 

r =  21,610 Km





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5 0
3 years ago
A pulley of radius 8.0 cm is connected to a motor that rotates at a rate 7000 rad s-1 and then decelerate uniformly at a rate of
zlopas [31]

Answer:

(a) α = - 1000 rad/s²

Negative sign represents deceleration.

(b) θ = 3581 rotations

(c) L = 1800 m

(d) a = - 80 m/s²  

Explanation:

(a)

using First equation of motion for angular motion:

ωf = ωi + αt

where,

ωf = Final Angular Speed = 2000 rad/s

ωi = Initial Angular Speed = 7000 rad/s

α = Angular Acceleration = ?

t = time = 5 s

Therefore,

2000 rad/s = 7000 rad/s + α(5s)

α = (2000 rad/s - 7000 rad/s)/5 s

<u>α = - 1000 rad/s²</u>

<u>Negative sign represents deceleration.</u>

(b)

Using second equation of motion:

θ = ωi t + (1/2)αt²

where,

θ = No. of Rotations = ?

Therefore,

θ = (7000 rad/s)(5 s) + (1/2)(- 1000 rad/s²)(5 s)²

θ = 35000 rad - 12500 rad

θ = (22500 rad)(1 rotation/2π rad)

<u>θ = 3581 rotations</u>

(c)

Length of String = L = (Circumference of Pulley)(θ)

L = [2π(0.08 m)][3581 rotations]

<u>L = 1800 m</u>

<u></u>

(d)

Tangential Acceleration = a = rα

a = (0.08 m)(-1000 rad/s²)

<u>a = - 80 m/s²</u>

4 0
3 years ago
[9]
ad-work [718]

Answer:

9.66 m/s 15° with +y

2.59 m/s 75° with +y

Explanation:

Momentum is conserved in the y direction.

mu₁ + mu₂ = mv₁ + mv₂

u₁ + u₂ = v₁ + v₂

10 m/s + 0 m/s = v₁ cos 15° + v₂ cos 75°

10 = v₁ cos 15° + v₂ cos 75°

Momentum is conserved in the x direction.

mu₁ + mu₂ = mv₁ + mv₂

u₁ + u₂ = v₁ + v₂

0 m/s + 0 m/s = v₁ sin 15° − v₂ sin 75°

0 = v₁ sin 15° − v₂ sin 75°

v₁ sin 15° = v₂ sin 75°

v₂ = v₁ sin 15° / sin 75°

Substitute.

10 = v₁ cos 15° + (v₁ sin 15° / sin 75°) cos 75°

10 = v₁ cos 15° + v₁ sin 15° / tan 75°

10 = v₁ (cos 15° + sin 15° / tan 75°)

v₁ ≈ 9.66 m/s

v₂ ≈ 2.59 m/s

7 0
3 years ago
14 km = m on conversions using a ladder method​
lozanna [386]
Using the metric units kilometer (km) and meter (m), as seen in the picture of the ladder method example. So, the answer would be 14km = 14000m

6 0
3 years ago
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