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Natasha_Volkova [10]
3 years ago
6

A 500 kg table with 4 legs rests on solid flat ground. I place 3 books on the center of the table with masses of 10 kg, 20 kg an

d 30 kg respectively. What is the normal force between the ground and one leg of the table?
Physics
1 answer:
BARSIC [14]3 years ago
6 0

Answer:

1,373.4 N

Explanation:

The mass of the table acts at the centre in addition to the books since that is the centre of gravity of the table.

Mass of books will be 10kg+20kg+30kg=60 kg

Total mass of table and books will be 500kg+60kg=560 kg

This mass is evenly distributed into the four legs hence 560kg/4 legs=140 kg per leg

Force is product of mass and acceleration due to gravity hence F=gm

Taking g as 9.81 m/s2 then

F=140*9.81=1,373.4 N

Therefore, rhe normal force is equivalent to 1,373.4 N

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Most ocean waves obtain their energy and motion from _____. the moon's gravitational attraction the sun plate movement the wind
IrinaVladis [17]
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Have a nice day!


3 0
3 years ago
Is a truck braking to avoid an accident moving at constant
frutty [35]

Answer:

its not moving at a constant velocity because it is slowing down

Explanation:

3 0
2 years ago
Read 2 more answers
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

5 0
2 years ago
A spacecraft is in a circular Earth orbit at an altitude of 6000 km . By how much will its altitude decrease if it moves to a ne
blagie [28]

Answer:

a) 2148 km = 2150 km

b) 840 km

Explanation:

The force keeping the satellite in circular motion is the force given by Netwon's gravitational law

Centripetal force = (mv²/r)

Force due to Newton's law of gravitation = (GMm/r²)

where m = mass of satellite

M = mass of the earth

G = Gravitational constant

v = velocity of the satellite

r = radius of circular orbit

(mv²/r) = (GMm/r²)

v² = (GM/r)

Meaning that the square of the velocity of orbit is inversely proportional to the radius of circular orbit. (Since G and M are constants)

v² = (k/r)

when v = v₀, r = 6000 + 6378 = 12,378 km (the radius of orbit = 6000 km + radius of the earth)

v₀² = (k/12,378)

K = 12378v₀²

When the velocity increases by 10%, v₁ = 1.1v₀, the square of the new velocity = (1.1v₀)² = 1.21v₀² and the new radius of orbit = r₁

1.21v₀² = (k/r₁)

r₁ = (k/1.21v₀²)

Recall, k = 12378v₀²

r₁ = 12378v₀² ÷ 1.21v₀² = 10,229.75 km

10,229.75 km = (10,229.75 - 6378) km altitude above the Earth's surface

New altitude of orbit = 3851.75 km

Decrease in altitude = 6000 - 3851.75 = 2148 km

b) The period of orbit is related to the radius of orbit through Kepler's Law

T² ∝ R³

T² = kR³

When the period of orbit is T₀, Radius of orbit = R₀ = (6000 + 6378) = 12378 km (Earth's radius = 6378 km)

T₀² = kR₀³

T₀² = k(12378)³

k = (T₀²) ÷ (12378)³

When the period reduces by 10%, T₁ = 0.90T₀ and the new radius of orbit = R₁

T₁² = kR₁³

(0.90T₀)² = kR₁³

0.81T₀² = kR₁³

R₁³ = (0.81T₀²) ÷ k

Recall, k = (T₀²)/(12378)³

R₁³ = (0.81T₀²) ÷ [(T₀²)/(12378)³]

R₁³ = 1,536,160,005,663.1

R₁ = ∛(1,536,160,005,663.1) = 11,538.4 km

New Altitude = R₁ - (Radius of the Earth)

= 11,538.4 - 6378 = 5160.4 km

Decrease in altitude = 6000 - 5160.4 = 839.6 km = 840 km

Hope this Helps!!!

3 0
3 years ago
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