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2 years ago

A satellite of 2,140 kg travels around the earth, of 5.97 - 1024

Physics

1 answer:

77julia77 [94]2 years ago

3 0

Answer:

8,961

Explanation:

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Gold has a density of 19300 kg/m³. Calculate the mass of 0.02 m³ of gold in kilograms.

hjlf

Answer:

Mass = 386 kg

Explanation:

Density = Mass / Volume

Mass = Density × Volume

Where D = 19300 kg/m³ , V = 0.02 m³

Putting the given in the above formula

Mass = 19300 × 0.02

Mass = 386 kg

7 0

2 years ago

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A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is f

Vinil7 [7]

Answer:

Solution is given in the attachments,below.

7 0

2 years ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

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2 years ago

Which transactions can increase the balance in a checking account? Check all that apply.

saul85 [17]

What all is the answers

4 0

2 years ago

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A projectile is launched at ground level with an initial speed of 53.0 m/s at an angle of 35.0° above the horizontal. It strikes

Gnom [1K]

Answer:

x = 108.5 m y = 45.37 m,

Explanation:

This is a projectile launching exercise, the horizontal distance it reaches is called the range and can be found by the expression

x = v₀ₓ t

the initial hairiness is

v₀ₓ = v₀ cos θ

x = v₀ cos θ t

let's calculate

x = 53 cos 35 2.50

x = 108.5 m

the height of the projectile can be calculated with

y =v₀ t - ½ g t²

y = vo sin tea t - ½ g t²

y = 53 sin 35 2.50 - ½ 9.8 2.5²

y = 45.37 m

7 0

2 years ago