When a mass of 8 kg is located on the Earth's surface, the magnitude of each of
the gravitational forces attracting the mass and the Earth toward each other is

   (mass) x (acceleration of gravity on Earth) =

      (8.0 kg) x (9.81 m/s²) =

      78.48 kg-m/s² = <em>78.48 newtons </em>(about 17 pounds 10.2 ounces)

6 0

4 years ago

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An atom of tungsten (W) has 74 protons. The atomic mass of tungsten is 183.84.

ASHA 777 [7]

Answer:

110

Explanation:

to find neutrons we have to subtract the atomic number from the atomic mass.

atomic mass = 183.84

atomic no = 74

183.84 - 74 = 109.84

round off = 110

3 0

4 years ago

A mass of 0.75 kilograms is attached to a spring/mass oscillator. A force of 5 newtons is required to stretch the spring 0.5 met

zlopas [31]

Answer:

b > 66.41 kg/s

Explanation:

The spring force F = -kx, where k = spring constant, the damping force f = -bv. The net force F' = F + f

F + f = ma

-kx - bv = ma

-kx -bdx/dt = md²x/dt².

Re-arranging the equation, we have

So, md²x/dt² + bdx/dt + kx = 0

Dividing through by m, we have

d²x/dt² + (b/m)dx/dt + (k/m)x = 0

This is a second-order differential equation. The characteristic equation is thus,

D² + (b/m)D + (k/m) = 0

Using the quadratic formula, we find D.

$D = \frac{-(b/m) +/- \sqrt{(b/m)^{2} - 4k/m} }{2}$

For an overdamped system,

$(b/m)^{2} - 4k/m} > 0$

$(b/m)^{2} > 4k/m}\\(b/m) > \sqrt{4k/m}} \\(b/m) > 2\sqrt{k/m}} \\b > 2\sqrt{km}}$

Now, k = F/x. Since the weight of the object causes the spring to stretch a distance of 0.5 m, k = mg/x where m = mass of object = 0.75 kg, g = 9.8 m/s² and x = x₀ =0.5 m.

Substituting k = mg/x into the inequality for b, we have

b > 2√{(mg/x₀)m}

b > 2√{(m²g/x₀)}

b > 2m√{g/x₀)}

b > 2 × 0.75 kg√{9.8 m/s²/0.5 m)}

b > 1.5 kg√{19.6/s²)}

b > 1.5 kg × 4.427/s

b > 66.41 kg/s

6 0

3 years ago

During constant linear acceleration, what changes uniformly? A. acceleration B. distance C. displacement D. velocity E. all of t

iogann1982 [59]

Answer:D-velocity

Explanation:

Given

Constant linear acceleration i.e. acceleration is constant

a=constant

and we know $\frac{\mathrm{d}V}{\mathrm{d} t}=a$

Thus velocity is changing uniformly to give constant acceleration

While displacement and distance are function of time thus they are not changing uniformly.

8 0

3 years ago