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djyliett [7]
3 years ago
11

How much gravitational force do two lead balls with a mass of 8 kilograms, the centers of mass of which are 17 cm apart, affect

each other?
Physics
1 answer:
Arisa [49]3 years ago
7 0

Answer:

1.48×10⁻⁷ Newtons

Explanation:

From the question,

According to newton's law of universal gravitation.

F = Gmm'/r²........................ Equation 1

F = gravitational force, G = gravitational constant, m = mass of the first ball, m' = mass of the second ball, r = distance between the balls.

Given: m = m' = 8 kg, r = 17 cm = 0.17 m,

Constant : G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

F =  (6.67×10⁻¹¹×8×8)/(0.17²)

F = 1.48×10⁻⁷ N

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What kind of image does the lens in a camera produce? real and upside down real and right-side up virtual and upside down virtua
Vlada [557]

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Explanation:

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7 0
2 years ago
Read 2 more answers
1) A uniform wooden beam, with mass of 120 and length L = 4 m, is supported as illustrated in the figure. If the static friction
Kobotan [32]

Answer:

1(a) 55.0°

1(b) 58.3°

2(a) 10.2 N

2(b) 2.61 m/s²

3(a) 76.7°

3(b) 12.8 m/s

3(c) 3.41 s

3(d) 21.8 m/s

3(e) 18.5 m

4(a) 7.35 m/s²

4(b) 31.3 m/s²

4(c) 12.8 m/s²

Explanation:

1) Draw a free body diagram on the beam.  There are five forces:

Weight force mg pulling down at the center of the beam,

Normal force Na pushing up at point A,

Friction force Na μa pushing left at point A,

Normal force Nb pushing perpendicular to the incline at point B,

Friction force Nb μb pushing up the incline at point B.

There are 3 unknown variables: Na, Nb, and θ.  So we're going to need 3 equations.

Sum of forces in the x direction:

∑F = ma

-Na μa + Nb sin φ − Nb μb cos φ = 0

Nb (sin φ − μb cos φ) = Na μa

Nb / Na = μa / (sin φ − μb cos φ)

Sum of forces in the y direction:

∑F = ma

Na + Nb cos φ + Nb μb sin φ − mg = 0

Na = mg − Nb (cos φ + μb sin φ)

Sum of torques about point B:

∑τ = Iα

-mg (L/2) cos θ + Na L cos θ − Na μa L sin θ = 0

mg (L/2) cos θ = Na L cos θ − Na μa L sin θ

mg cos θ = 2 Na cos θ − 2 Na μa sin θ

mg = 2 Na − 2 Na μa tan θ

Substitute:

Na = 2 Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)

0 = Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)

Na (1 − 2 μa tan θ) = Nb (cos φ + μb sin φ)

1 − 2 μa tan θ = (Nb / Na) (cos φ + μb sin φ)

2 μa tan θ = 1 − (Nb / Na) (cos φ + μb sin φ)

Substitute again:

2 μa tan θ = 1 − [μa / (sin φ − μb cos φ)] (cos φ + μb sin φ)

tan θ = 1/(2 μa) − (cos φ + μb sin φ) / (2 sin φ − 2 μb cos φ)

a) If φ = 70°, then θ = 55.0°.

b) If φ = 90°, then θ = 58.3°.

2) Draw a free body diagram of each mass.  For each mass, there are four forces.  For mass A:

Weight force Ma g pulling down,

Normal force Na pushing perpendicular to the incline,

Friction force Na μa pushing parallel down the incline,

Tension force T pulling parallel up the incline.

For mass B:

Weight force Mb g pulling down,

Normal force Nb pushing perpendicular to the incline,

Friction force Nb μb pushing parallel up the incline,

Tension force T pulling up the incline.

There are four unknown variables: Na, Nb, T, and a.  So we'll need four equations.

Sum of forces on A in the perpendicular direction:

∑F = ma

Na − Ma g cos θ = 0

Na = Ma g cos θ

Sum of forces on A up the incline:

∑F = ma

T − Na μa − Ma g sin θ = Ma a

T − Ma g cos θ μa − Ma g sin θ = Ma a

Sum of forces on B in the perpendicular direction:

∑F = ma

Nb − Mb g cos φ = 0

Nb = Mb g cos φ

Sum of forces on B down the incline:

∑F = ma

-T − Nb μb + Mb g sin φ = Mb a

-T − Mb g cos φ μb + Mb g sin φ = Mb a

Add together to eliminate T:

-Ma g cos θ μa − Ma g sin θ − Mb g cos φ μb + Mb g sin φ = Ma a + Mb a

g (-Ma (cos θ μa + sin θ) − Mb (cos φ μb − sin φ)) = (Ma + Mb) a

a = -g (Ma (cos θ μa + sin θ) + Mb (cos φ μb − sin φ)) / (Ma + Mb)

a = 2.61 m/s²

Plug into either equation to find T.

T = 10.2 N

3i) Given:

Δx = 3.7 m

vᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

t = 1.25 s

Find: v₀ₓ, v₀ᵧ

Δx = v₀ₓ t + ½ aₓ t²

3.7 m = v₀ₓ (1.25 s) + ½ (0 m/s²) (1.25 s)²

v₀ₓ = 2.96 m/s

vᵧ = aᵧt + v₀ᵧ

0 m/s = (-10 m/s²) (1.25 s) + v₀ᵧ

v₀ᵧ = 12.5 m/s

a) tan θ = v₀ᵧ / v₀ₓ

θ = 76.7°

b) v₀² = v₀ₓ² + v₀ᵧ²

v₀ = 12.8 m/s

3ii) Given:

Δx = D cos 57°

Δy = -D sin 57°

v₀ₓ = 2.96 m/s

v₀ᵧ = 12.5 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

c) Find t

Δx = v₀ₓ t + ½ aₓ t²

D cos 57° = (2.96 m/s) t + ½ (0 m/s²) t²

D cos 57° = 2.96t

Δy = v₀ᵧ t + ½ aᵧ t²

-D sin 57° = (12.5 m/s) t + ½ (-10 m/s²) t²

-D sin 57° = 12.5t − 5t²

Divide:

-tan 57° = (12.5t − 5t²) / 2.96t

-4.558t = 12.5t − 5t²

0 = 17.058t  − 5t²

t = 3.41 s

d) Find v

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (3.41 s) + 2.96 m/s

vₓ = 2.96 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (-10 m/s²) (3.41 s) + 12.5 m/s

vᵧ = -21.6 m/s

v² = vₓ² + vᵧ²

v = 21.8 m/s

e) Find D.

D cos 57° = 2.96t

D = 18.5 m

4) Given:

R = 90 m

d = 140 m

v₀ = 0 m/s

at = 0.7t m/s²

The distance to the opposite side of the curve is:

140 m + (90 m) (π/2) = 281 m

a) Find Δx and v if t = 10.5 s.

at = 0.7t

Integrate:

vt = 0.35t² + v₀

vt = 0.35 (10.5)²

vt = 38.6 m/s

Integrate again:

Δx = 0.1167 t³ + v₀ t + x₀

Δx = 0.1167 (10.5)³

Δx = 135 m

The car has not yet reached the curve, so the acceleration is purely tangential.

at = 0.7 (10.5)

at = 7.35 m/s²

b) Find Δx and v if t = 12.2 s.

at = 0.7t

Integrate:

vt = 0.35t² + v₀

vt = 0.35 (12.2)²

vt = 52.1 m/s

Integrate again:

Δx = 0.1167 t³ + v₀ t + x₀

Δx = 0.1167 (12.2)³

Δx = 212 m

The car is in the curve, so it has both tangential and centripetal accelerations.

at = 0.7 (12.2)

at = 8.54 m/s²

ac = v² / r

ac = (52.1 m/s)² / (90 m)

ac = 30.2 m/s²

a² = at² + ac²

a = 31.3 m/s²

c) Given:

Δx = 187 m

v₀ = 0 m/s

at = 3 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (3 m/s²) (187 m)

v = 33.5 m/s

ac = v² / r

ac = (33.5 m/s)² / 90 m

ac = 12.5 m/s²

a² = at² + ac²

a = 12.8 m/s²

5 0
3 years ago
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(
NeX [460]

Answer:

The transverse displacement is   y(1.51 , 0.150) = 0.055 m    

Explanation:

 From the question we are told that

     The generally equation for the mechanical wave is

                    y(x,t) = Acos (kx -wt)

     The speed of the transverse wave is v = 8.25 \ m/s

     The amplitude of the transverse wave is A = 5.50 *10^{-2} m

     The wavelength of the transverse wave is \lambda = 0540 m

      At t= 0.150s , x = 1.51 m

 The angular frequency of the wave is mathematically represented as

          w = vk

Substituting values  

         w = 8.25 * 11.64

        w = 96.03 \ rad/s

The propagation constant k is mathematically represented as

                  k = \frac{2 \pi}{\lambda}

Substituting values

                  k = \frac{2 * 3.142}{0. 540}

                   k =11.64 m^{-1}

Substituting values into the equation for mechanical waves

      y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03  * 0.150))

       y(1.51 , 0.150) = 0.055 m    

4 0
3 years ago
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