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Troyanec [42]
3 years ago
15

A rubber-wheeled 50 kg cart rolls down a 35 degree concrete incline. Coefficient of rolling friction between rubber and concrete

is = 0.02. (a) What is the cart's acceleration if rolling friction is neglected? (b) What is the cart's acceleration if rolling friction is included?
Physics
1 answer:
lapo4ka [179]3 years ago
8 0

Answer:

(a) 5.62 m/s²

(b) 5.46 m/s²

Explanation:

The given values are:

mass,

m = 50 kg

angle

= 35°

(a)

If friction neglected,

⇒ F_x=mgSin \theta=ma

⇒ a_x=gSin \theta

       =9.8 \ Sin35^{\circ}

       =5.62 \ m/s^2

(b)

If friction present,

⇒ F_x=mgCos \theta

⇒ F_x=mgSin \theta-\mu_r mgCos \theta

⇒ a=gSin \theta-\mu_rgCos \theta

      =9.8 \ Sin35^{\circ}-0.02\times 9.8 \ Cos30^{\circ}

      =5.46 \ m/s^2

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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
3 years ago
A truck traveling down the highway collides with a slower moving mosquito traveling in the same direction. Which of the followin
Ipatiy [6.2K]

Answer:

B. The truck and mosquito exert the same size force on each other.

Explanation:

Newton's third law (law of action-reaction) states that

"When an object A exerts a force (action) on an object B, then object B exerts an equal and opposite force (reaction) on object A"

In this case, we can call

object A = the truck

object B = the mosquito

Thereforce according to Newton's third law, the force exerted by the truck on the mosquito is equal in magnitude to the force exerted by the mosquito on the truck (and in opposite direction).

The reason for which the mosquito will experience much more damage is the fact that the mosquito's mass is much smaller than the truck's mass, and since the acceleration is inversely proportional to the mass:

a=\frac{F}{m}

the mosquito will experience a much larger deceleration than the truck, therefore much more damage.

6 0
3 years ago
What is the impulse needed to stop a 45-kg boy who is running at 6 m/s in 3 seconds?
Tju [1.3M]
Impulse = change of momentum
Impulse = 45 x 6 = 270 Ns
6 0
3 years ago
Assuming that each nucleus is roughlyspherical and that its mass is roughly equal to A (in atomic mass units {\rm u}), what is t
lara [203]

Answer:

ρ/ρ2 = 3 / R₀       the two densities are different

Explanation:

Density is defined as

       ρ = M / V

As the nucleus is spherical

       V = 4/3 π r³

Let's replace

      ρ = A / (4/3 π R₀³)

      ρ = ¾ A / π R₀³

b)

      ρ2 = F / area

The area of ​​a sphere is

     A = 4π R₀²

     ρ2 = F / 4π R₀²

     ρ2 = F / 4π R₀²

Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.

Let's look for the relationship of the two densities

     ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)

     ρ /ρ2 = 3 (A / F) (1 / R₀)

In this case it does not say that the nucleon number is A (F = A), the relationship is

     ρ/ρ2 = 3 / R₀

I see that the two densities are different

3 0
3 years ago
Many caterpillars construct cocoons from silk, one of the strongest naturally occurring materials known. Each thread is typicall
joja [24]

Answer:

a) N = 145,833,674.52174 = 1.458 × 10⁸ strands

b) diameter of single rope with the same effect = 2.415 cm

Explanation:

Hooke's law explains that stress is directly proportional to strain.

Stress ∝ Strain.

Stress = E × Strain

E = constant of proportionality = Young's Modulus = 4.0 ✕ 10⁹ N/m².

Stress = (Load/Total Cross sectional Area)

Load = a pair of 85 kg mountain climbers = 85 × 2 × 9.8 = 1666 N

Total Cross sectional Area = (Number of strands) × (Area of one strand) = A

Strain = (ΔL/L)

ΔL = 1.00 cm = 0.01 m

L = 11 m

Strain = (0.01/11) = 0.0009091

Stress = (Young's Modulus) × (Strain) = (4.0 ✕ 10⁹) × (0.0009091) = 3,636,363.64 N/m²

(Load/ total Area) = 3,636,363.64

Total area = (Load/3,636,363.64) = (1666/3,636,363.64) = 0.00045815 m²

Recall,

Total Cross sectional Area = (Number of strands) × (Area of one strand)

Area of one strand = (πd²/4)

diameter of one strand = 2 μm = (2×10⁻⁶) m

Area of one strand = (πd²/4)

= π × (2×10⁻⁶)² ÷ 4 = (3.142 × 10⁻¹²) m²

Total Cross sectional Area = (Number of strands) × (Area of one strand)

0.00045815 = N × (3.142 × 10⁻¹²)

N = 145,833,674.52174 = 1.458 × 10⁸ strands

b) If it was a single rope, the cross sectional Area would just be equal to the total cross sectional Area obtained in (a)

A = 0.00045815 m²

A = (πD²/4)

where D = diameter of the single rope

0.00045815 = (πD²/4)

D² = (4×0.00045815) ÷ π = 0.0005833347

D = 0.02415 m = 2.415 cm

Hope this Helps!!!

6 0
3 years ago
Read 2 more answers
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