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Eduardwww [97]
2 years ago
15

13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction

the ramp exerts on the sack and the rate of increase in the speed of sack at this instant.
y
*13–84. the 2-lb block is released from rest at a and slides down along the smooth cylindrical surface. if the attached spring has a stiffness k = 2 lb> ft, determine its unstretched length so that it does not allow the block to leave the surface until u = 60°.
a
prob. 13–82
prob. 13–84
2,fuuu 2,frur
Physics
1 answer:
marissa [1.9K]2 years ago
4 0
The second problem requires a figure to be answered. For the first problem
The acceleration of the sack is
1.5² - 0² = 2a(0.2)
a = 5.63 m/s2
The reaction of the ramp is
F = 8 kg (5.63 m/s2)
F = 45 N
Differentiate the kinematic equation involving time to get the rate of increase of the velocity.
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A force of 5000 n is applied outwardly to each end of a 5.0-m long rod with a radius of 34.0 mm and a young's modulus of 125 x 1
Blizzard [7]
Por definicion tenemos que
 (F/A) = E(∆/0)
 Sustituyendo los valores tenemos y despejando ∆:
 ∆ = (F/(πr2 × E))*0
 (5000×5)/(3.14×(34×10^−2)^2×(125×10^8))
 5.5×10^−6 m
 
4 0
2 years ago
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Consider a 2100-kg car cruising at constant speed of 70 km/h. Now the car starts to pass another car by accelerating to 110 km/h
Liula [17]

Answer: 51841.5 Watts

Explanation: Using the kinematic equation for the final velocity for a constant acceleration we have:

Vf=Vi+a*t

replacing the values the results is

a=(Vf-Vi)/t= (30.55 m/s-19.44 m/s)/5s= 2.22 m/s^2

Remenber that to convert the speed in Km/h to m/s we have to multiplier by the factor 0.277.

Finally to calculate the increment of power get the final velocity we have to use Neton second law to determine the Force applied to the car.

F=m* a=2100 Kg* 2.22 m/s^2= 4666.2 N

Then increment  power to accelerate is given by:

ΔPower= Force* Δ velocity= 4666.2 N* 11,11 m/s= 51841.5 Watts

6 0
2 years ago
A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts
likoan [24]
Refer to the figure shown below.

m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g,  the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where  =  kinetic coefficient of friction
T = the force in the hitch between the car and trailer.

Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N

Answer: The resistive force is 910 N

Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N

Answer: The force in the hitch is 1113 N

3 0
2 years ago
A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold
LUCKY_DIMON [66]

Answer:

(a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

Q=\dfrac{V}{t}

Put the value into the formula

Q=\dfrac{405}{90}

Q=4.5\ cm^3/s

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

Q=kiA

k=\dfrac{Q}{iA}

k=\dfrac{Ql}{Ah}

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}

k=8.6\times10^{-3}\ cm/s

The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

v=ki

v=\dfrac{kh}{l}

Put the value into the formula

v=\dfrac{8.6\times10^{-3}\times38}{17.5}

v=0.0187\ cm/s

The discharge velocity during the test is 0.0187 cm/s.

(b). We need to calculate the volume of solid in the ample

Using formula of volume

V_{s}=\dfrac{M_{s}}{V_{s}}

Put the value into the formula

V_{s}=\dfrac{4950\times10^{-3}}{2710}

V_{s}=1826.56\ cm^3

We need to calculate the volume of the soil specimen

Using formula of volume

V=A\times L

Put the value into the formula

V=\dfrac{\pi(17.5)^2}{4}\times17.5

V=4209.24\ cm^3

We need to calculate the volume of the voids

V_{v}=V-V_{s}

Put the value into the formula

V_{v}=4209.24-1826.56

V_{v}=2382.68\ cm^3

We need to calculate the seepage velocity

Using formula of velocity

Av=A_{v}v_{s}

v_{s}=\dfrac{Av}{A_{v}}

v_{s}=\dfrac{V}{V_{v}}\times v

Put the value into the formula

v_{s}=\dfrac{4209.24}{2382.68}\times0.0187

v_{s}=0.0330\ cm/s

The seepage velocity is 0.0330 cm/s.

Hence, (a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

8 0
2 years ago
"Filip, a finance executive, loves playing with numbers and hates sports. His manager, Daphne, asks him to represent their compa
avanturin [10]

Answer: Extrinsic motivation

Explanation:

This is an external motivation such as reward, desire to impress someone, prestige, etc which drives someone towards achieving a goal.

3 0
2 years ago
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