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Reika [66]
3 years ago
9

Two football players are pushing a 60kg blocking sled across the field at a constant speed of 2.0 m/s. The coefficient of kineti

c friction between the grass and the sled is 0.30. Once they stop pushing, how far will the sled slide before coming to rest
Physics
1 answer:
sertanlavr [38]3 years ago
8 0

Answer:

The sled slides 0.68m before rest.

Explanation:

Vi= 2 m/s

m= 60kg

μ= 0.3

N= m * g = 60kg * 9.8 m/s²= 588 N

Fr= μ * N

Fr= 176.4 N

a= Fr/m

a= -2.94m/s²

t= Vi/a

t= 0.68 seg

d= Vi*t - a*t²/2

d= 0.68m

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D is the answer most true
6 0
3 years ago
Determine the average normal stress developed in rod AB if the load has a mass of 55 kg . The diameter of rod AB is 8 mm.
Fudgin [204]

Answer:

The normal stress is 10.7[MPa]

Explanation:

The normal stress can be calculated with the following equation:

S_{norm} =\frac{F}{A} \\where:\\F= force [Newtons]\\A=area [m^2]\\S_{norm} = Normal stress [\frac{N}{m^{2} }] or [Pa]

The area of the rod can be calculated using the equation:

A=\frac{\pi }{4}*d^{2}  \\d=8[mm]=0.008[m]\\A=\frac{\pi }{4}*(0.008)^{2}  \\A=5.02*10^{-5} [m^{2} ]

The force is the result of the mass multiplied by the gravity.

F=55[kg]*9.81[m/s^{2} ] = 539.6[N]\\\\S_{norm} = 539.6/5.02*10^{-5} \\S_{norm} = 10.7*10^{6}[Pa] = 10.7[MPa]

3 0
3 years ago
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel
insens350 [35]

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

5 0
3 years ago
What causes a nucleus to become radioactive?
timurjin [86]

Answer:

Answer the strong nuclear force attracts and binds protons and makes the nucleus unstable

Explanation:

4 0
3 years ago
ad for the distance from the sun tothe earth is 1.5 x 10"m. how long does it take for light from ths sun to reach the eath? give
Alexandra [31]

Answer:

About 8.3 minutes

Explanation:

Use the formula for velocity as the distance covered by the light divided the time it takes: [tex]velocity=\frac{distance}{time}[/tex]

Use the information about the speed of light in vacuum: 300000000 \frac{m}{s} = 3*10^{8} \frac{m}{s}

and the information you are given regarding the distance between Sun and Earth: 1.5 * 10^{11} m

to solve the first velocity equation for the unknown time "t":

velocity=\frac{distance}{time} \\3*10^8\frac{m}{s} =\frac{1.5*10^11 m}{t} \\t=\frac{1.5*10^11 }{3*10^8} s= 500 s

we can convert second into minutes by dividing by 60: 500 s = 500/60 minutes = 8.3333... minutes

5 0
3 years ago
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