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iVinArrow [24]
3 years ago
11

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing, so the box will slide out if it is set moving.

The coefficients of kinetic and static friction between the box and the bed of the truck are 0.250 and 0.500, respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to 36.0m/s (≈ 80.5mph ) without causing the box to slide?
Physics
1 answer:
Anon25 [30]3 years ago
7 0

Answer:

t=7.33 s

Explanation:

According to Newton's second law:

\sum F=m*a

because we don't want the box to slide, the acceleration has to be zero.

\sum F=-F_{friction}+F_{truck}=0\\F_{truck}=F_{friction}\\F_{friction}=\µ*m*g=0.500*9.81*m\\F_{friction}=4.91*m

we know that:

F=m*a\\4.91m=m*a\\a=4.91m/s^2

Now having the acceleration, we can use the following formula.

v_f=a*t\\t=\frac{36.0m/s}{4.91m/s^2}\\\\t=7.33s

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dolphi86 [110]

Answer:

d) all of the above

Explanation:

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3 years ago
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Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a
Gnom [1K]

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F     (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg²  ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

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Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

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(ii) Firstly convert 54km/hr to m/s

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then apply the same formula as that in (i)

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-225 = -12.5s

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(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

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3 years ago
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Svetlanka [38]

Answer:

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KE = ½ (5 kg) (14 m/s)²

KE = 490 J

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