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Komok [63]
3 years ago
13

A parcel of mass 10 kg rests on a lorry.When the lorry accelerates at 1.5 m/s^2,the parcel is just about to slide backwards. Wha

t is the co-efficient of friction between the parcel and the lorry?​
Physics
1 answer:
Mademuasel [1]3 years ago
5 0

Answer:

The coefficient of static friction,  μ = 0.153

Explanation:

Given data,

The mass of the parcel, m = 20 kg

The acceleration of the lorry, a = 1.5 m/s²

The normal force acting on the parcel,

                                       f = m x g

                                         = 20 x 9.8

                                        = 196 N

The static frictional force acting on the lorry,

                                   F = 20 x 1.5

                                      = 30 N

The coefficient of static is

                     μ = F / f

                        = 30 / 196

                        = 0.153

Hence, the coefficient of static friction,  μ = 0.153

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A swimmer is swimming to the left with a speed of 1.0 m/s when she starts to speed up with constant acceleration. The swimmer re
kolezko [41]

Answer:

Correct answer: t = 2.86 seconds

Explanation:

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V² - V₀² = 2 a d    

where V is the final velocity (speed), V₀ the initial velocity (speed),

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We will calculate the acceleration from this formula

a = (V² - V₀²) / (2 d) = (2.5² - 1²) / (2 · 5) = (6.25 - 1) / 10 = 5.25 / 10

a = 0.525 m/s²

then we use this formula

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God is with you!!!

6 0
3 years ago
A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by
lana66690 [7]

Answer:

Angle: 48.19^o

Explanation:

<u>Two-Dimension Motion</u>

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

v_b\ cos\alpha

where v_b is the speed of the boy in still water and \alpha is the angle respect to the shoreline. If the river flows at speed v_s, we now set

v_b\ cos\alpha=v_s

\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}

\alpha=48.19^o

8 0
3 years ago
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