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Marta_Voda [28]

3 years ago

15

During an in-class demonstration of momentum change and impulse, Mr. H asks Jerome (102 kg) and Michael (98 kg) to sit on a larg

e 14-kg skate cart. Mr. H asks Suzie (44 kg) to sit on a second 14-kg skate cart. The two carts are placed on low friction boards in the hallway. Jerome pushes off of Suzie's cart. Measurements are made to determine that Suzie's cart acquired a post-impulse speed of 9.6 m/s. Determine the expected recoil speed of Jerome and Michael's cart.

1 answer:

Brums [2.3K]3 years ago

6 0

Answer:

2.6 m/s

Explanation:

mass of Jerome (Mj) = 102 kg

mass of Micheal (Mm) = 98 kg

mass of each cart (Mc) = 14 kg

recoil speed of Jerone and Micheal after push (Vi) = ?

mass of Suzie (Ms) = 44 kg

Suzie's speed after push (Vs) = 9.6 m/s

From the conservation of momentum, initial momentum = final momentum

Since they all are initially at rest the initial momentum is 0 and hence the final momentum is also 0.

Therefore for the final momentum we have

(Mj + Mm + Mc) Vi - (Ms + Mc) Vs = 0 ( the negative sign is because they move in opposite directions)

((102 + 98 + 14) x Vi ) - ((44 + 14) x 9.6) = 0

(214 x Vi) - (58 x 9.6) = 0

214 Vi = 556.8

Vi = 556.8 / 214 = 2.6 m/s

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3 years ago

A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

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Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$

vertical component: $F_{y}=Fsen\theta$

but from the given information we know that $F_{y}=12N$

so, equation these two $F_{y}=Fsen\theta$ and $F_{y}=12N$

$Fsen\theta =12N$

and we know the force $F=45N$ , thus:

$45sen\theta=12$

now we clear for $\theta$

$sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466$

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

$F_{x}=Fcos\theta$

$F_{x}=45cos(15.466)\\F_{x}=43.37N$

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

$W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J$

the work done is W=173.48J

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3 years ago

The charges that are free to move in a metallic conducting wire and that are responsible for the flow of electric current are

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Answer:Negatively charged particle called Free Electrons

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Current is the flow of charged particles called Free electrons. Electrons are free to move from one atom to another and we call them a sea of de-localized electrons. In absence of any externally applied emf, these electrons are randomly moving but with the onset of emf, these electrons flow in a particular direction.

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3 years ago

A cyclist going downhill is accelerating at 1. 2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is

mel-nik [20]

Answer:

$\boxed {\boxed {\sf v_i= 4 \ m/s}}$

Explanation:

We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.

$v_f= v_i + at$

The cyclist is acceleration at 1.2 meters per second squared. After 10 seconds, the velocity is 16 meters per second.

$v_f$ = 16 m/s
a= 1.2 m/s²
t= 10 s

Substitute the values into the formula.

$16 \ m/s = v_i + (1.2 \ m/s^2)(10 \ s)$

Multiply.

$16 \ m/s = v_i + (1.2 \ m/s^2 * 10 \ s)$

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We are solving for the initial velocity, so we must isolate the variable $v_i$ . Subtract 12 meters per second from both sides of the equation.

$16 \ m/s - 12 \ m/s = v_i + 12 \ m/s -12 \ m/s$

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The cyclist's initial velocity is <u>4 meters per second.</u>

6 0

3 years ago

A 4 kW vacuum cleaner is powered by an electric motor whose efficiency is 90%. (Note that the electric motor delivers 4 W of net

RoseWind [281]

Answer: $3.6\ kW$

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3 years ago