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Komok [63]
3 years ago
10

Part B: Quantitative

Physics
1 answer:
Feliz [49]3 years ago
4 0

Answer:

(a)

\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s

(b)

5600 N

Explanation:

Change in velocity, \triangle v is expressed as

\triangle v= v_f -v_i

Where v represent the velocity and subscripts f and i represent final and initial respectively. Given that the ball finally comes to rest, its final velocity is taken as 0. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

\triangle v=0-8=-8\ m/s

To find the change in momentum, m\triangle v then we substitute 7 kg for m and -8 m/s for \triangle v therefore \triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s

(b)

The impact force, F is equivalent to ma where m is the mass of the ball and a is the acceleration. Here, acceleration is given by dividing the change in velocity by time ie

a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}

Substituting t with 0.05 s then a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.01}=-800 m/s^{2}

Since F=ma then substituting m with 7 Kg we get that F=7*-800=-5600 N

Therefore, the impact force is equivalent to 5600 N

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↠Momentum = 4 × 3 kgm/s

↠Momentum = 12 kgm/s

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Read 2 more answers
A motorcycle, which has an initial linear speed of 8.0 m/s, decelerates to a speed of 2.2 m/s in 4.1 s. Each wheel has a radius
SVETLANKA909090 [29]

Answer:

Explanation:

Given

Initial linear speed v_1=8 m/s

initial angular velocity \omega _1=\frac{v_1}{r}=\frac{8}{0.6}=13.33 rad/s

Speed after 4.1 s is  v_2=2.2 m/s

\omega _2=\frac{2.2}{0.6}=3.66 rad/s

using \omega _2=\omega _1+\alpha t

where \alphais angular acceleration

3.66=13.33+\alpha \cdot 4.1

\alpha =-2.37 rad/s^2 i.e. clockwise

(b)angular displacement

\theta =\omega _1t+\frac{\alpha t^2}{2}

\theta =13.33\times 4.1-\frac{2.37\cdot 4.1^2}{2}

\theta =54.66-19.75

\theta =34.91 rad

3 0
3 years ago
To determine the coefficient of static friction between two materials, an engineer places a small sample of one material on a ho
Fudgin [204]

This question is incomplete, the missing image is uploaded along this answer.

Answer:

the coefficient of friction is 0.32

Explanation:

 Given the data in the question;

we make use of kinematic equation of motion;

ω = ω₀ + ∝t

we substitute

ω = ( 0 rad/s ) + ( 0.4 rad/s² )( 9.903 s )

ω = 3.9612 rad/s

The centripetal force acting on the sample is;

Fc = mrω²

from the image; r = 200 mm = 0.2 m

so we substitute

Fc = m(0.2 m ) ( 3.9612 rad/s )²

Fc = (3.13822 m/s²)m

we know that the frictional force between the two materials should be providing the necessary centripetal force to rotate the sample object;

f = Fc

μN = Fc

μmg = (3.13822 m/s²)m

μ = (3.13822 m/s²)m / mg

μ = (3.13822 m/s²) / g

acceleration due to gravity g = 9.8 m/s²

so

μ = (3.13822 m/s²) / 9.8 m/s²

μ = 0.32

Therefore, the coefficient of friction is 0.32

5 0
3 years ago
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