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svp [43]
1 year ago
8

in a one dimensional collision, a 4kg object and 6 kg onject have initial velocity. calculate the magnitude of impulse

Physics
1 answer:
Greeley [361]1 year ago
8 0

in a one dimensional collision, a 4kg object with 5ms^1 and 6 kg object with 2ms^1 have initial velocity, the magnitude of impulse is 12 , 18

given,

mass 1 = 4kg

mass 2 = 6kg

velocity 1 = 5ms^1

velocity 2 = 2ms^1

impulse 1 = 4*(5-2)

= 12

Impulse 2 = 6*(5-2)

= 18

Learn more about impulse here

brainly.com/question/16980676

#SPJ4

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3 years ago
Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f
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Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

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8 0
3 years ago
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Answer:

1250 J

Explanation:

Work is said to be done when a force causes an object to move over a distance. The amount of work done (W) is calculated by multiplying the force by the distance traveled.

That is;

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Where;

W = work done (J or N/m)

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Therefore, 1250Joules of work is done by the jack.

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3 years ago
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