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3 years ago

15

How much work must be done to stop a 1075-kg car traveling at 115 km/h ? Express your answer to two significant figures and incl

ude the appropriate units

1 answer:

jok3333 [9.3K]3 years ago

6 0

Answer:

Work done on the car will be 548440.9463 J

Explanation:

We have given mass of the car m = 1075 kg

As the car finally stops so final velocity of the car $v_f=0m/sec$

Initial velocity of car u = 115 km/hr $=115\times \frac{5}{18}=31.943m/sec$

Work done is equal to change in kinetic energy

So work done $=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$

= $=\frac{1}{2}m(v_f^2-v_i^2)=\frac{1}{2}\times 1075\times (0^2-31.943^2)=-548440.946J$

As the work done is negative so work is done on the car to stop it .

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Answer:

125 m

Explanation:

m = 3kg

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u = 0m/s

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using the formula,

v² = u² + 2as

50² = 0² + 2(10)(H)

2500 = 20H

H = 2500/20

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The work of energy theorem states that an increase in net work results in what?

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A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.

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Explanation:

You can calculate frequency by dividing velocity by wavelength

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3 years ago

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$

So we have:

$F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N$

So, the components of the resultant this time are:

$F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}$

Learn more about vector addition:

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#LearnwithBrainly

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3 years ago