2 years ago

Find the average speed of a walk to school and back if you took 12 minutes to walk there and 18 minutes to get back

Physics

1 answer:

nikdorinn [45]2 years ago

7 0

Answer:

12+ 18 divide by 2 is the average minutes

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A 85 kg lineman tackles a 90 kg receiver. The receiver is running 5.8 m/s, and the lineman is moving 4.1 m/s, at a right angle t

weeeeeb [17]

Answer:

3.59 m/s

Explanation:

We are given that

Mass of lineman,m=85 kg

Mass of receiver,m'=90 kg

Speed of receiver,v'=5.8 m/s

Speed of lineman,v=4.1 m/s

$\theta=90^{\circ}$

We have to find the their velocity immediately after the tackle.

Initial momentum, $P_i=\sqrt{p^2_1+p^2_2}=\sqrt{(85\times 4.1)^2+(90\times 5.8)^2}=627.6 kgm/s$

According to law of conservation of momentum

Initial momentum=Final momentum= $(m+m')V$

$627.6=(85+90)V=175V$

$V=\frac{627.6}{175}=3.59 m/s$

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3 years ago

En un momento dado , la nadadora de una prueba de natación de 100 m espalda está debajo de la cuerda falsa de salida. Indica a)

Virty [35]

Answer:

I only speak English

Explanation:

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2 years ago

Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an

Juli2301 [7.4K]

Answer:

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{42}{1.8} = 23.3333... \\$

We have the final answer as

Hope this helps you

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2 years ago

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Stephanie lives in California,and her cousin Thomas lives in France. which coordinate system would be most useful if the two wan

qwelly [4]

Answer:

Equatorial

Explanation:

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2 years ago

A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an

Vlada [557]

Answer:

Explanation:

Given

volume $V=16 ft^3$

Suppose base is square with side L

height of crate is h

Volume $V=L^2\times h$

$16=L^2\times h$

Cost of top and bottom area $c_1=3L^2$

Cost of Side area $c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}$

Total Cost $C=c_1+c_2$

Total Cost $C=3L^2+\frac{128}{L}$

Differentiate C w.r.t Length

$\frac{dC}{dL}=6L-\frac{128}{L^2}$

$L^3=\frac{128}{6}$

$L=2.75 ft$

$h=\frac{16}{2.75^2}=11.46 ft$

Dimensions are $L\times L\times h=2.75\times 2.75\times 11.46$

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2 years ago