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3 years ago

A dog exerts a force of 30N to move a wagon 2m in 5s. What is the power of the dog

Physics

1 answer:

Hunter-Best [27]3 years ago

6 0

Explanation:

power=f×v. recall= distances/ time

= f× d/t

= 30 × 2/5

=12watt

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Jenise is buying a car for $7,020. The TAVT rate is 9.1%.

djyliett [7]

Answer:

$7,658.82

Explanation:

Sales Tax Calculations:

Sales Tax Amount = Net Price x (Sales Tax Percentage / 100)

Total Price = Net Price + Sales Tax Amount

Net Price: $ 7,020.00

+Sales Tax (9.1%): $ 638.82

Total Price: $ 7,658.82

Therefore, the amount of tax that Jenise has to pay on her car is $7,658.82

8 0

3 years ago

slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a

Ostrovityanka [42]

Answer:

0.08 ft/min

Explanation:

To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth d and the other (lets call it l) is given by:

$l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft$

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section A at that point:

$A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}$

And the raising speed v of the water is given by:

$v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}$

where q is the water flow (1 cubic foot per minute).

7 0

3 years ago

What does the galaxy made of ?

ICE Princess25 [194]

Galaxies are sprawling systems of dust, gas, dark matter, and anywhere from a million to a trillion stars that are held together by gravity. Nearly all large galaxies are thought to also contain supermassive black holes at their centers.

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3 years ago

A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin

JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

$f_s = f(\dfrac{v}{v-(-v_s)})$