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3 years ago

Which statement accurately compares the heating and cooling rates for water and land?

Physics

2 answers:

AVprozaik [17]3 years ago

6 0

Answer:

C is the answer Hope I helped

Mama L [17]3 years ago

4 0

Water heats up and cools down at a slower rate then land

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a bowling ball and a baseball both roll across your foot at the same speed. The bowling ball hurts more

Maru [420]

Answer:

you are right bowling ball has more mass and it is denser than a baseball

Explanation:

5 0

3 years ago

Read 2 more answers

The maximum lift-to-drag ratio of the World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine

andre [41]

The related concepts to solve this problem is the Glide Ratio. This can be defined as the product between the height of fall and the lift-to-drag ratio. Mathematically, this expression can be written as,

$R = h (\frac{L}{D})_{max}$

Replacing,

$R = 5000ft (7.7)$

$R = 38500ft$

Converting this units to miles.

$R = 38500ft (\frac{1mile}{5280ft})$

$R = 7.2916miles$

Therefore the glide in terms of distance measured along the ground is 7.2916miles

3 0

4 years ago

An electrical motor spins at a constant 2695.0 rpm. If the armature radius is 7.165 cm, what is the acceleration of the edge of

djyliett [7]

Answer:

Acceleration will be $5706.77rad/sec^2$

So option (D) will be correct answer

Explanation:

We have given angular speed of the electrical motor $\omega =2695rpm$

We have to change this angular speed in rad/sec for further calculation

So $\omega =2695rpm=2695\times \frac{2\pi }{60}=282.2197rad/sec$

Armature radius is given r = 7.165 cm = 0.07165 m

We have to find the acceleration of edge of motor

Acceleration is given by $a=\omega ^2r=282.2197^2\times 0.07165=5706.77rad/sec^2$

So acceleration will be $5706.77rad/sec^2$

So option (D) will be correct answer

5 0

4 years ago

Which of these letters is the symbol for current in equations A: c B: i C: r D: t

True [87]

Answer: [B]: The letter, "<em /> I " ; for current; in units of "Amps" .
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4 0

3 years ago

A long distance runner running a 5.0km track is pacing himself by running 4.5km at 9.0km/h and the rest at 12.5km/h. What is his

sweet [91]

Answer:

9.26 km/h

Explanation:

Applying,

V' = D'/t'............... Equation 1

Where V' = Average speed, D' = Total distance, t' = total time.

Given: D' = 5 km

But,

v = d/t............ Equation 2

Where v = speed , d = distance, t = time

t = d/v............ Equation 3

Given: d = 4.5 km, v = 9 km/h, and d = 0.5 km, v = 12.5 km/h

Therefore,

t₁ = 4.5/9 = 0.5 hours

t₂ = 0.5/12.5

t₂ = 0.04 hours

Therefore,

V' = 5/(0.5+0.04)

V' = 5/0.54

V' = 9.26 km/h

6 0

3 years ago