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2 years ago

I traveled 6.00 km to school in 0.250 hours . What was my average speed

Physics

1 answer:

SOVA2 [1]2 years ago

5 0

Answer: 6.67 m/s

Explanation:

Recall that average speed is the total distance covered by a moving body divided by the total time taken. The SI unit is metres per second.

i.e Average speed = distance/time taken

Given that:

Distance covered = 6.00 km

Convert kilometers to meters

If 1 km = 1000 m

6.00 km = (6.00 x 1000) = 6000m

Time taken to school = 0.250 hours

Convert time in hours to seconds

If 1 hour = 60 minutes & 1 minute = 60 seconds

Then, 0.250 hours = (0.250 x 60 x 60)

= 900 seconds

Average speed = ?

Average speed = 6000m/900s

= 6.67 m/s

Thus, your average speed to school is 6.67 m/s

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Answer:

The correct answers are a and d

Explanation:

In this experiment it can be analyzed using Newton's second law

F = m a

m = F / m

The outside is supplied by the spring balance and is constant, therefore the acceleration of the system is also constant.

The acceleration can be found with the kinematic equations

x = v t + ½ a t²

As we start from rest the initial speed is zero

a = 2 x / t²

Therefore we need the reading of position and time.

Finally, the relationship between the balance reading and this acceleration of the mass of the system

Let's analyze the answers

a) True. It is one of the necessary quantities

b) False. With the equipment we cannot measure the speed directly

c) false. Acceleration is calculated

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The correct answers are a and d

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2 years ago

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A circular grill of diameter 0.25 m has an emissivity of 0.8. If the surface temperature is maintained at 150°C, determine the r

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Answer:

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

Explanation:

Power required to maintain the surface temperature at 150°C from 20°C

P= εσA(T^4-t^4)

P= power in watt

ε= emissivity

A= area of surface

T= 150°C= 423 K

t= 20°C= 303K

/sigma= 5.67×10^{-8} watt/m^2K^4

putting vales we get

$= 0.8\times5.67\times10^{-8} \pi\frac{0.25^4}{4}(423^4-303^4)$

P=52.51822 Watt

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

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2 years ago

A helicopter descends from a height of 600 m with uniform negative acceleration, reaching the ground at rest in 5.00 minutes. de

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Given:
h = 600 m, the height of descent
t = 5 min = 5*60 = 300 s, the time of descent.

Let a = the acceleration of descent., m/s².
Let u = initial velocity of descent, m/s.
Let t = time of descent, s.
The final velocity is v = 0 m/s because the helicopter comes to rest on the ground.
Note that u, v, and a are measured as positive upward.

Then
u + at = v
(u m/s) + (a m/s²)*(t s) = 0
u = - at
u = - 300a (1)

Also,
u*t + (1/2)at² = -h
(um/s)*(t s) + (1/2)(a m/s²)*(t s)² = 600
ut + (1/2)at² = 600 (2)
From (1), obtain
-300a +(1/2)(a)(90000) = -600
44700a = -600
a = - 1.3423 x 10⁻² m/s²

From (1), obtain
u = - 300*(-1.3423 x 10⁻²) = 4.03 m/s

Answer:
The acceleration is 0.0134 m/s² downward.
The initial velocity is 4.0 m/s upward.

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2 years ago

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Answer:

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Explanation:

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2 years ago

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Answer:

Explanation:

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2 years ago