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2 years ago

A student walks 4 block east, 7 blocks west, 1 blocks east then 2 blocks west in an hour. Her average speed is____ block/hour

Physics

1 answer:

Gemiola [76]2 years ago

4 0

Average speed = (total distance) / (total time)

Average speed = (4+7+1+2 blox) / (1 hour)

Average speed = 14 blocks/hour

I'm gonna go out on a limb here and take a wild guess:

I'm guessing that there's another question glued onto the end of this one, and it asks you to find either her displacement or her average velocity. I'm so sure of this that I'm gonna give you the solution for that too. If there's no more question, then you won't need this, and you can just discard it. I won't mind.

Average velocity = (displacement) / (time for the displacement)

"Displacement" = distance and direction from the start point to the end point, regardless of how she got there.

Displacement = (4E + 7W + 1E + 2W)

Displacement = (5E + 9W)

Displacement = 4 blocks west

Average velocity = (4 blocks west) / (1 hour)

Average velocity = 4 blocks/hour West

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The distance between the centers of the wheels of a motorcycle is 156 cm. The center of mass of the motorcycle, including the ri

Drupady [299]

Answer:

a = 9.86 m/s²

Explanation:

given,

distance between the centers of wheel = 156 cm

center of mass of motorcycle including rider = 77.5 cm

horizontal acceleration of motor cycle = ?

now,

The moment created by the wheels must equal the moment created by gravity.

take moment about wheel as it touches the ground, here we will take horizontal distance between them.

then, take the moment around the center of mass. Since the force on the ground from the wheels is horizontal, we need the vertical distance.

now equating both the moment

m g d = F h

d is the horizontal distance

h is the vertical distance

m g d = m a h

term of mass get eliminated

g d = a h

so,

$a = \dfrac{g\ d}{h}$

$a = \dfrac{9.8\times 0.78}{0.775}$

a = 9.86 m/s²

4 0

2 years ago

2. A liquid that is twice as dense as water is used in a barometer. With this barometer, atmospheric pressure would push the liq

Scrat [10]

Answer:

The reason why the height for the liquid is 5.18 m is because, for constant pressure, height is inversely proportional to density.

Explanation:

We know that pressure, P = ρgh where ρ = density of material, g = 9.8 m/s and h = height.

Since density of liquid = 2 × density of water = 2 × 1000 kg/m = 2000 kg/m³

Since atmospheric pressure = 101500 N/m²,

The height of liquid for this pressure is h = P/ρg = 101500/(2000 × 9.8) = 5.18 m

For the same pressure, the height for water is h = P/ρg = 101500/(1000 × 9.8) = 10.36 m

Since pressure = constant,

P = ρgh ⇒ P/ρg = h since P and g are constant, h ∝ 1/ρ

ρ₁h₁ = ρ₂h₂

So, the reason why the height for the liquid is 5.18 m is because, for constant pressure, height is inversely proportional to density.

8 0

2 years ago

A father racing his son has 1/2 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.4 m

Assoli18 [71]

answer:idk because idk how to work it out

5 0

2 years ago

A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc

adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

$f_{c_1$ = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

$f_{c_1$ = 3 × 10¹⁰ / 2( 0.0711 )

$f_{c_1$ = 3 × 10¹⁰ cm/s / 0.1422 cm

$f_{c_1$ = 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( ( $f_{c_1$ / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀ $e^{-\alpha _1l_{min$ = -100dB

we substitute

20log₁₀ $e^{-(310.7np/m)l_{min$ = -100dB

$l_{min$ = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

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2 years ago

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nadezda [96]

Answer: Mass

Explanation:

8 0

2 years ago