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3 years ago

Sudhir walks 0.40 km in a direction 60.0° west of north, then goes 0.50 km due west. What is his displacement?

Physics

1 answer:

Gemiola [76]3 years ago

8 0

Answer:

The displacement of Sudhir is 0.781 km.

Explanation:

Given;

initial distance, d₁ = 0.4 km = 400 m, N60.0°W

final distance, d₂ = 0.5 km

Make a sketch of Sudhir motion to form a right angled triangle;

(Check image uploaded).

Apply cosine rule to determine d "displacement"

d² = 500² + 400² - (2 x 500 x 400 x cos 120)

d² = 410,000 - (400,000 x -0.5)

d² = 410,000 - (-200,000)

d² = 410,000 + 200,000

d² = 610,000

d = √610000

d = 781.03 m

d = 0.781 km

Therefore, the displacement of Sudhir is 0.781 km.

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A car initially at rest accelerates at 10m/s^2. The car’s speed after it has traveled 25 meters is most nearly... A.) 0.0m/s B.)

STALIN [3.7K]

The car traverses a distance $x$ after time $t$ according to

$x=\dfrac12at^2$

where $a$ is its acceleration, 10 m/s^2. The time it takes for the car to travel 25 m is

$25\,\mathrm m=\left(5\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=\sqrt 5\,\mathrm s$

5 is pretty close to 4, so we can approximate the square root of 5 by 2. Then the car's velocity $v$ after 2 s of travel is given by

$v=\left(10\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)\approx20\dfrac{\rm m}{\rm s}$

which makes C the most likely answer.

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4 years ago

Elements in which family are most likely to have properties associated with

Elena-2011 [213]

Answer: As with all metals, the alkali metals are malleable, ductile, and are good conductors of heat and electricity. The alkali metals are softer than most other metals.

Alkaline earth metals

The alkaline earth elements are metallic elements found in the second group of the periodic table

Explanation:

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4 years ago

A centrifuge in a medical research laboratory rotates at an angular speed of 3,650 rev/min. When switched off, it rotates 46.0 t

erma4kov [3.2K]

Answer:

The constant angular acceleration of the centrifuge = -252.84 rad/s²

Explanation:

We will be using the equations of motion for this calculation.

Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.

First of, we write the given parameters.

w₀ = initial angular velocity = 2πf₀

f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s

w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s

θ = 46 revs = 46 × 2π = 289.14 rad

w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)

α = ?

Just like v² = u² + 2ay

w² = w₀² + 2αθ

0 = 382.38² + [2α × (289.14)]

578.29α = -146,214.4644

α = (-146,214.4644/578.29)

α = - 252.84 rad/s²

Hope this Helps!!!

6 0

4 years ago

Read 2 more answers

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c

seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

$A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2$

So the volume flow rate along the pipe is

$\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s$

We can use the similar logic to find the cross-section area at the refinery

$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

The radius of the pipe at the refinery is:

$A_r = \pi r^2$

$r^2 =A_r/\pi = 0.446/\pi = 0.141$

$r = \sqrt{0.141} = 0.377m$

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0

3 years ago

Ples help qustion in image

mash [69]

It is C) because that truck has more mass

8 0

3 years ago

Read 2 more answers