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2 years ago

8

One event in a high school track meet is the 400 meter run. If the runners of the 400 meter event start and end at the same loca

tion, what is their DISPLACEMENT?
A. 0

B. 200

C. 400

D. 800

1 answer:

Pavlova-9 [17]2 years ago

5 0

Easy it would be D the anwer

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A brick of mass 1.15 kg is attached to a thin (massless) cable and whirled around in a circle in a vertical plane. The circle ha

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Explanation:

It is given that,

Mass of the brick, m = 1.15 kg

Radius of the circle, r = 1.44 m

The cable will break if the tension exceeds 43.0 N

Let v is the maximum sped can have at the bottom of the circle before the cable will break. At the bottom of the circle, the net force is equal to the centripetal force along with the weight of the brick. So,

$T=\dfrac{mv^2}{r}+mg$

$\dfrac{T}{m}-g=\dfrac{v^2}{r}$

$v=\sqrt{(\dfrac{T}{m}-g)r}$

$v=\sqrt{(\dfrac{43}{1.15}-9.8)\times 1.44}$

v = 6.30 m/s

So, the maximum speed of the brick at the bottom of the circle before the cable will break is 6.3 m/s. Hence, this is the required solution.

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3 years ago

A gas-turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 and 1600 kPa. The working flui

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Answer:

a) 6498.84 kW

b) 0.51

c) 0.379

Explanation:

See the attached picture below for the solution

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2 years ago

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 40.0 N at the s

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Answer

acceleration due to gravity on Jupiter's moon,g' = 1.81 m/s²

weight of water melon on earth, W = 40 N

acceleration due to gravity on earth, g = 9.8 m/s²

a) Mass on the earth surface

$M = \dfrac{W}{g}$

$M = \dfrac{40}{9.8}$

M = 4.08 Kg

b) Mass on the surface of Lo

Mass of an object remain same.

Hence, mass of object at the surface of Lo = 4.08 Kg.

c) Weight at the surface of Lo

W' = m g'

W' =4.08 x 1.81

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2 years ago

A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt

Yuri [45]

The total power emitted by an object via radiation is:
$P=A\epsilon \sigma T^4$
where:
A is the surface of the object (in our problem, $A=1.25 m^2$
$\epsilon$ is the emissivity of the object (in our problem, $\epsilon=1$ )
$\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4)$ is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is $T=100^{\circ} C=373 K$

Substituting these values, we find the power emitted by radiation:
$P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW$
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3 years ago

A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf

kirza4 [7]

Answer: $v=\sqrt{\frac{FL}{m}}$

Explanation:

Given

Ball of mass m

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Let length of the cord be L m and moving at a speed of v m/s

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T=Centripetal Force

$F=T=\frac{mv^2}{L}$

$v=\sqrt{\frac{FL}{m}}$

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2 years ago