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3 years ago

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The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?

2 answers:

Hoochie [10]3 years ago

8 0

Answer:

A

Explanation:

v = change of X / change of T

v = 200/19.3

g100num [7]3 years ago

3 0

Answer:

A. 10.36 m/s

Explanation:

The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?

A. 10.36 m/s

B. 3960 m/s

C. 219.3 m

D. 0.0965 m/s

speed is the change in distance per time

speed is scalar quantity and hence as no direction but only magnitude. time and distance are also scalar quantity

200/19.3

speed=10.36m/s

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A baseball with a mass of 151 g is thrown horizontally with a speed of 40.3 m/s (90 mi/h) at a bat. The ball is in contact with

Ghella [55]

Answer:

A baseball (m= 149g) approaches a bat horizontally at a speed of 40.2 m/s (90 mi/h) and is hit straight back at a speed of 45.6m/s (102mi/h). If the ball is in contact with the bat for a time of 1.10ms, what is the average force exerted on the ball by the bat ? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.

Explanation:

Use the impulse equation (a form of Newton's 2nd Law): FΔt = Δ(mv) where Δ means "change in"

The change in momentum is mBB(vf - vi) = (.150 kg)(-46.9 m/s - 40.5 m/s)

Divide this by the time interval and you get F exerted by the bat in Newtons.

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6 0

3 years ago

Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum

asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0

3 years ago

The part of an atom that is mostly empty space is the

Ghella [55]

<h3><u>Answer;</u></h3>

Electron cloud

<h3><u>Explanation;</u></h3>

<em><u>An atom is the smallest particle of an element that can take part in a chemical reaction. Atom is made up of two parts ; that is the nucleus and the electron cloud. The nucleus contain subatomic particles; protons and neutrons, while the electron cloud contains the electrons.</u></em>
<em><u>The electron cloud is the largest part of the atom and is mostly an empty space. Most of an atom is a cloud of electrons surrounding a space called the nucleus with tiny protons and neutrons.</u></em>

5 0

3 years ago

Read 2 more answers

A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp

VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir $T_L=281K$

We know that efficiency of Carnot engine is given by

$\eta =1-\frac{T_L}{T_H}$

$0.26 =1-\frac{281}{T_H}$

$T_H=379.72K$

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so $T_H=379.72K$

So $0.35=1-\frac{T_L}{379.72}$

$T_L=246.818K$

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0

3 years ago

A train of mass 5.6 × 10^5kg is at rest in a station.at time t=0s, a resultant force acts on the train and it starts to accelera

lana66690 [7]

Answer:

420000N

Explanation:

Given parameters:

Mass of the train = 5.6 x 10⁵kg

Acceleration = 0.75m/s²

Unknown:

Resultant force = ?

Solution:

According to newton's second law, force is the product of mass and acceleration;

Force = mass x acceleration

Resultant force that acts on the train is given below;

Force = 5.6 x 10⁵kg x 0.75m/s² = 420000N

7 0

3 years ago