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aksik [14]
3 years ago
5

Which of the following best describes this Image?

Physics
2 answers:
Orlov [11]3 years ago
7 0

Answer: 4. Good accuracy

Explanation: Considering the center red dot to be the target and the dark dots to be the patch of the attempted targets.

The given figure has all the three attempts close enough to the target on the penultimate circular white patch, therefore we can say that the figure defines a good accuracy.

Accuracy is the degree of closeness to the targeted value (or spot, as here the case is) of an attempt (or a series of attempts) made.

Precision is defined as the degree of closeness of a series of attempts made i.e. the attempts taken must be close to each other having lesser deviation be it far from the target or closer to the target it does not matters.

ratelena [41]3 years ago
5 0

Good precision because he is hitting the same spot 3 times but we don't know if he is accurate. SO the answer is 2

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A potato is shot out of cylinder at an angle of 17 degrees above the horizontal with an initial speed of 20 m/s. What is its max
madreJ [45]

Answer:

Maximum height, h = 1.74 meters

Explanation:

It is given that,

A potato is shot out of the cylinder. It is a case of projectile motion. The potato makes an angle of 17 degrees above the horizontal.

Initial speed with which the potato is shot out, u = 20 m/s

We have to find the maximum height of the potato. The maximum height of a projectile (h) is given by the following formula as :

h=\dfrac{u^2sin^2\theta}{2g}

Where

\theta = angle between the projectile and the surface

g = acceleration due to gravity

h=\dfrac{(20\ m/s)^2sin^2(17)}{2\times 9.8\ m/s^2}

h = 1.74 m

or h = 1.74 meters

Hence, this is the required solution.

8 0
3 years ago
A low-luminosity star has a small and narrow ________, whereas a high-luminosity star has a large and wide one.
vova2212 [387]

A low-luminosity star has a small and narrow ​<u>habitable zone</u>, whereas a high-luminosity star has a large and wide one.

<h3>What is luminosity of a star?</h3>

The radiant power emitted by a light-emitting item over time is measured as luminosity, which is an absolute measure of radiated electromagnetic power (light).

The total quantity of electromagnetic energy released per unit of time by a star, galaxy, or other celestial object is referred to as luminosity in astronomy.

Learn more about low-luminosity star:
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4 0
1 year ago
Tres litros de oxigeno gaseoso a 15 grados centígrados y a presión atmosférica (1atm), se lleva a una presión de 10mm de Hg. ¿ c
lana [24]

Answer:

Three liters of gaseous oxygen at 15 degrees Celsius and at atmospheric pressure (1atm), is brought to a pressure of 10mm Hg. What will be the volume of the gas now if the temperature has not changed?

Explanation:

Given that, the temperature is constant

Then, using gay Lussac law

P1•V1 / T1 = P2•V2 / T2

Since temperature is constant

Then, T1 = T2 = T and they cancels out

So, we are left with

P1•V1 = P2•V2

Given that, .

Initial volume

V1 = 3 litres.

Initial pressure

P1 = 1atm = 101325 Pa

Final pressure

P2 = 10mmHg = 1333.22 Pa

Then, we want to find the final volume V2

Make V2 subject of formula.

V2 = V1•P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litres

So, the final volume is 288 litres.

In Spanish

Dado que la temperatura es constante

Luego, usando la ley gay de Lussac

P1 • V1 / T1 = P2 • V2 / T2

Como la temperatura es constante

Entonces, T1 = T2 = T y se cancelan

Entonces, nos quedamos con

P1 • V1 = P2 • V2

Dado que, .

Volumen inicial

V1 = 3 litros.

Presión inicial

P1 = 1atm = 101325 Pa

Presión final

P2 = 10 mmHg = 1333.22 Pa

Entonces, queremos encontrar el volumen final V2

Hacer V2 sujeto de fórmula.

V2 = V1 • P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litros

Entonces, el volumen final es de 288 litros

8 0
3 years ago
A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas
Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, \theta=33^{\circ}

Time for which the player's foot is in contact with it, \Delta t = 5.1\times 10^{-2}\ s

Part A,

The x component of the soccer ball's change in momentum is given by :

\Delta p_x=mv\ cos\theta

\Delta p_x=0.425\times 15\ cos(33)

p_x=5.34\ kg-m/s

The y component of the soccer ball's change in momentum is given by :

\Delta p_y=mv\ sin\theta

\Delta p_y=0.425\times 15\ sin(33)

p_y=3.47\ kg-m/s

Hence, this is the required solution.

3 0
3 years ago
Can electromagnetic waves be reflected
SashulF [63]

Answer:

Some types of electromagnetic waves, like radio waves, microwaves, infrared waves, visible light and ultraviolet waves, can be reflectedSome types of electromagnetic waves, like radio waves, microwaves, infrared waves, visible light and ultraviolet waves, can be reflected

Explanation:

7 0
2 years ago
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