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3 years ago

1.After striking the ball, it is important to ______.

Physics

2 answers:

egoroff_w [7]3 years ago

7 0

Answer:

1=A 2=3 3=Badmintion

Explanation:

I think I'm correct on these

mamaluj [8]3 years ago

6 0

1) follow through
2) hand eye coordination
3) basketball

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A toy helium balloon is initially at a temperature of T = 24o C. Its initial volume is 0.0042 m3 (It is a 10 cm radius sphere).

Anettt [7]

Answer:

$T_{f} = 335.780\,K\,(62.630\,^{\textdegree}C)$

Explanation:

Let assume that air behaves ideally. The equation of state of ideal gases is:

$P\cdot V = n\cdot R_{u}\cdot T$

Where:

$P$ - Pressure, in kPa.

$V$ - Volume, in m³.

$n$ - Quantity of moles, in kmol.

$R_{u}$ - Ideal gas constant, in $\frac{kPa\cdot m^{3}}{kmol\cdot K}$ .

$T$ - Temperature, in K.

Since there is no changes in pressure or the quantity of moles, the following relationship between initial and final volumes and temperatures is built:

$\frac{V_{o}}{T_{o}} = \frac{V_{f}}{T_{f}}$

The final temperature is:

$T_{f} = \frac{V_{f}}{V_{o}}\cdot T_{o}$

$T_{f} = 1.13\cdot (297.15\,K)$

$T_{f} = 335.780\,K\,(62.630\,^{\textdegree}C)$

7 0

4 years ago

Read 2 more answers

As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases

frozen [14]

Bohr's equation for the change in energy is
$\Delta E= \frac{hc}{\lambda}$
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
TRUE

b. As λ increases, f increases and ΔE increases.
FALSE

c. As λ increases, f increases and ΔE decreases.
FALSE

Answer:
As the wavelength increases, the frequency decreases and energy decreases.

3 0

4 years ago

Read 2 more answers

Hiran is standing beside the road when he hears a bird flying away from hip and chirping. The bird’s chirp has a frequency of 18

Murrr4er [49]

The frequency of bird chirping hear by hiran will be 1.77 kHz.

<u>Explanation:</u>

As per Doppler effect, the observer will feel a decrease in the frequency of the receiving signal if the source is moving away from the observer. So the shifted frequency is obtained using the below equation:

$f'=\frac{c}{c+v_{s} }f$

Here , c is the speed of sound, Vs is the velocity of source with which it is moving away. f is the original frequency of source and f' is the frequency shift heard by the observer.

As here, f = 1800 Hz, Vs= 6 m/s and c = 343 m/s, then

$f'=\frac{343}{343+6} \times 1800=1.77\ kHz$

So, the frequency of bird chirping hear by hiran will be 1.77 kHz.

4 0

3 years ago

(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is

Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = $\dfrac{1}{2}mv^2$

Using Conservation of energy

$KE_i + PE_i + KE_f + PE_f$

$0 + m g H = \dfrac{1}{2}mv^2 + 0$

$v = \sqrt{2gH}$

$v = \sqrt{2\times 9.8 \times 105}$

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0

3 years ago

If you carry out an experiment measuring the weight and mass of objects in one particular location on the earth, what relation w

Elina [12.6K]

The answer is letter C.

Weight (on Earth) is the force due to the mass of Earth attracting whatever mass is subject of discussion.

The force of attraction between any two masses is called Newton's Law of Universal Gravitation:

$G \frac{m_1m_2}{d^2}$

$G$ is simply a given constant.

If we're at the surface of Eath, $m_1$ refers to the mass of the Earth, $m_2$ to the mass of whatever is on the surface of Earth, and $d$ to the radius of Earth.

Normally, we define a constant $g$ to be equal to $G \frac{M}{r^2}$ ; in which $M$ is the mass of Earth and $r$ the radius of earth; $g$ happens to be around 9.8.

By that, we adapt the Law of Universal Gravitation to objects on the surface of Earth, we call that force Weight.

$W=mg$

As you can see, weight is directly proportional to mass, more mass implies more weight.

8 0

4 years ago