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const2013 [10]
3 years ago
9

Sabrina wants to get better at soccer. In order to do this, she has created a list of skills she will need to improve. Which ski

ll should get the most of her attention in order for her to excel at soccer?
Physics
1 answer:
Alex17521 [72]3 years ago
5 0
I will answer your question but i will not choose between your choices, if it has one, so i hope you consider it. In playing soccer, like all field related sport, one must consider its basic fundamentals fist. Second you must improve your stamina, agility and most of all your kicking skill, not just force but also be more intellectual in playing. Third is never give up on trying and work hard on what your are doing. 
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A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrof
Vinil7 [7]

The specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

<h3>What is the specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

Given data;

Mass of liquid sample of Alcohol  m₁ = 200-gram

The temperature of alcohol, T₁ =  -6°C.

Mass of liquid sample of water  m₂ = 400-gram

The temperature of the water, T₂=  20°C.

The specific heat capacity of the alcohol, S₁=?

The specific heat capacity of water is, S₂=4.19 kJ/kg.°C

As we know that;

<h3 />

\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72  \ kJ /kg ^0 C

Hence the specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

To learn more about the specific heat capacity, refer to the link brainly.com/question/2530523.

#SPJ1

6 0
2 years ago
Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

4 0
2 years ago
A 20.5 kg ball moving at 38.5 m/s on a horizontal, frictionless surface runs into a light spring of force
tekilochka [14]

Answer:

7.68 m

Explanation:

Kinetic energy in ball = elastic energy in spring

KE = EE

½ mv² = ½ kx²

mv² = kx²

x = v √(m / k)

x = (38.5 m/s) √(20.5 kg / 515 N/m)

x = 7.68 m

5 0
3 years ago
A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s.how much work did f
zhuklara [117]

Answer:

5.791244495 KNm

Explanation:

The height h is given by, h=42.6sin42.3^{o}

Potential energy, PE is given by

PE=mgh where m is mass of the woman, g is acceleration due to gravity whose value is taken as 9.81 m/s^{2} and h is already given hence substituting 77 Kg for m we obtain

PE=77*9.81*42.6sin42.3^{o}=21656.7095 Nm

PE=21.6567095 KNm

We also know that Kinetic energy is given by0.5mv^{2} where v is the velocity and substituting v for 20.3 we obtain

KE=0.5*77*20.3^{2}=15865.465 Nm

KE=15.865465 KNm

Friction work is the difference between PE and KE hence

Friction work=21.6567095 KNm-15865.465 Nm=5.791244495 KNm

8 0
3 years ago
A monochromatic light beam with a quantum energy value of 4.3 ev is incident upon a photocell. the work function of the photocel
Black_prince [1.1K]
Give me some answer choices and i will be happy to help
7 0
3 years ago
Read 2 more answers
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