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andrew-mc [135]

3 years ago

10

Let us remember your previous lesson on Physical Press Components! Directions: Analyze the folawna fitness components. Put a che

ck in the box to which Physical Fitness belongs to Components Slab-Related Health Related 1. Power 2. Agility 3. Balance 5. Muscular Strength

1 answer:

mart [117]3 years ago

8 0

Answer:

5 Muscular and strength

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A sled plus passenger with total mass m = 53.1 kg is pulled a distance d = 25.3 m across a horizontal, snow-packed surface for w

Alex Ar [27]

Answer:

Explanation:

Force of friction

F = μ mg

μ is coefficient of friction , m is mass and g is acceleration due to gravity .

If f be the force applied to pull the sled , the horizontal component of force should be equal to frictional force

The vertical component of applied force will reduce the normal force or reaction force from the ground

Reaction force R = mg - f sin28.3

frictional force = μ R where μ is coefficient of friction

frictional force = μ x (mg - f sin28.3 )

This force should be equal to horizontal component of f

μ x (mg - f sin28.3 ) = f cos 28.3

μ x mg = f μsin28.3 + f cos 28.3

f = μ x mg / (μsin28.3 + cos 28.3 )

a )

work done by pulling force = force x displacement

f cos28.3 x d

μ x mg d cos28.3 / (μsin28.3 + cos 28.3 )

b ) Putting the given values

= .155 x 53.1 x 9.8 x 25.3 cos28.3 / ( .155 x sin28.3 + cos 28.3 )

= 1796.76 / (.073 + .88 )

= 1885.37 J

c )

Work done by frictional force

= frictional force x displacement

= - μ x (mg - f sin28.3 ) x d

= - μ x mgd + f μsin28.3 x d

= - μ x mgd + μsin28.3 x d x μ x mg / (μsin28.3 + cos 28.3 )

d )

Putting the values in the equation above

- .155 x 53.1 x 9.8 x 25.3 +

.155 x .474 x 25.3 x .155 x 53.1 x 9.8 /( .155 x .474 + .88)

= -2040.67 + 149.92 / .95347

= -2040.67 + 157.23

= -1883.44 J .

6 0

3 years ago

An electron moves with a speed of 8.0 × 10^{6} m/s along the +x-axis. It enters a region where there is a magnetic field of 2.5

yanalaym [24]

Answer:

The magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N

Explanation:

Given;

speed of the electron, v = 8.0 × 10⁶ m/s

magnetic field strength, B = 2.5 T

angle of inclination of the field, θ = 60°

The magnetic force experienced by the electron in the magnetic field is given as;

F = qvBsinθ

where;

q is charge of electron = 1.6 x 10⁻¹⁹ C

B is strength of magnetic field

v is speed of the electron

Substitute the given values and solve for F

F = (1.6 x 10⁻¹⁹)( 8.0 × 10⁶)(2.5)sin60

F = 2.77 x 10⁻¹² N

Thus, the magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N

8 0

3 years ago

Liquid water can be separated into hydrogen gas and oxygen gas through electrolysis. 1 mole of hydrogen gas and 0.5 moles of

den301095 [7]

The temperature of the oxygen gas is 243.75 K.

Using ideal gas law to explain the answer, the absolute temperature of the gas will decrease if the number of moles of the gas increases and it will increase if the volume and/or pressure of the gas increases.

The reaction of the given elements;

$H_2 \ + \ \frac{1}{2} O_2 \ --->\ \ H_2O$

volume of the collected oxygen gas, V = 10 L

pressure of the gas, P = 1 atm

number of moles of the gas, n = 0.5

Using ideal law the temperature of the oxygen gas is calculated as follows;

$PV = nRT\\\\T = \frac{PV}{nR} \\\\where;\\R \ is \ the \ ideal \ gas \ constant = 0.08205 \ L.atm/K.mol\\\\T = \frac{1 \times 10 }{0.5 \times 0.08205} \\\\T = 243.75 \ K$

Thus, the temperature of the gas is 243.75 K.

Using ideal gas law to explain the answer. The absolute temperature of the oxygen gas is directly proportional to the product of its pressure and volume and inversely proportional to its number of moles. That is the absolute temperature of the gas will decrease if the number of moles of the gas increases and it will increase if the volume and/or pressure increases.

Learn more here: brainly.com/question/16617695

8 0

3 years ago

An object with a potential energy of 200 joules is dropped from a height of 2.00 meters. What is its kinetic energy when it reac

taurus [48]

200J

Hope this helps :D

4 0

4 years ago

a tire rolls from rest to an angular velocity of 6 radians per second. the angle it turns through is 12 radians. what is the ang

True [87]

Answer:

$\alpha =1.5\ rad/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega_i=0$

Final angular speed, $\omega_f=6\ rad/s$

Angular displacement, $\theta=12\ rad$

We need to find the angular acceleration of the tire. We can find it using the third equation of rotational kinematics. So,

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_f^2-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{6^2-0^2}{2\times 12}\\\\\alpha =1.5\ rad/s^2$

So, the angular acceleration of the tire is equal to $1.5\ rad/s^2$ .

7 0

3 years ago