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jek_recluse [69]

3 years ago

11

the proper execution of suggested physical fitness tests is important inachieving the accuracyof the assessment results of one's

phisical fitness.

1 answer:

ankoles [38]3 years ago

5 0

Answer:

Regular exercise and physical activity increase muscle strength, bone density, flexibility, and stability. Physical fitness can reduce your risk for and resilience to accidental injuries, especially as you get older.

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A commuter airplane starts from an airport and takes theroute. The plane first flies to city A, located 175 km away in a directi

bearhunter [10]

Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

rf = ra + rb + rc where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:

rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km

rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

$rf = \sqrt{rfX^2+rfY^2} = 245.45km$

And the angle measure from the y-axis is:

$\alpha =atan(rfX/rfY) = 21.45\°$

So the answer is 245.45km in a direction 21.45° west of north from city A

6 0

3 years ago

A cooling fan is turned off when it is running at 850 rev/min. It turns 1500 revolutions before it comes to a stop. (a) What was

8_murik_8 [283]

Answer

given,

cooling fan revolution = 850 rev/min

fan turns before revolution = 1500 revolutions

$\omega = 850 \dfrac{2\pi}{60}$

$\omega = 89\ rad/s$

θ = 1500 revolution

θ = 1500 x 2 x π

θ = 9424.78 rad

a) using equation of rotation

ω² = ω₀² + 2 α θ

ω = 0 because body comes to rest

0 = 89² + 2 x α x 9424.78

α = -0.42 rad/s²

b) time take for the fan to stop

ω = ω₀ + α t

0 = 89 - 0.42 t

$t = \dfrac{89}{0.42}$

t = 211.9 s

5 0

3 years ago

A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s

trasher [3.6K]

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum $p_{i}$ must be equal to the final momentum $p_{f}$ , and taking into account this is aninelastic collision:

Before the collision:

$p_{i}=mV_{o}+MU_{o}$ (1)

After the collision:

$p_{f}=(m+M)V_{f}$ (2)

Where:

$m=1380 kg$ is the mass of the car

$V_{o}=23 m/s$ is the velocity of the car, directed to the north

$M=1625 kg$ is the mass of the truck

$U_{o}=-26 m/s$ is the velocity of the truck, directed to the south

$V_{f}$ is the final velocity of both the car and the truck

$p_{i}=p_{f}$ (3)

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg}$ (6)

Finally:

$V_{f}=-3.49 m/s$ The negative sign indicates the direction of the velocity is to the south

8 0

3 years ago

lbvjy [14]

Answer:

Impulse = 322.5[kg*m/s], the answer is D

Explanation:

This method it is based on the principle of momentum and the amount of movement; and used to solve problems involving strength, mass, speed and time.

If units of the SI are used, the magnitude of the impulse of a force is expressed in N * s. however, when remembering the definition of the newton.

$N*S=(kg*m/s^{2} )*s = kg*m/s$

Now replacing the values on the following equation that express the definition of impulse

$Impulse = Force * Time\\\\Impulse = 215 * 1.5 = 322.5 [kg*m/s]$

4 0

3 years ago

Read 2 more answers

In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s

sladkih [1.3K]

Answer:

quality factor (Q) = 69.99

inductor = 1.591 x 10⁻⁴ H

capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R = 10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW) $= \frac{R}{2\pi L}$

$BW = \frac{R}{2\pi L}$

make L (inductor) the subject of the formula

$L = \frac{R}{2\pi *BW} = \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH$

$F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}$

make C (capacitor) the subject of the formula

$C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F$

quality factor (Q) $= \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99$

quality factor (Q) = 69.99

5 0

3 years ago