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3 years ago

Which of the following is a high impact activity? walking sprinting swimming sit-ups

Physics

2 answers:

Lena [83]3 years ago

8 0

sprinting is a high impact activity

Gnesinka [82]3 years ago

6 0

Answer: sprinting

Explanation:

The high impact activities are those activities which exerts direct force over the body. These activities includes the sports like football, running and gymnastics. These activities if done in excess can produce strain over the muscles and joints and may lead to injuries.

Sprinting is a running exercise which involves covering of a short distance in a limited period of time. It is likely to produce more strain over the body as compared to other activities. Therefore, it is an example of high impact activity.

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A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)

bezimeni [28]

Answer:

205 V

V $_{R}$ = 2.05 V

Explanation:

L = Inductance in Henries, (H) = 0.500 H

resistor is of 93 Ω so R = 93 Ω

The voltage across the inductor is

$V_{L} = - IwLsin(wt)$

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

= 11.0 V / 500 rad/s (0.500 H)

= 11.0 / 250

I = 0.044 A

Now

V $_{R}$ = IR

= (0.044 A) (93 Ω)

V $_{R}$ = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V $_{R}$ = V $_{R}$ cos (wt)

Putting V $_{R}$ = 4.092 V and w = 500 rad/s

V $_{R}$ = V $_{R}$ cos (wt)

= (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V $_{R}$ = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

= (4.092 V) (cos (500 rads/s)(0.00209))

= (4.092 V) (cos(1.045))

= (4.092 V)(0.501902)

= 2.053783

V $_{R}$ = 2.05 V

8 0

3 years ago

A rocket is fired at a 45° angle, what is the direction of the horizontal velocity vector at the peak height?

Bess [88]

Answer:

B: Horizontally to the left

Explanation:

Horizontal velocity is always constant throughout the entire trajectory of the rocket and acts in the horizontal direction in which the rocket was launched. This is because gravity only acts in the downwards vertical direction.

So in order words at peak height, horizontal velocity is in the horizontal direction in which the rocket was launched.

So if it was to the left, then direction is left but if right, then direction is right.

Looking at the options, the most appropriate will be:

Horizontally to the left

7 0

3 years ago

Knowledge and skills learned through socialization are an example of

ziro4ka [17]

Answer:

I think no.2 the answer

Because socialization and social resources are both for me

3 0

3 years ago

During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips i

Crank

Answer:

The mass of the mud is 3040000 kg.

Explanation:

Given that,

length = 2.5 km

Width = 0.80 km

Height = 2.0 m

Length of valley = 0.40 km

Width of valley = 0.40 km

Density = 1900 Kg/m³

Area = 4.0 m²

We need to calculate the mass of the mud

Using formula of density

$\rho=\dfrac{m}{V}$

$m=\rho\times V$

Where, V = volume of mud

$\rho$ = density of mud

Put the value into the formula

$m=1900\times4.0\times0.40\times10^{3}$

$m =3040000\ kg$

Hence, The mass of the mud is 3040000 kg.

4 0

3 years ago

The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo

a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

$t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s$

The tire angular velocity before stopping is:

$\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s$

Also its angular decceleration:

$\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2$

Using the following equation motion we can findout the angle it makes during the deceleration:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where $\omega$ = 0 m/s is the final angular velocity of the car when it stops, $\omega_0$ = 114rad/s is the initial angular velocity of the car $\alpha$ = 14.75 rad/s2 is the deceleration of the can, and $\Delta \theta$ is the angular distance traveled, which we care looking for:

$-114^2 = 2*(-14.75)*\Delta \theta$

$\Delta \theta = 440rad$ or 440/2π = 70 revelutions

4 0

3 years ago