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3 years ago

15

What is endurance? a). the ability to run faster b). a combination of balance and coordination c). how much you can stretch d).

the ability to exercise for longer periods of time

1 answer:

Alex787 [66]3 years ago

8 0

Answer:

D. the ability to exercise for longer periods of time

Explanation:

For example, when someone does endurance training, they are stretching their body's ability to do a certain exercise for longer times as opposed to increasing strength.

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SOMONE PLEASE HELP ASAP!!!!

kogti [31]

violet

it has the shortest wavelength

8 0

2 years ago

A 0.10kg hockey puck decreases it’s speed from 40m/s to 0m/s in 0.025s. Determine the force that it experiences

BigorU [14]

Vf = vi + at
0m/s = 40m/s + a(0.025s)
a = -1600m/s^2

Fnet = ma
Fnet = (0.10kg)(-1600m/s^2)
Fnet = -160 N

hope that helps

6 0

3 years ago

the fireman wishes to direct the flow of water from his hose to the fire at b. determine two possible angles u1 and u2 at which

Lelechka [254]

The two possible angles obtained by using the qudratic equation are;

θ $_{1}$ = 15.10° and θ2 = 73.51°

Given, speed of water = $v_{A}$ = 50ft/s

For the motion along x direction, time period can be calculated as follows:

$s_{x} = (v_{A}) _x_{} } t$

35 = (50 × cosθ) t

t = 0.64 / cosθ

For the motion in y direction, an equation can be obtained as follows:

$s_{y} = (v_{A})_{y} t +\frac{1}{2} (a_{y} )t^{2}$

$s_{y} = (-v_{A}$ $sin$ θ) $} t +\frac{1}{2} (a_{y} )t^{2}$

Plugging in the values we get:

$-20 = (-50_$ $sin$ θ) $} t +\frac{1}{2} (-32.2} )t^{2}$

-20 = -32tanθ - 10.304 $sec^{2}$ θ

Upon solving the above quadratic equation, we get,

tanθ = 0.27 , -3.38

Therefore,

tanθ $_{1}$ = 0.27

θ $_{1}$ = 15.10°

and, tanθ $_{2}$ = -3.38

θ $_{2}$ = 73.51

Learn more about quadratic equation here:

brainly.com/question/17177510

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8 0

11 months ago

Ball A (mass of 0.5 kg) going +2.0 m/s collides with a stationary ball B (mass of 0.4 kg).

ki77a [65]

The correct answer is 1.2 m/s

: mv+mv=mv+mv

(0.5kg)(2m/s)+(0.4kg)(0m/s)=(0.5kg)v+(0.4kg)(1m/s)

= 1kg*m/s=(0.5kg)v+0.4kg*m/s

=1kg*m/s-0.4kg*m/s=(0.5kg)v

=0.6kg*m/s=(0.5kg)v

to solve for v we divide both side by 0.5kg

v=1.2m/s

6 0

3 years ago

Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi

Nina [5.8K]

Answer:

$T_2 = 0.592$

Explanation:

Given

$T_1 = 1.48s$

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

$\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_{up}} = \frac{1+1}{C}$

$\frac{1}{C_{up}}= \frac{2}{C}$

Cross Multiply

$C_{up} * 2 = C * 1$

$C_{up} * 2 = C$

Make $C_{up}$ the subject

$C_{up} = \frac{1}{2}C$

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

$C_1 = 2 * C_{up}$

$C_1 = 2 * \frac{1}{2}C$

$C_1 = C$

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

$C_p = C+C$

$C_p = 2C$

So, the total capacitance (C2) is:

$\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}$

$\frac{1}{C_2} = \frac{5}{2C}$

Inverse both sides

$C_2 = \frac{2}{5}C$

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

$T\ \alpha\ C$

Convert to equation

$T\ =kC$

Make k the subject

$k = \frac{T}{C}$

$k = \frac{T_1}{C_1} = \frac{T_2}{C_2}$

$\frac{T_1}{C_1} = \frac{T_2}{C_2}$

Make T2 the subject

$T_2 = \frac{T_1 * C_2}{C_1}$

Substitute values for T1, C1 and C2

$T_2 = \frac{1.48 * \frac{2}{5}C}{C}$

$T_2 = \frac{1.48 * \frac{2}{5}}{1}$

$T_2 = \frac{0.592}{1}$

$T_2 = 0.592$

Hence, the time constance of (b) is 0.592 s

8 0

2 years ago