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Crank
2 years ago
11

It is a set of measure designed to determine the level of physical fitness?

Physics
1 answer:
JulsSmile [24]2 years ago
6 0

<em>The PFT or physical fitness test is a set of measures designed to determine the level of physical fitness.....</em>

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2
Karolina [17]
I’m pretty sure the answer would be D, sorry if it’s not correct!
5 0
3 years ago
Westinghouse and Edison fought what was known as the war of the currents. Eventually, Westinghouse triumped using Alternating Cu
DerKrebs [107]
The answer is B! :) Hope it helped
6 0
3 years ago
Read 2 more answers
A man performed 500 j of work moving a couch in his living room
Sergio039 [100]

Incomplete question. Missing part:

"it took him 10 seconds to move the couch. What was his power?"

Answer:

50 W

Explanation:

The power developed is given by:

P=\frac{W}{t}

where

W is the work done

t is the time taken

In this problem,

W = 500 J

t = 10.0 s

Substituting into the equation, we find

P=\frac{500}{10}=50 W

6 0
3 years ago
You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and
andreev551 [17]

Answer:

v_avg = 37 km/h

Explanation:

To find the average velocity in the complete trajectory you use the following formula:

v_{avg}=\frac{v_1+v_2}{2}   ( 1 )

v1: velocity in the first part of the trajectory = 70 km/h

v2: velocity in the second part of the trajectory = ?

You calculate v2 by using the following equation for a motion with constant velocity:

v_2=\frac{2.0km}{30min}*\frac{60min}{1h}=4\frac{km}{h}

you replace the values of v1 and v2 in (1) and you obtain:

v_{avg}=\frac{70km/h+4km/h}{2}=37\frac{km}{h}

hence, the average velocity is 37 km/h

6 0
2 years ago
Read 2 more answers
Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge
Serhud [2]

Answer:

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}=23.495N

Vector direction FC

\beta=35.91 degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

\alpha = tan^{-1}(\frac{3}{4}) = 36.86 degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2} Equation (1): Magnitude of the electric force of the charge qA over the charge qC

F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2} Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

k=8.99*10^9 \frac{N*m^2}{C^2}

q_{A}=2.70*10^{-4} C

q_{B}=-6.36*10^{-4} C

q_{C}=1.04*10^{-4} C

r_{AC} =3

r_{BC}=\sqrt{4^2+3^2} = 5

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N

F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N

Components of the FBC force at x and y:

F_{BCx}=23.785 *Cos(36.86)=19.03N

F_{BCy}=23.785 *Sin(36.86)=14.27N

Components of the resulting force acting on qC:

F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N

F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}= \sqrt{19.03^2+13.78^2} =23.495N

Vector direction FC

\beta = tan^{-1} (\frac{13.78}{19.03})=35.91 degrees: angle that forms FC with the horizontal

7 0
3 years ago
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