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posledela
4 years ago
9

Your performance on the pacer is used to measure which components of health- related fitness?

Physics
1 answer:
77julia77 [94]4 years ago
8 0
Your performance on the pacer test is used to measure cardiovascular components of health-related fitness
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Unit vector perpendicular to 4i+3j and -i-3j+2k​
RideAnS [48]

Answer:

idn

Explanation:how

7 0
3 years ago
What are three things you already know about the game of baseball?
Mazyrski [523]
Two teams of five trying to shoot in the hoop, the ball is moving around the court by players drippling or passing, the team with the basketball is called offense
4 0
3 years ago
Read 2 more answers
A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi
Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

\Delta x_1=0.50m-0.35m=0.15m

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

k\Delta x_1=m_1g

Solving for the spring constant k, we get:

k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m

Finally, we sum this to the unstretched length to obtain the length of the spring:

x_2=0.35m+0.088m=0.44m

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0
4 years ago
An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis
Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

W=\Delta K (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

W=\Delta K + E_{lost}

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane (\Delta K) and part is lost because of the air resistance (E_{lost}).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N

Now we can calculate the acceleration of the plane, by using Newton's second law:

a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

v^2- u^2 = 2aS

where

v=? is the final velocity

u=66.0 m/s is the initial velocity

a=1.69 m/s^2 is the acceleration

S=5.00 \cdot 10^2 m is the distance travelled

Solving for v, we find

v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s

8 0
3 years ago
Two light bulbs, A and B, are connected to a 120-V outlet (a constant voltage source). Light bulb A is rated at 60 W and light b
d1i1m1o1n [39]

Answer:

Bulb A has a greater resistance.

Explanation:

Electric power (P) = V²/R

P = V²/R................ Equation 1

Where P = power, V = Voltage, R = Resistance.

Make R the subject of the equation

R = V²/P ................ Equation 2

For Bulb A,

Given: V = 120 V, P = 60 W.

Substitute into equation 2

R = 120²/60

R = 240 Ω

For bulb B

Given: V =120 V, P = 100 W.

Substitute into equation 2

R = 120²/100

R = 14400/100

R = 144 Ω

Hence Bulb A has a greater resistance.

5 0
3 years ago
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