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gtnhenbr [62]
2 years ago
10

Why when i take "The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it c

ontinues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible." i don't fall flat on my face?
Physics
1 answer:
timurjin [86]2 years ago
6 0

Answer:

NOOOOOOO NOT THE PACER TEST!!

Explanation:

MEMORIES OF MIDDLE SCHOOL I'D LIKE TO FORGET

also. LOL "Remember to run in a straight line" BRUH WE KNOW XDXD

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PLEASE HELP!
lutik1710 [3]

Answer:

Thats her fault.........................b

Explanation:

6 0
2 years ago
Isss in the pic pls help
defon

Answer:

reacts with metals: acid

reacts with nonmetal: base

taste sour: acid

can cause burns: both

conduct electricity: both

taste bitter: base

5 0
2 years ago
What two types of evidence are used to classify orgianisms?
jeyben [28]
One is their traits and their characterists that they have in common 
6 0
3 years ago
Read 2 more answers
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
3 years ago
A U-shaped tube, open to the air on both ends, contains mercury. Water is poured into the left arm until the water column is 10.
Masja [62]

Answer:

0.368 cm

Explanation:

x = distance by which the mercury rise

d = depth of the water = 10 cm = 0.10 m

ρ = density of water = 1000 kgm⁻³

ρ' = density of mercury = 13600 kgm⁻³

P₀ = atmospheric pressure

Using equilibrium of pressure on both side

P₀ + ρ g d = P₀ + ρ' g (2x)

(1000) (0.10) = (13600) (2x)

x = 0.00368 m

x = 0.368 cm

8 0
2 years ago
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