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3 years ago

Speed training increases one's maximum velocity. a. True b. False

Physics

2 answers:

Harrizon [31]3 years ago

3 0

That is absolutely true.

docker41 [41]3 years ago

3 0

Yup! it's True.It's true

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iris [78.8K]

Answer:

the answer is a time your welcome

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3 years ago

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A student travels 4.0m East, 8.0m South, 4.0m west, and finally 8.0m north. What is her displacement? What distance did she trav

erica [24]

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2 years ago

Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.

scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

Pressure at sea level = 1 atm = 101300 Pa

we know

P = ρ g h

$h = \dfrac{P}{\rho\ g}$

$h = \dfrac{101300}{1.3\times 9.8}$

h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

at x = 0 air density = 0

at x= h ρ_l = ρ_sl

assuming density is zero at x - distance

$\rho_x = \dfrac{\rho_{sl}}{h}\times x$

now, Pressure at depth x

$dP = \rho_x g dx$

$dP = \dfrac{\rho_{sl}}{h}\times x g dx$

integrating both side

$P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx$

$P =\dfrac{\rho_{sl}\times g h}{2}$

now,

$h=\dfrac{2P}{\rho_{sl}\times g}$

$h=\dfrac{2\times 101300}{1.3\times 9.8}$

h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

6 0

3 years ago

An airplane travels 2100 km at a speed of 1000 km/h. It then encounters a headwind which slows it to 800 km/h for the next 1300

Sidana [21]

I don’t get it I need help

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3 years ago

Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau

Tamiku [17]

Answer:

$7.0\cdot 10^{-13}C$

Explanation:

The magnitude of the electrostatic force between two charged objects is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.

In this problem, we have:

$r=0.070 cm =7\cdot 10^{-4} m$ is the distance between the charges

$q_1=q_2=q$ since the charges are identical

$F=9.0\cdot 10^{-9}N$ is the force between the charges

Re-arranging the equation and solving for q, we find the charge on each drop:

$F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C$

8 0

2 years ago