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Vsevolod [243]
3 years ago
5

When you ride your bike around a corner at 10 m/s, you are accelerating?

Physics
2 answers:
oee [108]3 years ago
8 0

Answer:

yes because acceleration is a vector quantity and as u turn around a corner you are changing direction.

Explanation:

WINSTONCH [101]3 years ago
5 0

When you ride your bike around a corner at 10 m/s, you are accelerating. Acceleration is caused by any forces. Sliding friction keeps you in the seat when a car goes around a corner. If you throw a ball into the air, Earth exerts a force on the ball.

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Heat likes to remain <br> ONMFRIU ← Whats that unscrambled?
OlgaM077 [116]
<span>So the question is how does heat prefer to remain and to unscramble the letters ONMFRIU. The unscrambled letters mean: UNIFORM. The heat likes to remain uniform because thermodynamic systems always tend to reach thermal equilybrium after some period of time that is specific for each system. </span>
5 0
3 years ago
What are the ways of heat transfer?
AnnZ [28]

Answer:

Conduction, Convection and Radiation

Explanation:

Hope this helps!

7 0
3 years ago
Read 2 more answers
How large a band of frequencies does each television broadcasting channel get ?
k0ka [10]

Answer:

Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)

8 0
2 years ago
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A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

\mu = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N

Magnitude of frictional force is 628.022466 N

F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2

v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s

Initial speed of the player is 8.61 m/s

4 0
3 years ago
Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin
Vedmedyk [2.9K]

Answer:

The correct answer is 2.8*10^{-5}ms^{-1}

Explanation:

The formula for the electron drift speed is given as follows,

u=I/nAq

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is 8.93/63.5 molcm^{-3}. Converting this number to m³ using very elementary unit conversion we get 140384molm^{-3}. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}

if we convert the area from mm³ to m³ we get A=80*10^{-6}m^{2}.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}

which is our final answer.

6 0
3 years ago
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