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liq [111]
2 years ago
6

When joshua brakes his speeding bicycle to a stop, kinetic energy is transformed to select one:?

Physics
2 answers:
Harrizon [31]2 years ago
8 0

Answer:

the answer is heat. pleas give me a thanks if the answer is right

photoshop1234 [79]2 years ago
7 0
Potential energy, I’m pretty sure I don’t know but we was learning this in science and this is all I remember that potential energy is the moment energy reaches to a stop...
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4. If you swing wide to the left before turning right, another driver may try to pass you on the right. True or False
olchik [2.2K]

Answer:

True.

Explanation:

Don't turn wide to the left as you start the turn. A driver behind may think you are turning left and try to pass you on the right. You may crash into the other vehicle as you complete your turn.

Instead, slowly give yourself and others more time to avoid problems, keep the rear of the vehicle close to the curb. This will stop other drivers from passing you on the right. This is called (button Hook)

If you are driving a truck or bus that cannot make the right turn without swinging into the other lane, turn wide as you complete the turn.

8 0
2 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
2 years ago
Two spherical objects have equal masses and
ehidna [41]

to your question is 54 cm
6 0
2 years ago
24 POINTS!!!!!!!!!!!!!!!
TEA [102]
Potential Energy (Initial one) = m * g * h
P.E. = 60 * 9.8 * 10
P.E. = 5880

Kinetic Energy (Final One) = 1/2 mv²
K.E. = 1/2 * 60 * (10)²
K.E. = 6000/2
K.E. = 3000

Lost Energy = 5880 - 3000 = 2880 J

In short, Your Answer would be 2880 Joules

Hope this helps!
7 0
2 years ago
Read 2 more answers
g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts,
Stolb23 [73]

Answer:

523269.9\ \text{N/m}

Explanation:

q = Charge

r = Distance

q_1=25\ \text{C}

r_1=3000\ \text{m}

q_2=40\ \text{C}

r_2=850\ \text{m}

The electric field is given by

E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}

The electric field at the aircraft is 523269.9\ \text{N/m}

4 0
3 years ago
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