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3 years ago

Which has greater kinetic energy, a car traveling at 40 mph or a half-as-massive car traveling at 80 mph?

Physics

1 answer:

ziro4ka [17]3 years ago

8 0

Answer:

The 80 mph car

Because the formula says 1/2 mass but for the velocity it is squared

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A water-balloon launcher with mass 5 kg fires a 1 kg balloon with a velocity of

Mashcka [7]

1.6 m/s west is the answer

3 0

3 years ago

On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole

ohaa [14]

The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
$F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\$
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. $\alpha$ is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
$g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)
$
Our final equation is:
$g=g_0-g_0(a-1)cos^2(\alpha)$
Colatitude is:
$\alpha_c=90^\circ-\alpha$
The answer is:
$g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)$

5 0

3 years ago

Convert 2 kg to cg give you answer in SI

BARSIC [14]

1 kg=100000 cg

2 kg=200000 cq

If mass is the quantity then kg is the S.I

2 kg=2kg

6 0

3 years ago

A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

3 0

3 years ago

calculate the period of a wave whose frequency is 5 Hertz and whose wavelength is one centimeter give your answer in a decimal f

olga2289 [7]

The period of the wave is the reciprocal of its frequency.

1 / (5 per second) = 0.2 second .

The wavelength is irrelevant to the period. But since you
gave it to us, we can also calculate the speed of the wave.

Wave speed = (frequency) x (wavelength)

= (5 per second) x (1cm) = 5 cm per second

4 0

3 years ago