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TEA [102]
4 years ago
5

I start going 10 m/sIf I slow down at 2 m/s/s how far will I go before I stop

Physics
1 answer:
klasskru [66]4 years ago
5 0

Answer:

25 m

Explanation:

Given:

v₀ = 10 m/s

v = 0 m/s

a = -2 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (10 m/s)² + 2 (-2 m/s²) Δx

Δx = 25 m

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Calculate the speed of a dolphin that is observed swimming 75 meters in 5.0 seconds.
GarryVolchara [31]

Answer:

15 meters/second

Explanation:

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3 years ago
Can someone tell me the answer to this?
larisa86 [58]

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3 years ago
Read 2 more answers
How much work in joules must be done to stop a 920-kg car traveling at 90 km/h?
tankabanditka [31]

Answer:

Work done, W = −287500 Joules

Explanation:

It is given that,

Mass of the car, m = 920 kg

The car is travelling at a speed of, u = 90 km/h = 25 m/s

We need to find the amount of work must be done to stop this car. The final velocity of the car, v = 0

Work done is also defined as the change in kinetic energy of an object i.e.

W=\Delta K

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 920\ kg(0-(25\ m/s)^2)

W = −287500 Joules

Negative sign shows the work done is done is opposite direction.

3 0
3 years ago
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

8 0
3 years ago
A rotating toy above a baby's crib makes one
marshall27 [118]

Answer:

(a) The angular displacement of the toy is 18.85 rad.

----

(b) The toy's angular velocity is 6.284 rad/min.

----

(c) if the counterclockwise rotation is in negative direction, the angular acceleration will be positive when the toy is turned off and vice versa.

If my answer was helpful please mark as a brainlist

4 0
3 years ago
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