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notka56 [123]
2 years ago
15

A force of 6.0 N gives a 2.0 kg block an acceleration of 3.0

Physics
1 answer:
RideAnS [48]2 years ago
8 0
Answer: I think it’s 0N
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8. An airplane is flying at 200 m/s when it touches the ground at the airport. It has a constant negative acceleration, and slow
Temka [501]

Answer:

Explanation:

Given

Initial velocity u = 200m/s

Final velocity = 4m/s

Distance S = 4000m

Required

Acceleration

Substitute the given parameters into the formula

v² = u²+2as

4² = 200²+2a(4000)

16 = 40000+8000a

8000a = 16-40000

8000a = -39,984

a = - 39,984/8000

a = -4.998m/s²

Hence the acceleration is -4.998m/s²

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2 years ago
Why do elements within a group have similar chemical properties
CaHeK987 [17]
Elements<span> in the same </span>group<span> in the periodic table </span>have similar chemical properties<span>. This is because their atoms </span>have<span> the same number of electrons in the highest occupied energy level. </span>Group<span> 1 </span>elements<span> are reactive metals called the alkali metals.</span>Group<span> 0 </span>elements<span> are unreactive non-metals called the noble gases.
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4 0
3 years ago
If you lived by the sea, what effect of the moon would you see? Describe the effect.
dolphi86 [110]

Answer:

the changes in waves

Explanation:

the moon has its own gravitational pull thus making waves  and the rising tides

8 0
3 years ago
The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of
motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

F\propto \dfrac{mgv^2}{r}

F=\dfrac{kmgv^2}{r}

mg is the weight of the car

r is the radius of the curve

v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

Radius of the curve, r = 650 ft

Speed of the car, v = 40 mph

640=\dfrac{k(2600)(40)^2}{650}

k = 0.1

Case 2.

Radius of the curve, r = 750 ft

Speed of the car, v = 30 mph

F=\dfrac{0.1\times 2600\times (30)^2}{750}

F = 312 N

Hence, this is the required solution.

6 0
2 years ago
What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each
Arturiano [62]

\huge\mathfrak\red{✔Answer:-}

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

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3 years ago
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