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Vika [28.1K]
3 years ago
13

If a constant fotce of 490 newtons is applied to an object and it proceeds to accelerate in the opposjte direction at 9.8 meters

per second, every second, what is its mass?
Physics
1 answer:
Sloan [31]3 years ago
8 0

Answer:

50 kg

Explanation:

f=ma

Force = mass x acceleration

490 = 9.8 x m

m = 50 kg

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What would the speed of this wave be?
valentinak56 [21]

Answer:

c) 100,000 m/s

Explanation:

You need to take the same wave length from the top graph and bottom one, so let's take half a wave length then in the top one that is 0.005, but in the bottom one it's 2000/4 = 500 because they are smaller and there are 4 half waves before you get to 2000, whereas in the top one there is 1 half wave before you get to 0.005 on the graph.

Now use speed = distance / time

speed = 500 / 0.005 = 100 000 m/s

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It takes Serina 0.68 hours to drive to school. Her route is 34 km long. What is Serina's average speed on her drive to school?
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Speed = distance / time

Speed = 34 km / 0.68 hrs

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3 years ago
A car accelerates at a rate of 4 m/s in 5 s. Calculate the change of velocity PLS HELP
garri49 [273]

Answer:

Explanation:

a=Δv/Δt

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4 years ago
Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the
Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

F = G \frac{m_1m_2}{r^2}

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

F = m_2a

Therefore, we can see that

a(r) = G \frac{m_1}{r^2}

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}

a(R) is actually the constant acceleration at sea level

and

a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}

Therefore:

a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}

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4 years ago
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What kind of materials would you build your house out of to stay cool in a hot,
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