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2 years ago

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A testable question is a question that _____. A. might be asked on a science test B. can be answered with either a yes or no C.

you're testing to see if it is a good question D. can be answered by gathering observations

1 answer:

Cloud [144]2 years ago

4 0

Answer:

D. can be answered by gathering observations.

Explanation:

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Stolb23 [73]

Asteroid 1 has more mass because the same force exerted caused this asteroid to move less therefore having more mass

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2 years ago

Read 2 more answers

What can you caculate if you know the frequency and speed of a wave

Mila [183]

Frequency and speed = wavelength.

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3 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 3.00 kg , up an incline of constant slope an

Sever21 [200]

Answer:

v = 8.45 m/s

Explanation:

given,

mass = 3 kg

angle = 30.0°

vertical distance = 3.3 m

μ = 0.06

according to conservation of energy

KE(loss) = PE(gain) + Work done (against\ friction)..............(1)

frictional Force

$F_f = \mu N$

$F_f = \mu m g cos \theta$

work against friction

W = F d

$W = \mu m g cos \theta \times h l sin\theta$

$W = \dfrac{\mu m g \times h}{tan\theta}$

Potential energy

PE = mgh

$\dfrac{1}{2}mv^2= \dfrac{\mu mgh}{tan \theta}+ mgh$

$\dfrac{1}{2}v^2= \dfrac{0.06 \times 9.81 \times 3.3}{tan 30^0}+ 9.8\times 3.3$

v = 8.45 m/s

the minimum speed is equal to 8.45 m/s

6 0

2 years ago

The atomic radii of Mg2+ and F- ions are 0.072 and 0.133 nm, respectively.(a) Calculate the force of attraction between these tw

timurjin [86]

Answer:

$1.09527\times 10^{-8}\ N$

Explanation:

$q_1$ = Mg ion = $+2q$

$q_2$ = F ion = $-q$

q = Charge of electron = $1.6\times 10^{-19}\ C$

r = Distance between ions = $0.072+0.133\ nm$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electrical force is given by

$F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N$

The attractive force is $1.09527\times 10^{-8}\ N$

8 0

3 years ago

A cyclist starts at rest and accelerates at 0.5 m/s2 south for 20 s. What is the cyclist’s final velocity

inn [45]

The cyclist's final velocity is 10 m/s.

From the question,

We are to determine the cyclist's final velocity.

<h3>Linear motion</h3>

From one of the equations of motion for linear motion, we have

$v=u+at$

Where

v is the final velocity

u is the initial velocity

a is the acceleration

and t is the time

From the given information,

The cyclist starts at rest, this means the initial velocity is 0 m/s

That is,

u = 0 m/s

Also

a = 0.5 m/s²

and t = 20 s

Putting the parameters into the equation, we get

$v = 0 + 0.5 \times 20$

$v = 0 + 10$

$v = 10 \ m/s$

Hence, the cyclist's final velocity is 10 m/s.

Learn more on linear motion here: brainly.com/question/19365526

8 0

1 year ago