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Westkost [7]
3 years ago
12

What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when release

d?
Physics
1 answer:
yawa3891 [41]3 years ago
4 0
To answer this question you have to know weight, shape, and material
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what is the momentum of a 10000kg truck whose velocity is 20m/s and what velocity must a 5000kg truck attain in order to have th
Mice21 [21]

Answer:

40m/s

Explanation:

P=mv

m1v1=m2v2

(20)(10000)=(5000)v

200000=5000(v)

v=40m/s

4 0
4 years ago
Read 2 more answers
The density of a substance is 3.4 g cm-3. Its relative density relative to another substance is 2.0. what is the density of the
ankoles [38]

Answer:

1.7 g/cm³

Explanation:

Given that:

Density of substance = 3.4 g/cm³

Relative density to another substance = 2

Density of second substance=?

Let density of second substance = x

Relative density = density of substance / density of second substance

Relative density = density of substance / x

2.0 = 3.4g/cm³ / x

2 * x = 3.4 g /cm³

x = 3.4 g/cm³ ÷ 2

x = 1.7 g/cm³

5 0
3 years ago
If the sun is 400 times bigger than the moon, how couild the moon possibly cover the sun during a solar eclipse?​
AlekseyPX

Explanation:

the Moon passes between Earth and the Sun Even though the Moon is much smaller than the Sun, because it is just the right distance away from Earth, the Moon can fully block the Sun's light from Earth's perspective This completely blocks out the Sun's light

3 0
3 years ago
Read 2 more answers
What is the wavelength (in nm) of radiation that has an energy content of 1.0x10^2 kJ/mol?
hammer [34]
<span>Energy = h nu, where nu is the frequency
h = 6.63 x 10^-34 J-s, Planck's constant
So nu = E/h = 1 x 10^5 J /h = 0.15 x 10^29 / s
nu lambda = c, the speed of light.
lambda = wavelength = c / nu =3 x 10^8 / 0.15 x 10^29 = 20 x 10^-21 m.

this can possibly be a gamma ray. Gamma rays are very penetrating. It's both matter and an energy. They are electromagnetic radiation that results from a radioactive material. 

</span><span>
</span>
7 0
3 years ago
Read 2 more answers
A marble on a frictionless track, starting from point A in the drawing, is projected down the curved runway. (This means that th
EleoNora [17]

Answer:

v = 4.4 m / s

Explanation:

Unfortunately, the exercise scheme does not appear. Let's analyze the problem the marble leaves point A with an initial velocity, goes down and then rises to a given height where its velocity is zero, in the whole trajectory they tell us that the resistance is zero, so we can use the conservation relations of the enegy.

Starting point. Point A

          Em₀ = K + U = ½ m v2 + mg y_a

point B.

          Em_f = U = m g y

the energy is conserved

         Em₀ = Em_f

         ½ m v² + mg y_a = m g y

        ½ m v² = m g (y -y_a)

         v = \sqrt {2g ( y - y_a)}

         In the exercise the diagram is not seen, but the height of point A must be known, suppose that y_a = 4 m

       v = \sqrt{ 2 \  9.8 ( 5 -4)}

       v = 4.4 m / s

4 0
3 years ago
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