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nadezda [96]
4 years ago
15

I need to find 1).a,b,c

Physics
1 answer:
Aleksandr [31]4 years ago
5 0
Let's cut through the weeds and the trash
and get down to the real situation:

                  A stone is tossed straight up at  5.89 m/s .
                  Ignore air resistance.


Gravity slows down the speed of any rising object by  9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for

                     (5.89 m/s / 9.8 m/s²)  =  0.6 seconds.

At that timer, he has run out of upward gas.  He is at the top
of his rise, he stops rising, and begins to fall.

His average speed on the way up is  (1/2) (5.89 + 0) = 2.945 m/s .

Moving for 0.6 seconds at an average speed of  2.945 m/s,
he topped out at

                    (2.945 m/s) (0.6 s) =  1.767 meters above the trampoline.

With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.

   (0.6 sec up) + (0.6 sec down)  =  1.2 seconds until he hits rubber again.



 
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Cientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below.
scZoUnD [109]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back.

Explanation:

3 0
3 years ago
A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it
Delvig [45]

Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = 5.93\times 9.81

T cos θ = 58.17

T sin \theta =\dfrac{mv^2}{r}

T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}

T sin^2 \theta =56.93

sin^2 \theta = 1 - cos^2 \theta

T (1 - cos^2 \theta) =56.93

T (1 - (\dfrac{58.17}{T})^2) =56.93

T² - 56.93T - 3383.75 = 0

T =  93.22 N

cos \theta = \dfrac{58.17}{93.22}

θ = 51.39°

6 0
3 years ago
the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp
Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0
3 years ago
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BARSIC [14]

The correct answer is D. An average city

Explanation:

A neutron star differs from others due to its massive density, this means a lot of matter is compressed in a small area. Indeed, neutron stars have a mass of around 1.4 to 2.8 times the mass of the sun. But these are considerably small as they only measure around 20 kilometers, which is the size of an average city. Additionally, neutron stars are this dense because they are the result of a regular star exploding, which leads to a super-dense core, or neutron star. In this context, the mass of a neutron star is compressed to the size of an average city.

7 0
3 years ago
When do sunspots disappear?
dedylja [7]

Answer:

In 5 years or so, the sun will be awash in sunspots and more prone to violent bursts of magnetic activity.

Explanation

once the magnetic field weakens the area and cold plasma enters the area of the sunspot

8 0
3 years ago
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