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Sati [7]
3 years ago
11

Restate newton's first law in terms of acceleratin

Physics
1 answer:
Yanka [14]3 years ago
6 0

Well, I never heard it expressed that way before,
but it might go something like this:

               "Inertia is that property of all matter by which it
                tends
to maintain motion without any acceleration
                until
acted upon by an external force."
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find the speed of a rolling ball that travels a distance of 10 m over the top of a smooth table in 2.0 seconds
umka21 [38]
The speed is between 5-15
4 0
3 years ago
Read 2 more answers
When thorium (90 protons) ejects a beta particle, the resulting nucleus has
Sav [38]

Answer:

none of the above

Explanation:

The actual answer is '91 protons'. In fact, the beta decay of the thorium-234 is the following:

{}_{90}Th \rightarrow {}_{-1}e+{}_{91}Pa

where inside the nucleus of Thorium (90 protons), a neutron turns into an electron (the beta particle) + a proton. Therefore, the resulting nucleus (which is Protoactinium) has a total of 90+1 = 91 protons.

So, the correct answer would be '91 protons'.

7 0
3 years ago
Just need to know net force, unbalanced or balanced and accelerated or not accelerated :)
jonny [76]

Answer:

9. 45 N North East

Unbalanced

Accelerate

10. 45 N right and 6 N down

Unbalanced

Accelerate

Explanation:

Hope this helps!

5 0
3 years ago
A spherical balloon is inflating with helium at a rate of 72 ft2/min . How fast is the​ balloon's radius increasing at the insta
Yanka [14]

The volume of the balloon is given by:

V = 4πr³/3

V = volume, r = radius

Differentiate both sides with respect to time t:

dV/dt = 4πr²(dr/dt)

Isolate dr/dt:

dr/dt = (dV/dt)/(4πr²)

Given values:

dV/dt = 72ft³/min

r = 3ft

Plug in and solve for dr/dt:

dr/dt = 72/(4π(3)²)

dr/dt = 0.64ft/min

The radius is increasing at a rate of 0.64ft/min

The surface area of the balloon is given by:

A = 4πr²

A = surface area, r = radius

Differentiate both sides with respect to time t:

dA/dt = 8πr(dr/dt)

Given values:

r = 3ft

dr/dt = 0.64ft/min

Plug in and solve for dA/dt:

dA/dt = 8π(3)(0.64)

dA/dt = 48.25ft²/min

The surface area is changing at a rate of 48.25ft²/min

7 0
3 years ago
A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-
solong [7]

Answer:

 N₁ = 393.96 N   and  N = 197.96 N

Explanation:

In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop

Lowest point, we write Newton's second law n for the y-axis

          N -W = m a

where the acceleration is ccentripeta

          a = v² / r

           

          N = W + m v² / r

          N = mg + mv² / r

         

we can use energy to find the speed at the bottom of the circle

starting point. Highest point where the ball is released

           Em₀ = U = m g h

lowest point. Stop curl down

           Em_{f} = K = ½ m v²

           Emo = Em_{f}

           m g h = ½ m v²

           v² = 2 gh

we substitute

             N = m (g + 2gh / r)

            N = mg (1 + 2h / r)

let's calculate

          N₁ = 5 9.8 (1 + 2 17.6 / 5)

          N₁ = 393.96 N

headed up

we repeat the calculation in the longest part of the loop

          -N -W = - m v₂² / r

            N = m v₂² / r - W

             N = m (v₂²/r  - g)

we seek speed with the conservation of energy

           Em₀ = U = m g h

final point. Top of circle with height 2r

             Em_{f} = K + U = ½ m v₂² + mg (2r)

              Em₀ =   Em_{f}

            mgh = ½ m v₂² + 2mgr

             v₂² = 2 g (h-2r)

we substitute

            N = m (2g (h-2r) / r - g)

            N = mg (2 (h-r) / r 1) = mg (2h/r  -2 -1)

             N = mg (2h/r  - 3)

            N = 5 9.8 (2 17.6 / 5 -3)

            N = 197.96 N

Directed down

3 0
3 years ago
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