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sattari [20]
3 years ago
15

What would be the acceleration in a body moving with uniform velocity and why​

Physics
1 answer:
ozzi3 years ago
5 0

Answer:

If a body is moving with uniform velocity, acceleration is zero. So net force acting on it will be also zero. But the body has energy due to its constant motion.

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Explain Thomsons model of an atom<br><br><br>please its aurgent fast​
Aleks04 [339]

Answer:

Thomson's model showed an atom that had a positively charged medium, or space, with negatively charged electrons inside the medium. After its proposal, the model was called a "plum pudding" model because the positive medium was like a pudding, with electrons, or plums, inside.

3 0
2 years ago
g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
statuscvo [17]

Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

8 0
3 years ago
A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T.
sineoko [7]

Answer:10842.33m/s

Explanation:

F=qvBsine

V=f/(qBsine)

V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)

V=10842.33m/s

5 0
3 years ago
which of the following describes water changing from liqyid to solid, please give me a proof for this question please i really n
Fittoniya [83]
The answers A, this is because Ice is originally water and when water goes below it's freezing point it turns into ice
4 0
3 years ago
Read 2 more answers
What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular
monitta

Answer:

a_total = 2 √ (α² + w⁴) ,   a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

          a_total² = a_T²2 + a_{c}²

where

the centripetal acceleration is

  a_{c} = v² / r = w r²

tangential acceleration

   a_T = dv / dt

angular and linear acceleration are related

         a_T = α  r

we substitute in the first equation

       a_total = √ [(α r)² + (w r² )²]

       a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

        w = w₀ + α t

        w = 0 + 1.0 2

        w = 2.0rad / s

       

we substitute

        a_total = r √(1² + 2²) = r √5

        a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

         a_total = 2,236 m

7 0
3 years ago
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