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maks197457 [2]
2 years ago
13

In Figure 4.27 four particles form a square of edge length

Physics
1 answer:
Shalnov [3]2 years ago
8 0

Answer:

so sorry

don't know but please mark me as brainliest please

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A particle moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.
mrs_skeptik [129]

The particle's acceleration is 5.1 m/s²

<h3>What is Acceleration ?</h3>

Acceleration can be defined as the rate at which velocity is changing. It is a vector quantity and it is measured in m/s²

Given that a particle is moving along a straight line with constant acceleration has a velocity of 2.35 m/s at t=3.42 s, and a velocity of -8.72 m/s at t=5.59s

The given parameters are;

  • V1 = 2.35 m/s
  • V2 = - 8.72 m/s
  • T1 = 3.42s
  • T2 = 5.59s

Acceleration a = ΔV ÷ ΔT

a = (2.35 + 8.72) / (5.59 - 3.42)

a = 11.07 / 2.17

a = 5.1 m/s²

Therefore, the particle's acceleration is 5.1 m/s²

Learn more about Acceleration here: brainly.com/question/9069726

#SPJ1

4 0
1 year ago
A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.1 m/s , a typical raindrop speed. how fast
Airida [17]

here we will use the momentum conservation

initial total momentum = final total momentum

P_i = P_f

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now plug in all data here

m(0) + 50m(8.1) = mv + 50mv

405 = 51 v

v = 7.94 m/s

so the final speed will be 7.94 m/s

6 0
2 years ago
What current is used to power the United States power grid?
alexdok [17]
The answer is Alternating Current
3 0
2 years ago
Read 2 more answers
Glycerin is poured into an open U-shaped tube until the height in both sides is 20cm. Ethyl alcohol is then poured into one arm
Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining \rho_A: Ethyl alcohol density and \rho_G: Glycerin density , we can write:

\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3

Simplifying:

\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)

On the other hand:

h_1 + h_2 = \Delta h + h_3

Rearranging:

h_1 - \Delta h = h_3 - h_2 (2)

Replacing  (2) in (1):

\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)

Rearranging:

\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h

Data:

h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}

\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}}  = \Delta h

7.5 cm =  \Delta h

7 0
3 years ago
Picture question please answer do in 5 mins
alexandr402 [8]

Answer:

Can't see anything, please share clearly

4 0
3 years ago
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