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vovangra [49]
1 year ago
11

As a train accelerates away from a station, it reaches a speed of 4.9 m/s in 5.4 s. if the train's acceleration remains constant

, what is its speed after an additional 5.0 s has elapsed?
Physics
1 answer:
romanna [79]1 year ago
3 0

The speed of the train after an additional 5 s has elapsed is 9.45 m/s

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

  • a is the acceleration
  • v is the final velocity
  • u is the initial velocity
  • t is the time

<h3>How to determine the acceleration in 5.4 s</h3>
  • Initial velocity (u) = 0 m/s
  • Final velocity (v) = 4.9 m/s
  • Time (t) = 5.4 s
  • Acceleration (a) =?

a = (v – u) / t

a = (4.9 – 0) / 5.4

a = 4.9 / 5.4

a = 0.91 m/s²

<h3>How to determine the final velocity after 5 s</h3>
  • Initial velocity (u) = 4.9 m/s
  • Acceleration (a) = 0.91 m/s²
  • Time (t) = 5 s
  • Final velocity (v) = ?

a = (v – u) / t

0.91 = (v – 4.9) / 5

Cross multiply

v – 4.9 = 0.91 × 5

v – 4.9 = 4.55

Collect like terms

v = 4.55 + 4.9

v = 9.45 m/s

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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Answer:

Along the Atlantic and Gulf Coasts of Florida, the land surface is also sinking. If the oceans and atmosphere continue to warm, sea level along the Florida coast is likely to rise one to four feet in the next century. Rising sea level submerges wetlands and dry land, erodes beaches, and exacerbates coastal flooding.

Explanation:

6 0
3 years ago
The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov
victus00 [196]

The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

=0.10368kg m²

l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

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5 0
1 year ago
The steps taken in scientific studies often follow similar patterns. which answer correctly describes the order in which the ste
scZoUnD [109]
OPTION A WOULD BE YOUR ANSWER
7 0
3 years ago
Read 2 more answers
Please help me with this physics prooblem
zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

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3 years ago
A 7.00 kg bowling ball is held 2.00 m above the ground. Using g- 9.8 m/s^2, how much energy does the bowling ball have due to it
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It should be B.137 just took the test.
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