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ira [324]
3 years ago
15

An object is moving with an initial velocity of 6.5m/s. It is then subject to a constant acceleration of 2.1m/s^2 for 14s. How f

ar will it have traveled during the time of its acceleration?
Physics
1 answer:
alina1380 [7]3 years ago
4 0

Answer:

Distance travel s = 296.8 m

Explanation:

Given:

Initial velocity u = 6.5 m/s

Constant acceleration a = 2.1 m/s²

Time T = 14 s

Find:

Distance travel s

Computation:

s = ut + (1/2)at²

s = (6.5)(14) + (1/2)(2.1)(14)²

s = 91 + 205.8

Distance travel s = 296.8 m

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Answer:

The duration  is  T  =72 \  years /tex]Explanation:From the question we are told that     The  distance is  [tex]D  =  35 \ light-years = 35 *  9.46 *10^{15} = 3.311 *10^{17} \  m

  Generally the time it would take for the message to get the the other civilization is mathematically represented as

         t =  \frac{D}{c}

Here c  is the speed of light with the value  c =  3.08 *10^{8} \  m/s

=>     t =  \frac{3.311 *10^{17} }{3.08 *10^{8}}

=>     t =  1.075 *10^9 \ s

converting to years

           t =  1.075 *10^9 *  3.17 *10^{-8}

              t =  1.075 *10^9 *  3.17 *10^{-8}

            t =  34 \ years

Now the total time taken is mathematically represented as

      T  =  2*  t  +  2 + 2

=>   T  =  2* 34  +  2 + 2

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4 0
3 years ago
An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f
vladimir1956 [14]

Answer:

V = 0.0806 m/s

Explanation:

given data

mass quarterback = 80 kg

mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

M(quarterback) ×  V = m(football) × v (football)   ........................1

put here value we get

80 ×  V  = 0.43  × 15

V = 0.0806 m/s

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3 years ago
The graph shows the number of beans eaten by a
il63 [147K]

Answer:

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Explanation:

6 0
4 years ago
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A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside
neonofarm [45]

Answer:

\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}, level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}

\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt}  = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}

By replacing all known variables:

\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}

The positive sign of the rate of change of the tank level indicates a rising behaviour.

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Answer:

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