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jeyben [28]
3 years ago
15

A car from rest moves a distance s(m) in t second, where s=3t³+t²/4 determine(i)the initial acceleration of the car (ii)the acce

leration of the car at the end of 3rd second​
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
8 0

Answer:

0.5 , 54.5

Explanation:

for acceleration we should derivate the equation 2 times

x=3t³+t²/4

v=9t²+t/2

a=18t+1/2

a(0)=0.5

a(3)=54.5

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Which of the following most<br> directly limits the number of<br> organisms in a community?
astraxan [27]

Answer:

It can either be food, mating, or competition with other organisms for resources.

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the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. the s
Travka [436]
When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat. 
3 0
3 years ago
3. A cat walks 0.220km North, then 0. 120 km South in a time of 400 seconds. whats the displacement and average velocity?
Andru [333]

Answer:

The rate at which velocity changes with respect to a change in time is called. acceleration.

Explanation:

4 0
3 years ago
How much energy is needed to generate 0.71 x 10-16 kg of mass?
AleksAgata [21]

Answer:

6.39 J of energy is needed to generate 0.71 * 10⁻¹⁶ kg mass

Explanation:

According to the Equation: E = mc²

where the mass, m = 0.71 * 10⁻¹⁶ kg

the speed of light, c = 3 * 10⁸ m/s

The amount of energy needed to generate a mass of 0.71 * 10⁻¹⁶ kg is calculated as follows:

E = (0.71 * 10⁻¹⁶) (3 * 10⁸)²

E = 0.71 * 10⁻¹⁶ * 9 * 10¹⁶

E = 0.71 * 9

E = 6.39 J

6 0
3 years ago
An electron is confined to a one dimensional region, bounded by an infinite potential. If the energy of the electron in its firs
OLga [1]

Answer:

The energy in its ground state is 10 meV.

Explanation:

It is given that,

The energy of the electron in its first excited state is 40 meV.

Energy of the electron in any state is given by :

E=\dfrac{n^2\pi^2h^2}{8mL^2}

For ground state, n = 1

E_1=\dfrac{\pi^2h^2}{8mL^2}.............(1)

For first excited state, n = 2

40=\dfrac{2^2\pi^2h^2}{8mL^2}.............(2)

Dividing equation (1) and (2), we get :

\dfrac{E_1}{40}=\dfrac{1}{4}

E_1=10\ meV

So, the energy in its ground state is 10 meV. Hence, this is the required solution.

4 0
3 years ago
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