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3 years ago

A motorcycle accelerates from 10. m/s to 25 m/s in 5.0 seconds. What is the average acceleration of the bike?

Physics

2 answers:

I am Lyosha [343]3 years ago

5 0

Answer:

Explanation: Option (A) is the correct answer.

The data is given as follows.

Initial velocity $(v_{i})$ = 10 m/s, final velocity $(v_{f})$ = 25 m/s.

Initial time $(t_{i})$ = 0 seconds, final time $(v_{t})$ = 5 seconds.

The formula to calculate average acceleration is as follows.

Average acceleration = $\frac{v_{f} - v_{i}}{t_{f} - t_{i}}$

= $\frac{(25 - 10)m/s}{(5 - 0)s}$

= $\frac{15}{5}$

= 3 m/s/s

Thus, we can conclude that the average acceleration of the bike is 3 m/s/s.

valentinak56 [21]3 years ago

3 0

If a motorcycle accelerates from 10 m/s to 15 m/s in 5 seconds, the average acceleration of the bike is 3 m/s/s. This only means that the motorcycle’<span>s velocity will increase 3 m/s every second. You need to divide 15 m/s which is the difference of 10 and 25 to 5 seconds.</span>

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______ is/are usually the best conductor/s. a. Metals c. Air b. Water d. Minerals Please select the best answer from the choices

nataly862011 [7]

Metals are the best conductors.

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3 years ago

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A 5.7 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0s, the bl

Genrish500 [490]

Answer:

Explanation:

Given

mass of block $m=5.7\ kg$

at $t=0 s$

displacement is $x=-0.7\ m$

velocity $v=-0.8\ m/s$

acceleration $a=2.7\ m/s^2$

suppose $x=A\cos (\omega t+\phi )$ is the general equation of SHM

where A=amplitude

$\omega$ =natural frequency of oscillation

therefore velocity and acceleration is given by

$v=-A\omega \sin (\omega t+\phi )$

$a=A\omega ^2\cos (\omega t+\phi )$

for t=0

$-0.7=A\cos (\phi )---1$

$v=-0.8=-A\omega \sin(\phi)---2$

$a=2.7=-A\omega ^2\cos(\phi )----3$

divide 1 and 3 we get

$\omega ^2=\frac{27}{7}$

$\omega =\sqrt{\frac{27}{7}}$

Now square and 1 and 2 we get

$(0.7)^2+(\frac{0.8}{\omega })^2=A^2$

$A^2=0.49+0.166$

$A=0.81\ m$

3 0

3 years ago

A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface $\Phi$ it is possible to calculate the magnitude $\frac{\Phi}{A}=B\ cos \theta$ .

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux $\Phi$ is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is ( $m^{2}$ ). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with $\theta$ being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as: