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3 years ago

a runner starts from rest and has an acceleration of 3 m/s^2. How fast is she running after 1.1 seconds

1 answer:

8 0

Acceleration x time = velocity

Since you're given acceleration and time, just plug the values into the equation.

3 $\frac{m}{s^{2}}$ x 1.1 s = ?

Solve that equation, and remember your velocity should be in m/s.

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A brick with a mass of 20 kg and a weight of 196 N is released on a frictionless plane at an angle of 30°. What is the accelerat

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Answer:4.9m/s^2

Basically answer A

Explanation:

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3 years ago

You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand at 30 m/s. What is the kinetic energy (KE) of the volley

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3 years ago

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Which element easily loses one electron to form a positive ion?

olga nikolaevna [1]

Answer/Explanation: As mentioned above, the characteristic chemical property of a metal atom is to lose one or more of its electrons to form a positive ion. However, certain metals lose electrons much more readily than others. In particular, cesium (Cs) can give up its valence electron more easily than can lithium (Li).

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3 years ago

A 235.0 g metal block absorbs 2.44 × 103 J of heat to raise its temperature by 35 K. What is the specific heat of the metal? Sho

andrew-mc [135]

When an object absorbs an amount of energy equal to Q, its temperature raises by $\Delta T$ following the formula
$Q=m C_s \Delta T$
where m is the mass of the object and $C_s$ is the specific heat capacity of the material.

In our problem, we have $Q=2.44 \cdot 10^3 J$ , $m=235.0 g$ and $\Delta T=35 K$ , so we can re-arrange the formula and substitute the numbers to find the specific heat capacity of the metal:
$C_s = \frac{Q}{m \Delta T}= \frac{2.44 \cdot 10^3 J}{(235.0 g)(35 K)}=0.297 J g^{-1} K^{-1}$

3 0

3 years ago

A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge

Kisachek [45]

Answer:

$\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r$

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

$\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

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3 years ago