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creativ13 [48]
3 years ago
10

A car accelerates from 10m/s to 20m/s over a distance of 80m. What is its acceleration?

Physics
1 answer:
Lina20 [59]3 years ago
5 0

Answer:

0.125m/s^2

Explanation:

20-10=10

10 divided by 80=0.125m/s^2

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There are four springs stretched by the same mass.
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Spring C stretches 100 cm.

Explanation:

The spring constant is simply the stiffness of the spring. The higher the spring constant the more stiff the spring is.

Spring constant shows the force needed to stretch a spring from it's equilibrium position. If a material requires more force to cause it to stretch, it will have a high spring constant.

          According to hooke's law "the force needed to extended an elastic material is directly proportional to its extension"

          F = ke

   k is the spring constant

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We see that the spring that stretches by 100 is the less stiff compared to other springs. It has the smallest spring constant.

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A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.
denpristay [2]

Answer:

Approximately 5.5\; \rm N, assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let q_1 and q_2 denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, q_1 = 20 \times 10^{-6}\; \rm C  and q_2 = 15 \times 10^{-6}\; \rm C.

Let r denote the distance between these two point charges. In this question, r = 0.7\; \rm m.

Let k denote the Coulomb constant. In standard units, k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}.

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}.

Substitute in the values and evaluate:

\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}.

8 0
3 years ago
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