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3 years ago

An 800 N box is pushed across a level floor for a distance of 5.0 m with a force of 400 N. How much work was done on this box.

Physics

1 answer:

AveGali [126]3 years ago

4 0

Answer: 2000 J.

Explanation: Since work is force*displacement, we just have to multiply the force by the distance: w = f*d = 400 N*5.0 m = 2000 J.

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What is the value of acceleration due to gravity at heght h=4re

Norma-Jean [14]

Answer: Formula for Acceleration Due to Gravity

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Explanation:

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3 years ago

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

Solve:40KMEN

Oksanka [162]

The spider will cross the driveway <u>25.71 in seconds</u>.

Why?

It's a conversion problem, so, in order to solve it, we need to convert the units of the given information.

Let's convert the given speed (in cm/s) to m/s.

We know that:

$1m=100cm$

So, converting we have:

$14\frac{cm}{s}*\frac{1m}{100cm}=0.14\frac{m}{s}$

Then, calculating the time, we have:

$distance=speed*time\\\\3.6m=0.14\frac{m}{s}*time\\\\time=\frac{3.6m}{0.14\frac{m}{s}}=25.71s$

We have that the spider will cross the driveway in 25.71 seconds.

Have a nice day!

8 0

3 years ago

A metallic sheet has a large number of slits, 5.0 mm wide and 20 cm apart, and is used as a diffraction grating for microwaves.

san4es73 [151]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and constructive interference, that is to say everything that refers to an overlap of two or more equal frequency waves, which when interfering create a new pattern of waves of greater intensity (amplitude) whose cusp is the antinode.

Mathematically its definition can be given as:

$d sin\theta =m\lambda$