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3 years ago

What is the distance a truck can travel if it moves 7 m/s for 20 seconds?

2 answers:

8 0

If it moves 7 m/s and never slows down or speeds up, and it keeps doing this for 20 seconds, then it will cover 7 meters in the 1st second, 7 meters in the 2nd second, 7 meters in the 3rd second, 7 meters in the 4th second, 7 meters in the 5th second, and so on and so forth, twenty times. All together, it'll cover

7+7+7+7+7+7+7+7+7+7+7+7+7+7+7+7+7+7+7+7 meters

What I did there was: I wrote down ' 7 ' twenty times.

Now, there's a much easier way to write it.

In arithmetic, a little dot like this --> · <-- means "times", so we can write

20 · 7 and then we can work it out with our calculator or our pencil.

20 · 7 = <em> 140 meters</em> .

artcher [175]3 years ago

6 0

if it moves 7m/s, that means every second it goes 7m. Now we just multiply by the time (20 seconds) and end up with 140m.

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Tres personas, A, B, C, jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección ho

Yuri [45]

Answer:

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

Explanation:

Si la caja debe hallarse en equilibrio físico, entonces se debe satisfacer la siguiente ecuación:

$F_{A} + F_{B} + F_{C} = 0$ (1)

Si sabemos que $F_{A} = -3$ y $F_{B} = 5$ , entonces el valor de la fuerza que debe ejercer la persona C debe ser:

$F_{C} = -F_{A}-F_{B}$

$F_{C} = -(-3)-5$

$F_{C} = -2$

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

3 0

3 years ago

The potential energy of a particle as a function of position will be given as U(x) = A x2 + B x + C, where U will be in joules w

satela [25.4K]

Answer:

F = - 2 A x - B

Explanation:

The force and potential energy are related by the expression

F = - dU / dx i ^ -dU / dy j ^ - dU / dz k ^

Where i ^, j ^, k ^ are the unit vectors on the x and z axis

The potential they give us is

U (x) = A x² + B x + C

Let's calculate the derivatives

dU / dx = A 2x + B + 0

The other derivatives are zero because the potential does not depend on these variables.

Let's calculate the strength

F = - 2 A x - B

3 0

3 years ago

What is one advantage of doing a field experiment instead of a laboratory experiment

Flauer [41]

It mimics the real world accurately

Explanation:

Experiments conducted in the field clearly presents the real world at it is to the scientist. Hardly can any part be controlled precisely and this gives a near to perfect scenario.

In the laboratory, for example, an organism is isolated from its environment and might not fully display its natural instinct and physiological capabilities.
Most laboratory set up are driven towards a model instead of real life settings.
The laboratory is more controlled and less varied and might truly represent the real world. It will only portray a part of the real world and series of further tests might have to be carried out to have a better model.

Learn more:

Experiment brainly.com/question/5096428

#learnwithBrainly

5 0

3 years ago

A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.602 and t

diamong [38]

Answer:

11.98 N

Explanation:

Normal force = mg = 2.03 * 9.81

coeff of static friction must be overcome for the book to begin moving

.602 = F / (2.03 * 9.81) = 11.98 N

5 0

2 years ago

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C = 50 - x cm

We know that ,electric field is given as

$E=K\dfrac{q}{r^2}$

$K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\$

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0

3 years ago