Ask question

Ask question

All categories

Artyom0805 [142]

2 years ago

5

Ben and Dan both weigh 600 N. They are doing pull-ups together. Each pull-up is 0.5 meters of distance. How much work do they ea

ch do for every pull-up? 300 J 600 J 1200 J 1500 J

1 answer:

Natasha2012 [34]2 years ago

5 0

Answer:

300 J

Explanation:

Work = (Force)*(distance) = 600 N ∗ 0.5 m = 300 J

You might be interested in

A 5cm tall object is placed 4cm in front of a converging lens that has a focal length of 8cm. Where is the image located in ____

OverLord2011 [107]

Answer:

a. -8 cm

Explanation:

$d_{o}$ = distance of the object = 4 cm

$d_{i}$ = distance of the image = ?

$f$ = focal length of the converging lens = 8 cm

using the lens equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$

$\frac{1}{4} + \frac{1}{d_{i}} = \frac{1}{8}$

$d_{i}$ = - 8 cm

4 0

3 years ago

The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha

katovenus [111]

Answer:

(1) passed through the foil

Explanation:

Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.

When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.

While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.

And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.

5 0

3 years ago

Read 2 more answers

“I know what the atomic number of this atom is, but I don’t know what the number of electrons is,” a friend says. How would you

liq [111]

Once the atomic number of an atom is known, the number of electrons can be deduced depending on if the atom is an ion or a neutral one.

<h3>Atomic number</h3>

The atomic number of an atom is the number of protons in the nucleus of the atom.

For atoms that are neutral, that is, no net charges, the number of protons is always equal to the number of electrons. In other words, the positive charges always balance the negative charges in neutral atoms.

Thus, if the atomic number of a neutral atom is 6, for example, the proton number will also be 6. Since the proton must balance the electron, the number of electrons will also be 6.

More on atomic numbers can be found here; brainly.com/question/17274608

8 0

2 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

3 years ago

What causes tendonitis?

Irina18 [472]

overuse of a muscle Answer:

Explanation:

6 0

3 years ago

Read 2 more answers