1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Katen [24]
3 years ago
11

If you were told an atom was an ion, you would know the atom must have a ______?

Physics
1 answer:
andreyandreev [35.5K]3 years ago
3 0

Answer:

charge

Explanation:

7r0I and its etc. ,"!×_/;

You might be interested in
A certain unbalanced force gives a 5kg object an acceleration of 15 m/s2. What would be the acceleration if the same force was a
stira [4]

Answer:

a2 = 2.5 m/s2

Explanation:

F1 = m1 a1 We use the same force so F1 = F2

= 5kg × 15m/s2 F2 = m2 a2

= 75N a2 is required

a2 = F2 / m2

= 75N / 30 kg

= 2.5 m/s2

7 0
4 years ago
The velocity of an object is the distance it travels per unit time. Suppose the velocity of a gilding bird is measured to be 52.
Elanso [62]

Answer:

d=7.115s

Explanation:

What problem says can be written mathematically as:

v=\frac{d}{t}

Where:

v=Velocity\\t=Time\\d=Distance

The problem itself it's really simple, we only need to replace the data provided in the previous equation, but first, let's convert the units of the velocity from cm/s to m/s because we have to work with the same units and working in meters is the most apropiate action, because is the base unit of length in the International System of Units:

52\frac{cm}{s} *\frac{1m}{100cm} =0.52\frac{m}{s}

Now, we can replace the data in the equation and find the time it will take the bird to travel 3.7 m:

0.52=\frac{3.7}{t}

Solving for t, multiplying by t both sides, and dividing by 0.52 both sides:

t=\frac{3.7}{0.52} =7.115384615s\approx7.115s

5 0
3 years ago
A weight trainer lifts a 90.0-kg barbell from a stand 0.90 m high and raises it to a height of 1.75 m. What is the increase in t
nirvana33 [79]

Answer:

∆PE = 749.7 J

At 0.9 m high, PE = 793.8 J

At 1.75 m high, PE = 1543.5 J

7 0
2 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
3 years ago
Indiana Jones is in a temple searching for artifacts. He finds a gold sphere with a radius of 2 cm sitting on a pressure sensiti
Vlad1618 [11]

Answer:

Volume of Sand = 0.4 m³

Radius of Sand Sphere = 0.46 m

Explanation:

First we need to find the volume of gold sphere:

Vg = (4/3)πr³

where,

Vg = Volume of gold sphere = ?

r = radius of gold sphere = 2 cm = 0.02 m

Therefore,

Vg = (4/3)π(0.2 m)³

Vg = 0.0335 m³

Now, we find mass of the gold:

ρg = mg/Vg

where,

ρg = density of gold = 19300 kg/m³

mg = mass of gold = ?

Vg = Volume of gold sphere = 0.0335 m³

Therefore,

mg = (19300 kg/m³)(0.0335 m³)

mg = 646.75 kg

Now, the volume of sand required for equivalent mass of gold, will be given by:

ρs = mg/Vs

where,

ρs = density of sand = 1602 kg/m³

mg = mass of gold = 646.75 kg

Vs = Volume of sand = ?

Therefore,

1602 kg/m³ = 646.75 kg/Vs

Vs = (646.75 kg)/(1602 kg/m³)

<u>Vs = 0.4 m³</u>

Now, for the radius of sand sphere to give a volume of 0.4 m³, can be determined from the formula:

Vs = (4/3)πr³

0.4 m³ = (4/3)πr³

r³ = 3(0.4 m³)/4π

r³ = 0.095 m³

r = ∛(0.095 m³)

<u>r = 0.46 m</u>

4 0
3 years ago
Other questions:
  • A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 205 Hz. A person on the pl
    5·1 answer
  • After a group of researchers conducts a new experiment that has never been conducted before, what is the best way for other scie
    6·2 answers
  • A car traveling south is 200 kilometers from its starting point after 2 hours. What is the average velocity of the car?
    7·2 answers
  • Which statements about acceleration are true?
    7·2 answers
  • Some fabrics build up better than others because
    9·2 answers
  • An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect
    10·1 answer
  • Which equation represents a neutralization reaction?
    10·1 answer
  • 20 points plz help i dont really understand how to do this
    13·1 answer
  • 2. Mrs. Stern is standing still on rollerblades on a frictionless floor in the middle of the A-gym while
    7·1 answer
  • At which latitude would tropical rain forests be most likely?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!