Ask question

Ask question

All categories

3 years ago

10

a baseball was thrown with an initial velocity of 14 m/s at an angle above the horizontal . It remained in the air for 2 s. Whic

h of the following quantities was constant while the baseball was in the air?

1 answer:

liberstina [14]3 years ago

3 0

Answer:

Explanation:acceleration

You might be interested in

An example of a constant acceleration is

serious [3.7K]

An example is free fall ,

8 0

3 years ago

Read 2 more answers

A material through which a current does not move easily is a(n) _____.

barxatty [35]

<span>A material through which a current does not move easily is called
an insulator.

Technically, charges CAN move through an insulator, but they lose
a lot of energy doing it, so the current that flows through the insulator
is very very small, usually too small to even measure.

Another way to look at it: Insulators have high resistance.
</span>

6 0

3 years ago

A student uses the right-hand rule as shown. an illustration with a right hand with fingers curled and thumb pointed up. what is

Afina-wow [57]

The direction of the magnetic field in front of the wire closest to the student is on the left. The direction is found by the right-hand rule.

<h3>What is the right-hand rule?</h3>

The right-hand rule is a popular mnemonic for remembering how axes in three-dimensional space are oriented.

The fact that the three axes of three-dimensional space have two different orientations gives birth to the majority of the many left-hand and right-hand rules.

Using the right-hand rule, we can recall this diagram. Your thumb points in the direction of the magnetic force pushing on the moving charge

If you point your pointer finger in the direction of the positive charge and then your middle finger in the direction of the magnetic field.

To learn more about the right-hand rule refer to the link;

brainly.com/question/9750730

6 0

2 years ago

Help with these three

Elenna [48]

The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
$g = 9.81 \frac{m}{s {}^{2} }$
let's do it for this case:
$f_{g} = m \times g \\ f _{g} = 65kg \times 9.81 \frac{m}{s {}^{2} } = 637.65$
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
$f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} } \\ f = 325n$
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere

8 0

3 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago