A skydiver is in free fall. What is the only force acting upon the skydiver.

A 23.0 kg iron weightlifting plate has a volume of 2920 cm3 . what is the density of the iron plate in g/cm3?

yanalaym [24]

The first thing you should know for this case is that density is defined as the quotient between mass and volume:
D = M / V
In addition, you should keep in mind the following conversion:
1Kg = 1000g
Substituting the values we have:
D = (23.0 * 1000) / (2920) = 7.88 g / cm ^ 3
answer
the density of the iron plate is 7.88 g / cm ^ 3

8 0

2 years ago

A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top

elixir [45]

Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

8 0

2 years ago

What are the units of acceleration?

Marina86 [1]

Acceleration is any change in speed or direction of motion.

The dimension of speed is [length/time],
so a change is [length/time²].

Popular units include [meter/second²] and [feet/second²] .
________________________

Direction almost always boils down to an angle, (which technically
has no dimensions), so a change in direction is [angle/time] .

Popular units include [radian/second] and [degree/second] .

5 0

3 years ago

A roller coaster's velocity at the top of the hill is 10 m/s. Two seconds later it reaches the bottom of the hill with a velocit

svet-max [94.6K]

your answer is 8 m / sec²

5 0

2 years ago

At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There

katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

$\tau_{net} = I \alpha$ (1)

where

$\tau_{net}$ is the net torque acting on the object in rotation

I is the moment of inertia of the object

$\alpha$ is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

$\alpha = \frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time interval

So we can rewrite eq.(1) as

$\tau_{net}=I\frac{\Delta \omega}{\Delta t}$

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

$\tau_{net}=0$

From the previous equation, this implies that

$\Delta \omega =0$

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

$\tau_{net}=0$

Therefore, this means that

$\Delta \omega =0$

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0

3 years ago