Ask question

Ask question

All categories

3 years ago

9

A 5.36 kg object falls freely (ignore air resistance), after being dropped from rest. Determine the initial kinetic energy (in J

), the final kinetic energy (in J), and the change in kinetic energy (in J) for the following.
(a) first meter of fallinitial kinetic energy? Jfinal kinetic energy? Jchange in kinetic energy? J(b) second meter of fallinitial kinetic energy? Jfinal kinetic energy? Jchange in kinetic energy

1 answer:

matrenka [14]3 years ago

6 0

Answer: 52.53

Explanation:

m*g = 5.36k * 9.8N/kg = 52.53 N. = Wt.

of object.

a. KE = 0 J. = Initial KE

V^2 = Vo^2 + 2g*h

V^2 = 0 + 19.6*1 = 19.6

V = 4.43 m/s.

KE = 0.5m*V^2 = 2.68*4.43^2 = 52.59 J.

b. V^2 = Vo^2 + 2g*h

V^2 = 4.43^2 + 19.6*1 = 39.22

V = 6.26 m/s.

KEo = 2.68*4.43^2 = 52.59 J.

KE = 2.68*6.26^2 = 105 J.

You might be interested in

what is the initial velocity of a go-kart traveling at a uniform acceleration of 0.5 m/s^2 for 5s as it slows down to a stop?

timurjin [86]

The initial velocity of go-kart is 2.5 m/s.

<u>Explanation:</u>

Here, the uniform acceleration of go-kart is given as 0.5 m/s². Also the time required by it to stop is also given as 5 s. As acceleration is the measure of change in velocity per unit time.

In this case, the velocity should be changed from a value to zero to come to rest. So the initial velocity will be positive value and final velocity is zero.

As we know the values of acceleration, final velocity and time, the initial velocity can be easily determined as follows.

$Acceleration = \frac{Final velocity -Initial velocity}{Time}$

Since, final velocity is zero, acceleration is 0.5 m/s² and time is 5 s, then,

$-0.5=\frac{-\text {Initial velocity}}{5}$

Initial velocity = 0.5 × 5 = 2.5 m/s.

So the initial velocity of go-kart is 2.5 m/s.

8 0

3 years ago

If a net force acting on a moving object causes its speed to increase, what

Klio2033 [76]

Answer:

It increases.

Explanation:

If a force acts in the same direction as the object's motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed.

7 0

3 years ago

An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.

I am Lyosha [343]

Answer:

The density of this object is approximately $1.36\; {\rm kg \cdot L^{-1}}$ .

The density of the oil in this question is approximately $0.600\; {\rm kg \cdot L^{-1}}$ .

(Assumption: the gravitational field strength is $g =9.806\; {\rm N \cdot kg^{-1}}$ )

Explanation:

When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

$\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}$ .

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was $(63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}$ . Hence, the weight of water that this object displaced would be $47.0 \; {\rm N}$ .

The mass of water displaced would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}$ .

The volume of that much water (which this object had displaced) would be:

$\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}$ .

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately $4.793\; {\rm L}$ .

The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately $4.793\; {\rm L}$ .

The weight of oil displaced would be equal to the magnitude of the buoyancy force: $63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}$ .

The mass of that much oil would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}$ .

Hence, the density of the oil in this question would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

7 0

2 years ago

Adding or removing thermal energy to or from a substance does not always cause its temperature

julia-pushkina [17]

Change in thermal energy not always cause it's temperature change. It is the situation when water reaches either at 0 C or 100 C then thermal energy doesn't cause change in temperature instead it changes the state of matter.

In short, Your Answer would be "True"

Hope this helps!

6 0

3 years ago

Read 2 more answers

What is the force of an object with a mass of 65 kg and an an unknown acceleration?

Alik [6]

We know, F = m.a

F = 65 * a

Where, F = force

a = unknown acceleration

8 0

3 years ago