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3 years ago

How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m?(gravity is 9.8)

1 answer:

8 0

First, calculate the initial velocity of the dog given with the vertical height and the acceleration due to gravity which is calculated through the equation,
2ad = Vo²
Substituting the known values,
2(9.8 m/s²)(1.2 m) = V₀²
V₀ = 4.85 m/s
The kinetic energy is solved through the equation,
KE = 0.5mv²
Substituting the known values to the latest equation,
KE = 0.5 (7.2 kg)(4.85 m/s)²
KE = 17.46 J
Thus, the kinetic energy is 17.46 J.

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If a point has 40 J of energy and the electric potential is 8 V, what must be the charge?

Alekssandra [29.7K]

If a point has 40 J of energy and the electric potential is 8 V, the charge must be: A. 5 C

Given the following the details;

Energy = 40 Joules

Electric potential = 8 Volts

To find the quantity of charge;

Mathematically, the quantity of charge with respect to electric potential is given by the formula;

$Quantity \; of \; charge = \frac{Energy}{Electric \; potential}$

Substituting the values into the formula, we have;

$Quantity \; of \; charge = \frac{40}{8}$

Quantity of charge = 5 Coulombs

Therefore, the quantity of charge must be 5 Coulombs.

Find more information: brainly.com/question/21808222

8 0

3 years ago

Read 2 more answers

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

3 years ago

What is the best unit to use when measuring the mass of a mineral sample?

Ksivusya [100]

Measuring density: Measure the mass (in grams) of each mineral sample available to you. The mass of each sample is measured using a balance or electronic scale. Record mass on a chart.

8 0

3 years ago

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

densk [106]

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

$F_f = \mu m g$