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3 years ago

A bird has a mass of 0.8 kg and flies at a speed of 11.2 m/s. How much kinetic energy does the bird have?

2 answers:

riadik2000 [5.3K]3 years ago

7 0

The kinetic energy of an object is directly proportional to its mass and the square of its velocity

KE = 1/2 (mv²)

KE = Kinetic Energy
m = mass in kg
v = velocity in m/s

Given:

m = .8 kg
v = 11.2 m/s

Substitute:

KE = 1/2 (.8)(11.2²)
KE = 50.18 J

scZoUnD [109]3 years ago

5 0

Answer:

50.2 like the other dude said

Explanation:

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a bus takes to reach from station A to station B and then 3 hour to return from station B to station A find the average velocity

Anton [14]

Answer:

you have probably missed some details in the question.

3 0

3 years ago

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a

BlackZzzverrR [31]

Answer:

E = 8.5 * 10^6 V/m

Explanation:

In general we have the following relation between the Electric Field and the Elecric Potential:

$\int\limits^2_1 {\vec{E} \cdot\vec{dl}} = V_{2} -V_{1}$

Due to the vector nature of the electric filed, we can only know the mean Electric field E across the membrane, and take it out from the integral, that is:

E = (ΔV)/L

Where L is the thickness of the membrane and ΔV is the potential difference.

Therefore:

E = 8.53933*10^6 V/m

rounding to the first tenth:

E = 8.5 * 10^6 V/m

3 0

3 years ago

Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a

Verizon [17]

Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.

Reasons:

The energy given to the block by the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = $0.5 \cdot k \cdot x^2$ = kinetic energy of

block = $0.5 \cdot m \cdot v^2$

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

The force of the weight of the block on the string, $F = m \cdot g \cdot sin(\theta)$

The energy given to the block = $0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)$ = The kinetic energy of block as it leaves the spring = $\mathbf{0.5 \cdot m \cdot v^2}$

Which gives;

$0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2$

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and c are constants

The graph of the equation a·x² + c·v² = b is an ellipse

Therefore;

As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.

Please find attached a drawing related to the question obtained from a similar question online

The possible question options are;

As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x

The relationship is no longer linear and v will be more for the same value of x

The relationship is still linear, with lesser value of v

The relationship is still linear, with higher value of v

The relationship is still linear, but vary inversely, such that as x increases, v decreases

Learn more here:

brainly.com/question/9134528

6 0

2 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$