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2 years ago

A 0.12 kg bird is flying at a constant speed of 7.8 m/s. what is the birds conetic energy?

Physics

2 answers:

Contact [7]2 years ago

8 0

Answer:

3.6504 Joules is the birds kinetic energy.

Explanation:

Mass of the bird, m = 0.12 kg

Speed by which bird is flying , v = 7.8 m/s

The kinetic energy of the flying bird is given by :K.E

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

$K.E.=\frac{1}{2}\times 0.12 kg\times (7.8 m/s)^=3.6504 Joules$

3.6504 Joules is the birds kinetic energy.

lana [24]2 years ago

7 0

KE=1/2mv^2 - equation for kinetic energy
KE=(1/2)(0.12 kg)((7.8 m/s)^2 - plug it into the formula
KE=(0.06 kg)(60.84 m/s) - multiply 1/2 to the mass and square the speed
KE= 3.7 J - answer
Hope this helps

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The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a

Amanda [17]

Answer:

amount of energy = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = $\frac{excess initial cost}{\Delta M}$

payback period = $\frac{32 + 66}{520.34}$

payback period = 0.188 year

8 0

2 years ago

Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

matrenka [14]

(a) does an electric field exert a force on a stationary charged object?
Yes. The force exerted by an electric field of intensity E on an object with charge q is
 $F=qE$
As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with charge q and speed v is
 $F=qvB \sin \theta$
where $\theta$ is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

(c) does an electric field exert a force on a moving charged object?
Yes, The intensity of the electric force is still
 $F=qE$
as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

(d) does a magnetic field do so?
Yes. As we said in point b, the magnetic force is
$F=qvB \sin \theta$
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

(f) does a magnetic field do so?
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
 $F=qE$
So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
 $F=qvB \sin \theta$
is different from zero because v is different from zero.

6 0

2 years ago

Suggest reasons why poaching for subsistence is likely to be less damaging to the biodiversity of an area than poaching for prof

dlinn [17]

An example for ruining a biodiversity is fishing. The two factors that have contributed to increased fishing in deep ocean waters in recent years are the human population growth and decreased fishing opportunities inshore. Increase population growth increases the demand for food which also leads to increase in fish demand. Because the fish demand is high, inshore fishing opportunities decrease that is why deep ocean waters is the new venue for fishing. This may sound absurd but poaching for subsistence is likely to be less damaging to he biodiversity of an area than poaching for profit. Because the people do not care anymore to the biodiversity that they interrupted just to get back more profit. They do not care what must be taken from it like getting bigger fishes and leaving the smaller ones behind to maintain productivity.

5 0

2 years ago

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

Ad libitum [116K]

Answer:

$F=(3i+3.6j)\ N$

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, $u=(1i+0j)\ m/s$

After 8 seconds, final velocity of the puck, $v=(6i+6j)\ m/s$

Let the x and y component of force is given by $F_x\ and\ F_y$ .

x component of force is given by :

$F_x=m\times \dfrac{v-u}{t}$

$F_x=4.8\times \dfrac{6-1}{8}$

$F_x=3\ N$

y component of force is given by :

$F_y=m\times \dfrac{v-u}{t}$

$F_y=4.8\times \dfrac{6-0}{8}$

$F_y=3.6\ N$

So, the component of the force is $F=(3i+3.6j)\ N$ . Hence, this is the required solution.

7 0

2 years ago

Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed

makkiz [27]

Using the given formula with v0=56 ft/s and h=40 ft
h = -16t2 + v0t
40 = -16t2 + 56t
16t2 - 56t + 40 = 0
Solving the quadratic equation:
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32
We have two possible solutions
t1 = (56+24)/32 = 2.5
t2 = (56-24)/32 = 1
So initially the ball reach a height of 40 ft in 1 second.

3 0

2 years ago