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4 years ago

10

What kinds of birds visits you a feeder at different times a year

1 answer:

likoan [24]4 years ago

6 0

Purple finches, chickadees,cardinals,blue jays,woodpeckers,red polls,juncos,goldfinches,and red crossbills.

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The velocity of a particle traveling in a straight line is given by v = (6t - 3t²) m/s, where t is in seconds. If s = 0 when t =

qaws [65]

Answer

given,

v = (6 t - 3 t²) m/s

we know,

$v = \dfrac{dx}{dt}$

$a = \dfrac{dv}{dt}$

position of the particle

$dx = v dt$

integrating both side

$\int dx = (6t - 3 t^2)\int dt$

x = 3 t² - t³

Position of the particle at t= 3 s

x = 3 x 3² - 3³

x = 0 m

now, particle’s deceleration

$a = \dfrac{dv}{dt}$

$a = \dfrac{d}{dt}(6t - 3 t^2)$

a = 6 - 6 t

at t= 3 s

a = 6 - 6 x 3

a = -12 m/s²

distance traveled by the particle

x = 3 t² - t³

at t = 0 x = 0

t = 1 s , x = 3 (1)² - 1³ = 2 m

t = 2 s , x = 3(2)² - 2³ = 4 m

t = 3 s , x = 0 m

total distance traveled by the particle

D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s

D = 2 + 4 + 2 = 8 m

average speed of the particle

$v_{avg} = \dfrac{distance}{time}$

$v_{avg} = \dfrac{8}{3}$

$v_{avg} =2.67\ m/s$

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3 years ago

An astronaut weighs 8.00x10^2 newtons on the surface of Earth. What is the weight of the astronaut 6.37x10^6 meters above the su

Orlov [11]

800/2^2=2.oo10^10^2N

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4 years ago

Why does the mass of a candle decrease after burning?

34kurt

Because when a candle burns the wax is burning off so as its heated more and more wax is lost and if more wax is lost... Obviously it means that the candles mass will be lower :)

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4 years ago

As a result of mitosis, the cels of a Molecular organism sure which of these properties select to correct answers. A. All cells

KATRIN_1 [288]

Answer:

C. All cells have identical genetic information.

4 0

3 years ago

Read 2 more answers

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc

____ [38]

Answer:

Part a)

$a = 0.36 m/s^2$

Part b)

$a = -1.29 m/s^2$

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

$F_x = Fcos25$

$F_x = 80 cos25 = 72.5 N$

$F_y = Fsin25$

$F_y = 80 sin25 = 33.8 N$

now in vertical direction we can use force balance for the box to find the normal force on it

$F_n + F_y = mg$

$F_n = (25)(9.81) - 33.8$

$F_n = 211.45 N$

now kinetic friction on the box opposite to applied force due to rough floor is given as

$F_k = \mu F_n$

$F_k = (0.300)(211.45) = 63.44 N$

now the net force on the box in forward direction is given as

$F_{net} = F_x - F_k$

$F_{net} = 72.5 - 63.44 = 9.065 N$

now the acceleration of the box is given as

$a = \frac{F_{net}}{m}$

$a = \frac{9.065}{25} = 0.36 m/s^2$

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

$F_x = 72.5 N$

$F_y = 33.8 N$

now force balance perpendicular to inclined plane is given as

$F_n + F_y = mgcos\theta$

$F_n = (25)(9.81)cos10 - 33.8 = 207.7 N$

now the friction force opposite to the motion on the box is given as

$f_k = \mu F_n$

$f_k = (0.300)(207.7) = 62.3 N$

now the net pulling force along the inclined plane is given as

$F_{net} = F_x - F_k - mgsin10$

$F_{net} = 72.5 - 62.3 - (25)(9.81)sin10$

$F_{net} = -32.38 N$

now the box will decelerate and it is given as

$a = \frac{F_{net}}{m}$

$a = \frac{-32.38}{25} = -1.29 m/s^2$

7 0

3 years ago