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3 years ago

6

PLEASE HELP WILL GIVE BRAINLIEST (I NEED TWO PICKED) PLZzZZZZ AND THxS

1 answer:

vovangra [49]3 years ago

5 0

Answer:

A and B

Explanation:

They contain Chlorophyll

They are relatively tall

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What type of machine is wire cutter pliers?

Sonja [21]

It seems that you have missed the necessary options for us to answer this question, but anyway, here is the answer. The type of machine that a wire cutter pliers is classified is a simple machine. When we say simple machine, this is the type of machine that is considered basic wherein you need to apply force for it to function. Hope this helps.

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3 years ago

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The temperature of a smelting furnace is found to be 2000 degree Celsius.find the temperature on Fahrenheit scale

Dima020 [189]

Answer:

3632.

Explanation:

Based on the formula:

2000×(9/5)+32=3632

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2 years ago

A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an a

kondor19780726 [428]

RESULT: Twalk= 3.64 hr

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2 years ago

A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done to compress the spring initially= $\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J$

b.

By conservation law of energy

Initial energy of spring=Kinetic energy of object

$37.8=\frac{1}{2}(3)v^2$

$v^2=\frac{37.8\times 2}{3}$

$v=\sqrt{\frac{37.8\times 2}{3}}$

v=5.02 m/s

c.Work done by friction on the incline, $w_{friction}=P.E-spring \;energy$

$W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J$

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3 years ago

A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction

finlep [7]

consider the motion in x-direction

$v_{ox}$ = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

$a_{x}$ = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = $v_{ox}$ t + (0.5) $a_{x}$ t²

100 = $v_{ox}$ (4.60)

$v_{ox}$ = 21.7 m/s

consider the motion along y-direction

$v_{oy}$ = initial velocity in y-direction = ?

Y = vertical displacement = 0 m

$a_{y}$ = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = $v_{oy}$ t + (0.5) $a_{y}$ t²

0 = $v_{oy}$ (4.60) + (0.5) (- 9.8) (4.60)²

$v_{oy}$ = 22.54 m/s

initial velocity is given as

$v_{o}$ = sqrt(( $v_{ox}$ )² + ( $v_{oy}$ )²)

$v_{o}$ = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

6 0

3 years ago