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raketka [301]
3 years ago
10

What do glucose starch and cellulose have in common

Physics
1 answer:
Andreas93 [3]3 years ago
6 0

Answer:

Glucose, starch and cellulose are all carbohydrates.

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Kinetic Energy Assignment: Lab Report
kompoz [17]

Answer:

so what do we gotta do?

Explanation:

is there any book or homework?

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3 years ago
Which of the following is one way of preventing poisoning?
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I think its number 1

7 0
3 years ago
In the sum →A+→B=→C, vector →A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the +x direction, and vector
icang [17]

Answer:

Explanation: Ok, first caracterize the two vectors that we know.

A = ax + ay = (12*cos(40°)*i + 12*sin(40°)*j) m

now, see that C is angled 20° from -x, -x is angled 180° counterclockwise from +x, so C is angled 200° counterclockwise from +x

C = cx + cy = (15*cos(200°)*i + 15*sin(200°)*j) m

where i and j refers to the versors associated to te x axis and the y axis respectively.

in a sum of vectors, we must decompose in components, so: ax + bx  = cx and ay + by = cy. From this two equations we can obtain B.

bx= (15*cos(200°) - 12*cos(40°)) m = -23.288 m

by = (15*sin(200°) - 12*sin(40°)) m = -12.843 m

Now with te value of both components of B, we proceed to see his magnitude an angle relative to +x.

Lets call a to the angle between -x and B, from trigonometry we know that tg(a) = by/bx, that means a = arctg(12.843/23.288) = 28.8°

So the total angle will be 180° + 28.8° = 208.8°.

For the magnitude of B, lets call it B', we can use the angle that we just obtained.

bx = B'*cos(208.8°) so B' = (-23.288 m)/cos(208.8°) =  26.58 m.

So the magnitude of B is 26.58 m.

7 0
3 years ago
Which of the following shows kinetic energy being converted into potential energy?
Komok [63]
B) a rock being tossed high into the air
3 0
3 years ago
An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be
lord [1]

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0
3 years ago
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