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3 years ago

which cell structure is responsible for taking in and breaking down foreign particles and damaged organelles?

Physics

1 answer:

Klio2033 [76]3 years ago

4 0

Answer:

lysosomes

Explanation:

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2 Ca + O 2 → 2 CaO What is the product of the reaction

scoray [572]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The product in a chemical reaction is written on the Right side of the arrow, so

the product formed here in given reaction is :

CaO (Calcium Oxide)

5 0

2 years ago

Two rods are made from materials that have different Young's moduli. The rods are constructed to have the same length and same c

ANTONII [103]

Answer:

This is true,the rod with smaller elastic modulus will stretch more than larger elastic modulus.

Explanation:

σ=E*ε

ε=δ/L

σ=E*δ/L

δ=(σ*L)/E

σ=F/A

δ=(F*L)/(A*E)

As Force,Area and Length is same

δ∞1/E

From the expression as E increase δ will be small,so there will be more stretch for smaller elastic modulus.

5 0

3 years ago

A student rides a bicycle for 15 miles in 3 hours. What is the student's speed? What else would you need to know for the velocit

kaheart [24]

Answer:

5 miles per hour

Explanation:

if you divide 15 by 3 you get 5, therefore the student is going 5 miles per hour.

3 0

3 years ago

A 1400 kg car traveling at 17.0 m/s to the south collides with a 4700 kg truck that is at rest. The car and truck stick together

STatiana [176]

Answer:

Final velocity = 7.677 m/s

KE before crash = 202300 J

KE after crash = 182,702.62 J

Explanation:

We are given;

m1 = 1400 kg

m2 = 4700 kg

u1 = 17 m/s

u2 = 0 m/s

Using formula for inelastic collision, we have;

m1•u1 + m2•u2 = (m1 + m2)v

Where v is final velocity after collision.

Plugging in the relevant values;

(1400 × 17) + (4700 × 0) = (1400 + 1700)v

23800 = 3100v

v = 23800/3100

v = 7.677 m/s

Kinetic energy before crash = ½ × 1400 × 17² = 202300 J

Kinetic energy after crash = ½(1400 + 1700) × 7.677² = 182,702.62 J

8 0

3 years ago

How much charge does a 9.0 v battery transfer from the negative to the positive terminal while doing 39 j of work?

sdas [7]

The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
$W=q \Delta V$
where q is the charge and $\Delta V$ is the potential difference.

In our problem, the work done is W=39 J while the potential difference of the battery is $\Delta V = 9.0 V$ , so we can find the charge transferred by the battery:
$q= \frac{W}{\Delta V}= \frac{39 J}{9.0 V}=4.33 C$

3 0

3 years ago