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2 years ago

13

BRAINLIEST FOR YOU IF YOU ANSWER

1 answer:

luda_lava [24]2 years ago

5 0

Answer:

d

Explanation:

i think

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develop a plan to test the hardness of a sample of feldspar using the following items glass plate copper penny and streak plate

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2 years ago

If the baseball and the plastic ball were moving at the same speed which ball would hit a bat harder

Lerok [7]

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2 years ago

CAN SOMEONE PLEASE ANSWER NUMBER 10 FOR ME WITH AN EXPLANATION HURRY I HAVE LIMITED TIME PLEASE!!!

lara31 [8.8K]

Answer: B

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4 0

2 years ago

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

tresset_1 [31]

Answer:

a) Acceleration of the car is given as

$a_{car} = -21 m/s^2$

b) Acceleration of the truck is given as

$a_{truck} = 10.15 m/s^2$

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

$m_1v_1 + m_2v_2 = (m_1 + m_2)v$

now we have

$1400 (6.32) + 2900(0) = (1400 + 2900) v$

$v = 2.06 m/s$

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

$a_{car} = \frac{v_f - v_i}{t}$

$a_{car} = \frac{2.06 - 6.32}{0.203}$

$a_{car} = -21 m/s^2$

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

$a_{truck} = \frac{v_f - v_i}{t}$

$a_{truck} = \frac{2.06 - 0}{0.203}$

$a_{truck} = 10.15 m/s^2$

5 0

2 years ago

wo plates with area 6.00×10−3 m2 are separated by a distance of 1.50×10−4 m . If a charge of 5.40×10−8 C is moved from one plate

liq [111]

Answer:

V = 152.542 volts

Explanation:

Given data:

area of plates $6.00\times 10^{-3} m^2$

distance between the plates is $1.50\times 10^{-4} m$

charge = $5.40\times 10^{-8} c$

we know that capacitance is given as

$C = \frac{\epsilon A}{d}$

$C = \frac{8.85\times 10^{-12} 6\times 10^{-3}}{1.50\times 10^{-4}}$

$C = 3.54\times 10^{-10} F$

potential difference is given as

$V =\frac{Q}{C} = \frac{5.40\times 10^{-8}}{3.54\times 10^{-10}}$

V = 152.542 volts

3 0

3 years ago