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2 years ago

7

Ejbsjca bdsj dbcsc j ecsabkfskbj jcbaskjbb

1 answer:

alexandr1967 [171]2 years ago

3 0

Sorry but I can't help you cuz the words you use makes no sens if you are going to ask a question pls do and dont joke around about it thank you very much and have a great day! ^^

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A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go ?

AnnZ [28]

So, this is a problem where the accleration is not provided, since it is implied. The only acceleration is acceleration due to gravity (9.8 m/s)

The equation we will use for this problem is $V^2 =V_{0}^2 + 2a (X-X_0)$

V is the final velocity, V₀ is the initial velocity, a is the acceleration, X is the final height, and X₀ is the starting height.

We can assume that the ball starts on the ground since no height is given, so now we plug our numbers in.

We will use 0 as the final velocity, since the ball will stop moving upwards when it is the highest. We will use -9.8 since that is the acceleration due to gravity and we will use 22m/s as V₀ since that is the starting velocity.

$V^2 =V_{0}^2 + 2a (X-X_0)\\0^2 = 22^2 + 2\times-9.8(X-0)\\0=484-19.6x\\-484=-19.6x\\24.69387755 = x\\x\approx24.69$

So, the ball will go 24.69 meters up

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3 years ago

The lithosphere contains rocks, soils, and minerals.

sergij07 [2.7K]

Answer: TRUE

Explanation:

Hope this helped :))

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3 years ago

Read 2 more answers

Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo

Bingel [31]

Answer:

$t=14.678\times 10^{-3}s$

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

$Q=Q_o(1-e^{-\frac{t}{RC}})$

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

$C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})$

or

$0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}$

or

$e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1$

taking log both sides, we get

$ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})$

or

$-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)$

or

$t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}$

or

$t=14.678\times 10^{-3}s$

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3 years ago

Which does not decrease your chances of being injured

Scorpion4ik [409]

C the answer is c please give brainliest

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3 years ago

What is a small portable device that counts every step taken throughout the day?

Valentin [98]

Answer: B

Explanation: i learned it last year

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2 years ago