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aev [14]
2 years ago
11

A car weighs 14500 N. What is the mass?

Physics
2 answers:
Anarel [89]2 years ago
6 0

Answer:

1478.08 kg

Explanation:

Weight = mass * gravitational acceleration

14500 = mass * 9.81

mass = 1478.08 kg

Weight is depending on a gravitational acceleration . It's not depending on a mass. Mass doesnt chnage when we move object to one planet to another planet but. Gravitational acceleration is varying planet to planet .

Ira Lisetskai [31]2 years ago
5 0
The correct answer would be 1478.08 kg ! Hope this helps!
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Imagine two people standing at placemark A and placemark E, looking at each other across the fault. Which of the following state
lord [1]

Answer:

To both observers, the land opposite them is moving to the right.

Explanation:

I have this class too. and there is also a quizlet with all the answers to the rest of the other questions. Trust me its right .

https://quizlet.com/261219090/oce-1001-chapter-2-flash-cards/

5 0
2 years ago
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0
2 years ago
Nitrogen (n2) gas within a piston–cylinder assembly undergoes a compression from p1 = 20 bar, v1 = 0.5 m3 to a state where v2 =
Bingel [31]

Part a)

As we know that

P_1V_1^{1.35} = P_2V_2^{1.35}

here we know that

P1 = 20 bar

V1 = 0.5 m^3

V2 = 2.75 m^3

from above equation

20* 0.5^{1.35} = P * (2.75)^{1.35}

P = 2 bar

so final state pressure will be 2 bar

Part b)

now in order to find the work done

W = \int PdV

W = \int \frac{c}{V^{1.35}}dV

W = c\frac{V^{-0.35}}{-0.35}

W = \frac{P_1V_1 - P_2V_2}{0.35}

W = \frac{20* 0.5 - 2 * 2.75}{0.35}* 10^5 = 12.86 * 10^5 J

3 0
2 years ago
Read 2 more answers
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The question with the complete options:

Look at these two sentences about Undeposited Funds. 1. By posting to Undeposited Funds, you can create a single bank deposit for multiple payments, making it easy ___________. 2. When receiving a payment, make sure _________________. Which of the options below correctly fills in the blanks? A.)1. To match your bank register with your bank statement; 2. the Deposit to account is Undeposited Funds

B) 1. To match your bank register with your bank statement; 2. the Deposit to account is Checking

C)1. To match your expenses with your bank statement; 2. the Deposit to account is Uncategorized asset

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Answer: The correct option is A (1. To match your bank register with your bank statement; 2. the Deposit to account is Undeposited Funds)

Explanation: Undeposited funds is a type of account created to keep funds that are not yet deposited in the individuals account. It's a default account which is used by online marketers to keep funds until they are ready to be paid.

By posting to Undeposited Funds, you can create a single bank deposit for multiple payments, making it easy to match your bank register with your bank statement. When receiving a payment, make sure the Deposit to account is Undeposited Funds.

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1000 kcal because you only get 10% of the energy of the thing you eat
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