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Nastasia [14]
3 years ago
7

10. A 25 kg apple cart is being pushed with an applied force of 115 N. The coefficient of friction between the ground and the ca

rt is 0.35. What is the acceleration of the cart?
Physics
1 answer:
ziro4ka [17]3 years ago
8 0

Answer:

1.1 m/s²

Explanation:

From the question,

F -mgμ = ma.................... Equation 1

Where F = applied force, m = mass of the apple cart, g = acceleration due to gravity, μ =  coefficient of friction., a = acceleration of the apple cart.

Given: F = 115 N, m = 25 kg,  μ  = 0.35

Constant: g = 10 m/s²

Substitute these values into equation 2

115-(25×10×0.35) = 25×a

115-87.5 = 25a

25a = 27.5

a = 27.5/25

a = 1.1 m/s²

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2 years ago
The "Garbage Project" at the University of Arizona reports that the amount of paper discarded by households per week is normally
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P(X\geq a)=P(z\geq \frac{a-\mu}{\sigma})

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7 0
3 years ago
Convert the following dB to decimal a. -12 b. 3 c. 10 d. 0
stiks02 [169]

Answer:

for -12db

 \frac{V1}{V2}=0.251\\\\\frac{P1}{P2}=0.0631

for 3db

\frac{V1}{V2}=\sqrt{2}\\\\\frac{P1}{P2}=2

for 10db

\frac{V1}{V2}=3.16\\\\\frac{P1}{P2}=10

for 0db

\frac{V1}{V2}=1\\\\\frac{P1}{P2}=1

Explanation:

The decibel is a logaritmic value given by:

db=10*log(\frac{P1}{P2})=20*log(\frac{V1}{V2})

we use 10 for power values and 20 for other values such voltages or currents.

\frac{V1}{V2}=10^{\frac{db}{10}}\\\\\frac{P1}{P2}=10^{\frac{db}{20}}

for -12db

\frac{V1}{V2}=10^{\frac{-12}{10}}=0.251\\\\\frac{P1}{P2}=10^{\frac{-12}{20}}=0.0631

for 3db

\frac{V1}{V2}=10^{\frac{3}{10}}=\sqrt{2}\\\\\frac{P1}{P2}=10^{\frac{3}{20}}=2

for 10db

\frac{V1}{V2}=10^{\frac{10}{10}}=3.16\\\\\frac{P1}{P2}=10^{\frac{10}{20}}=10

for 0db

\frac{V1}{V2}=1\\\\\frac{P1}{P2}=1

4 0
3 years ago
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