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Alexandra [31]
2 years ago
11

A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc

eleration of the cart? A. 8.6 meters/second2 B. 9.0 meters/second2 C. 9.5 meters/second2 D. 10.5 meters/second2 E. 0.8 meters/second2
Mathematics
2 answers:
Snezhnost [94]2 years ago
6 0

Answer:

letter a

Step-by-step explanation:

arsen [322]2 years ago
4 0

Answer:

option A

a = 8.65 m/s²

Step-by-step explanation:

Given that,

force applied on a cart (forward direction) = 19N

frictional force experience by cart (backward direction) = 1.7N

mass of the cart = 2 kg

Frictional force always opposes applied force, so the Resultant force on the cart would have to be 19N - 1.7N.

Formula to use

Resultant force = ma

plug values in the formula

19 - 1.7 = 2(a)

17.3 = 2(a)

a = 8.65 m/s²

so the acceleration of the cart is 8.65m/s²

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The basic consumer right includes "Right to be informed" that states the availability of information required for weighing alternatives, and protection from false and misleading claims in advertising and labeling practices.

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7 0
2 years ago
In one baseball season Peter hit twice the difference of the number of home runs Alice hit and 6. altogether,they hit 18 home ru
Viktor [21]
<span>In a baseball season:
Peter hit = 2 (x - 6)
Total = 19 hits

X= Alex’s hit
Y = Peter’s Hit

=> Y = 2(x - 6)
=> x + y = 18
Let’s start from the second given equation:
=> x + y =18
=> y = 18 – x
Now, let’s use the first given equation to solve:
=> 18 – x = 2 (x - 6)
=> 18 – x = 2x – 12
=> -3x = -30
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Now, Let’s try:
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=> y = 2 (10 - 6)
=> y = 2 (4)
=> y = 8

Thus, Peter hit 8 and Alice hit 10.</span>



4 0
2 years ago
A recent poll of 124 randomly selected residents of a town with a population of 310 showed that 93 of them are opposed to a new
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The sample size is 124.
93 of them are opposed to new shopping center.

So, 
n = 124
p = \frac{93}{124}=0.75

The point estimate of the population proportion = p = 0.75
q = 1 - p = 0.25

Margin of error (E) can be calculated by:

E= Z_{c}  \sqrt{ \frac{pq}{n} }

Using the values, we get:

E=1.645 \sqrt{ \frac{0.75*0.25}{124} }=0.06

Therefore, the margin of error is approximately 0.06 or 6%.
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