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1 year ago

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Austin's truck has a mass of 2000 kg when traveling at 22.0 m/s, it brakes to a stop in 4.0 s. show that the magnitude of the br

aking force acting on the truck is 11,000 n

2 answers:

olga2289 [7]1 year ago

5 0

Since F=m•a, you want to show that a = -5.5

Andrei [34K]1 year ago

5 0

Since the truck was in motion before the brakes were applied, it will decelerate within the 4s before coming to a stop. Hence the acceleration is $-a\:ms^{-2}$ .

When the truck comes to a stop, It will have a final velocity of $v=0\:ms^{-1}$ .

Also the initial velocity is 22.0 m/s. This means, $u=22.0\:ms^{-1}$ and time, $t=4s$ .

We can use the relation,

$v=u+at$

to determine the acceleration of the truck.

Let us now plug in all the values to obtain,

$0=22.0+(-a)(4)$

$\Rightarrow -22.0=-4a$

$\Rightarrow \frac{-22.0}{-4}=a$

$\Rightarrow a=5.5 ms^{-2}$

Using the relation,

$F=ma$

the magnitude of the braking force is

$F=2000 \times 5.5 N=11000N$

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Answer:

$a = - \frac{3}{4}$

Step-by-step explanation:

The correct solution of the given equation is as follows :

We have,

$\frac{1}{64} = 16^{2a}$

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⇒ $2^{(2 \times (-3))} = 2^{8a}$

⇒ $2^{- 6} = 2^{8a}$

Comparing the power of equal base, we get

- 6 = 8a

⇒ $a = - \frac{6}{8}$

⇒ $a = - \frac{3}{4}$

Therefore, before equating the powers of two terms the base of the terms should be equal and here this is the error. (Answer)

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1 year ago

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Suppose a continuous probability distribution has an average of μ=35 and a standard deviation of σ=16. Draw 100 times at random

yulyashka [42]

Answer:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, we have that the mean is $\mu*n$ and the standard deviation is $s = \sigma \sqrt{n}$

In this problem, we have that:

$\mu = 100*35 = 3500, \sigma = \sqrt{100}*16 = 160$

This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.

X = 4000

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{4000 - 3500}{160}$

$Z = 3.13$

$Z = 3.13$ has a pvalue of 0.9991

X = 3000

$Z = \frac{X - \mu}{s}$

$Z = \frac{3000 - 3500}{160}$

$Z = -3.13$

$Z = -3.13$ has a pvalue of 0.0009

0.9991 - 0.0009 = 0.9982

So the correct answer is:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

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2 years ago

Harriet is cultivating a strain of bacteria in a petri dish. Currently, she has 103 bacteria in the dish. The bacteria divide ev

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Based on the conditions given above, the number of bacteria at any time t (in hours) is calculated by the equation,
at = (a1)(2^t/2)

where a1 is the initial number of bacteria and at is the number at any time t. Substituting the givens,
a6 = (103)(2^6/2) = 824

Thus, there are 824 bacteria after 6 hours.

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1 year ago

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Gala2k [10]

We know that

[length of a circumference ]=2*pi*r
for r=3 ft

[length of a circumference ]=2*pi*3-----------> 18.84 ft

<span>The angular velocity of the particle is 18 radians per minute.

if 2pi radians (full circle)------------------> 18.84 ft
18 radians----------------------------------> X
X=18*18.84/(2*pi)---------> X=54 ft
so
</span>The linear velocity of the particle is 54 ft/minute

the answer is
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1 year ago

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Eli needs pieces of ribbon cut into strips that are 2 1\2 feet long. The roll of ribbon is 7 1\2 feet long. Which expression cou

julsineya [31]

Answer: There are 3 strips that can be cut from the roll of ribbon.

Step-by-step explanation:

since we have given that

Length of a ribbon is given by

$7\frac{1}{2}\ ft\\\\=\frac{15}{2}\ ft$

Length of pieces of ribbon cut into strips is given by

$2\frac{1}{2}\ ft\\\\=\frac{5}{2}\ ft$

So, we need to find the number of strips that can be cut is given by

$\text{ Number of strips }=\frac{\text{Length of roll}}{\text{ Length of strip}}\\\\=\frac{\frac{15}{2}}{\frac{5}{2}}\\\\=\frac{15\times 2}{2\times 5}\\\\=\frac{15}{5}\\\\=3$

Hence, there are 3 strips that can be cut from the roll of ribbon.

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2 years ago

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