A 40 N force is applied to an object with a mass of 0.1 kg on a frictionless surface. What is the acceleration of the object?

Answers

Answer 1

Answer:

The acceleration of the object is 400 m/s²

What is the acceleration of the object?

To find the acceleration of an object when a 40 N force is applied to it with a mass of 0.1 kg on a frictionless surface, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

The formula for Newton's second law of motion is:

F = ma

where:

F is the net force,

m is the mass of the object,

a is the acceleration.

In this case, the net force is 40 N and the mass of the object is 0.1 kg.

Substituting these values into the formula, we get:

40 N = (0.1 kg) a

Solving for a, we get:

a = 400 m/s²

Therefore, the acceleration of the object is 400 m/s²

Read more about acceleration of object at: brainly.com/question/605631

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Answer 2

Answer:

Answer:

400

Step-by-step explanation:

40 divided by .1

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What is the slope of the line passing through the given points.
(-3,2), (6,-1)

Answers

Answer:

-1/3

Step-by-step explanation:

Since we are given two points, we can find the slope using

m = ( y2-y1)/(x2-x1)

= ( -1-2)/( 6- -3)

= ( -1-2) /( 6+3)

= -3/9

= -1/3

Answer:

Slope (m) = $(-1)/(3)$ = -0.33333333...

Step-by-step explanation:

Slope (m) = ΔY -1

ΔX 3

(m)= -1/3

In a random sample of 500 college students, 23% say that they read or watch the news every day. Develop a 90% confidence interval for the proportion of all students who read or watch the news on a daily basis. Interpret your results. If you wanted to develop a 95% confidence interval with a margin of error of .01, how many students would need to be surveyed?

Answers

Answer:

The 90% confidence interval is $0.199 < p < 0.261$

The sample size to develop a 95% confidence interval is $n = 2032$

Step-by-step explanation:

From the question we are told that

The sample size is n =500

The sample proportion is $\^ p = 0.23$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $(\alpha )/(2)$ is

$Z_{(\alpha )/(2) } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{(\alpha )/(2) } * \sqrt{(\^ p (1- \^ p))/(n) }$

=> $E = 1.645 * \sqrt{(0.23 (1- 0.23))/(500) }$

=> $E = 0.03096$

Generally 90% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.23 -0.03096 < p < 0.23 + 0.03096$

=> $0.199 < p < 0.261$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $(\alpha )/(2)$ is

$Z_{(\alpha )/(2) } = 1.96$

The margin of error is given as $E = 0.01$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{(\alpha )/(2) }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n = [(1.96 )/(0.01) ]^2 *0.23 (1 - 0.23 )$

=> $n = 2032$

Will mark Brainlest (from a deck of cards,pemba withdraw a card at random what is the probability that the card is queen) step by using formula

Answers

Answer:

1/13

Step-by-step explanation:

Total cards = 52

Number of Queen = 4

Probability of the chosen card to be queen

$=(Number \ of \ queen)/(total \ number \ of \ cards)\n\n=(4)/(52) \n\n= (1)/(13)$

Answer:

1/13

Step-by-step explanation:

there are total no of 52 cards

out of that there are 4 queen

propability = tatal no of favorable outcomes / total no of possible outcomes

=4 / 52

=1/13

Solve the system of equation by elimination. x+2y= -7
-x-2y=8

A. (8,7)
B. No solution
C. (-7,8)
D. (-8,7)

If you can, please explain step by step:)

Answers

Answer: D

Step-by-step explanation:

D. (-8,7) just do it

Solve the equation:

a+20=11

Answers

Answer:

Step-by-step explanation:

a +20=11

-20-20

a= -9

subtract 20 on each side, you get a=-9

Consider a colony of E.Coli bacteria that is growing exponentially. A microbiologist finds that, initially, 1,000 bacteria are present and 50 minutes later there are 10,000 bacteria. a) Find expression for the number of bacteria Q(t) after t minutes. b) When will there be 1,000,000 bacteria?

Answers

Answer:

a) $Q(t)=Q_(0)e^((ln10/50)t)$

b) 150min

Step-by-step explanation:

a) Using the formula for colony growth $Q(t)=Q_(0)e^(kt)$ we need to find the specific value of k, to do this you'll use the time, initial and final number of colonies provided in your problem.

$Q(t)=Q_(0)e^(kt)\n\n10000=1000e^(k(50)) \n\n(10000)/(1000)=e^(50k) \n\n10=e^(50k)\n ln(10=e^(50k))\n\nln10=50k\n\nk=ln10/50\n\nQ(t)=Q_(0)e^((ln10/50)t)\n\n$

b) Once we have our expression we only have to use our final and initial number of bacteria to find t.

$Q(t)=Q_(0)e^((ln10/50)t)\n\n\n1000000=1000e^(0.046t)\n 1000=e^(0.046t)\nln( 1000=e^(0.046t))\nln1000=0.046t\nt=ln1000/0.046\nt=150 minutes$

I hope you find this information useful! good luck!

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