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ioda
3 years ago
10

A group of students conducted several trials of an experiment to study Newton’s second law of motion. They concluded that tripli

ng the mass required tripling the net force applied. What quantity were the students holding constant?
net force

acceleration <-- MY ANSWER

number of trials

mass
Mathematics
1 answer:
padilas [110]3 years ago
7 0

Answer:

Acceleration

Step-by-step explanation:

The Newton's second law of motion can be expressed as;

F = ma

Where F is the force acting on an object, m is the mass of the object and a is the acceleration of the object.

For an object of mass 2kg and acceleration of 5 m/s^{2}, we have;

F_{1} = 2 × 5 = 10 N

If we triple the mass,

m = 2 × 3 = 6 kg

Therefore,

F_{2} = 6 × 5 = 30 N

Thus, at constant acceleration, F_{2} = 3 × F_{1}

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Put the correct answer in each blank.
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Answer:

H or (3.5,-3)

number two:

B and (3.5,5)

Step-by-step explanation:

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3 years ago
4 - (-3) simplify the expression
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Answer:

7

Step-by-step explanation:

4-(-3)=4+3 (negative times negative is positive)

4+3=7

Hope that helps!

6 0
3 years ago
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Question 5 of 10
suter [353]
C. Similar -SAS is the answer I think
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3 years ago
Abby, Billy and Cathy fire one shot each at a target. The probability that Abby will hit the target is 1/5. The probability that
Tems11 [23]

Answer:

1.  0.0167

2. 0.2

3. 0.6

Step-by-step explanation:

Probabilities on Independent Events

If A and B are independent events (the occurrence of A doesn't affect the occurrence of B and vice-versa), then the probability that both events occur is

\displaystyle P(A\bigcap B)= \ P(A).P(B)

Being P(A) and P(B) the individual probability of each independent event

The probability that A does not occur is  

\displaystyle P(\bar{A})=1-P(A)

The probability that B does not occur is

\displaystyle P(\bar{B})=1-P(B)

The probability that C does not occur is

\displaystyle P(\bar{C})=1-P(C)

We have 3 independent events. We know that because they fire together, no mutual affectation can happen

.

The probability that Abby will hit the target is 1/5.  

\displaystyle P(A)=\frac{1}{5}

The probability that Billy will hit the target is 1/4.

\displaystyle P(B)=\frac{1}{4}

The probability that Cathy will hit the target is 1/3

\displaystyle P(C)=\frac{1}{3}

Part 1.

The probability that all three shots hit the target is

\displaystyle P(A\bigcap B\bigcap C)=\frac{1}{5}.\frac{1}{4}.\frac{1}{3}

\displaystyle P(A\bigcap B\bigcap C)=\frac{1}{60}

The probability that all three shots hit the target is

\displaystyle P= \frac{1}{60}=0.0167

Part 2.

The probability that only Cathy's shot hits the target is computed assuming Abby and Billy won't succeed

.

\displaystyle P(\bar{A}\bigcap \bar{B}\bigcap C)=P(\bar{A})\  P(\bar{B})\ P(C)=(1-\frac{1}{5})\ (1-\frac{1}{4})\ (\frac{1}{3})=\frac{4}{5}.\frac{3}{4}.\frac{1}{3}=\frac{1}{5}=0.2

3.

The probability that at least one shot hits the target is when one of them succeeds, two of them succed or all of them succeed

\displaystyle P(A\bigcap \bar{B}\bigcap \bar{C})+P(\bar{A}\bigcap {B}\bigcap \bar{C})+P(\bar{A}\bigcap\bar{B}\bigcap C)+P(A\bigcap B\bigcap \bar{C})+P(A\bigcap \bar{B}\bigcap C)+P(\bar{A}\bigcap B\bigcap C)+P(A\bigcap B\bigcap C)

But it's easier to find the negated probability of the above, i.e. we compute the probability that NO ONE hits the target and subtract it from 1

\displaystyle P=1-P(\bar{A}\bigcap \bar{B}\bigcap \bar{C})

P=1 -(1-\frac{1}{5})(1-\frac{1}{4})(1-\frac{1}{3})=1-\frac{4}{5}.\frac{3}{4}.\frac{2}{3}=1-\frac{2}{5}=\frac{3}{5}

P=0.6

6 0
3 years ago
What is the distance between two points with coordinates (12,0) and (0,35)
Rudiy27

Answer:

caca I I I ii I I I I I I I I I I I I I I I.

6 0
3 years ago
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