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GuDViN [60]
3 years ago
5

An object with a mass of 120 kilograms is moving at a velocity of 60 m/s. What is its momentum? A. 2 kg-m/s B. 0.5 kg-m/s C. 7,2

00 kg-m/s D. 3,600 kg-m/s
Mathematics
2 answers:
MakcuM [25]3 years ago
4 0

Answer:

The correct answer is C) 7,200.

Step-by-step explanation:

In order to find the answer for this, start by using the formula for momentum.

Mo = M*V

Mo = 120 * 60

Mo = 7,200

liq [111]3 years ago
3 0

Answer: C 7200 kg-m/s

Step-by-step explanation:

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Which of the following is the weakest r-squared value? Which one is the strongest?
Setler [38]

Answer:

Weakest value = 0.0196

Strongest value = 0.5929

Step-by-step explanation:

Given:

Choices

0.691

-0.14

-0.24

-0.77

Computation:

0.691² = 0.477481

-0.14² = 0.0196

-0.24² = 0.0576

-0.77² = 0.5929

Weakest value = 0.0196

Strongest value = 0.5929

7 0
3 years ago
What's bigger 7% or 0.7?
Dennis_Churaev [7]

7%

is the bigger number bc 0.7 is a decimal an it like ex.... 0.50th its like half of one whole percentage

do u see wat I mean


3 0
3 years ago
Evaluate -4(12) please help
alex41 [277]

Answer:

You are going to get -48

Step-by-step explanation:

For some of the reason, this seemed like multiplication

But anyway, hope this helps, The other person Put down an answer that maybe is incorrect and they posted it like a file like they just don't care like a BOT

3 0
2 years ago
Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)
Mnenie [13.5K]
One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
(x-h)^2+(y-k)^2=r^2
the center is (h,k) and the radius is r

and the distance formula is
distance between (x_1,y_1) and (x_2,y_2) is
D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}


r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle


so
(x+2)^2+(y-3)^2=18
(x-(-2))^2+(y-3)^2=(\sqrt{18})^2
(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2

the radius is 3\sqrt{2}
center is (-2,3)

find distance between (8,4) and (-2,3)

D=\sqrt{(8-(-2))^2+(4-3)^2}
D=\sqrt{(8+2)^2+(1)^2}
D=\sqrt{10^2+1}
D=\sqrt{100+1}
D=\sqrt{101}




r=3\sqrt{2}≈4.2
D=\sqrt{101}≈10.04

do r<D

(8,4) is outside the circle

6 0
3 years ago
Not sure how to do this
Sergio039 [100]
45 i believe it is i may be wrong!!!
5 0
2 years ago
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