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alisha [4.7K]
3 years ago
15

A woman lifts her 100-newton child up one meter and carries her for a distance of 50 meters to the child's bedroom. How much wor

k does the woman do?
Physics
2 answers:
tankabanditka [31]3 years ago
5 0
100 J

Please mark me brainliest it would be greatly appreciated haha
garik1379 [7]3 years ago
5 0

Answer:

100 J

Explanation:

magic

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Every force has a specific and identifiable cause called a(n)...............
goblinko [34]
1 Bc I just did it and got it right
8 0
2 years ago
Two soccer players kick the same 2-kg ball at the same time in opposite directions one kicks with a force of 15 n the other kick
Illusion [34]
Because the direction of the kicks are opposite, the net force between the applied forces is their difference.
                        Fn = F₂ - F₁
Substituting,
                        Fn = 15 N - 5 N
                        Fn = 10 N

From Newton's second law of motion,
                       Fn = m x a
where m is mass and a is acceleration. Manipulating the equation so that we are able to calculate for a,
                       a = Fn / m

Substituting,
                      a = (10 N) / 2 kg
                       a = 5 m/s²

<em>ANSWER: 5 m/s²</em>
7 0
3 years ago
Read 2 more answers
Please help guyss!!!
Savatey [412]

Explanation:

The distance traveled north ward  = 2 miles

       distance traveled south ward  = 3miles

 Displacement is the distance traveled in a specific direction. So, it depend on the start and finish point.

       The displacement  = 3 miles  - 2miles  = 1mile southward

Distance is the total length of path;

        Distance  = 3 miles + 2 miles = 5miles

5 0
3 years ago
A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 4.0 m/s, skat
Cerrena [4.2K]

Answer:

a)t=59.817652833125294s \approx 59.8s

b)D_1=894.01m

Explanation:

From the question we are told that

Speed of opposing player V_2=4.0m/s

First player chase his opponent aftert=2.80t

Acceleration of  first player a=0.14 m/s2

Let time of catch be  t_c

a)Generally the Equation for distance covered is mathematically given as follows

Distance to catch First opponent

D_1=\frac{1}{2}(0.14)t^2

D_1=0.07t^2

Distance to covered Second opponent

D_2=(2.8+t)*4

Generally when first opponent catch the second opponent it is represented mathematically as

D_2=D_1

0.07t^2=(2.8+t)*4

0.07t^2=11.2+4t

0.07t^2-4t-11.2

t=59.817652833125294s \approx 59.8s

b)Generally the the total time traveled by the first opponent is mathematically given as

D_1=\frac{t^2}{4}

D_1=\frac{59.8^2}{4}

D_1=894.01m

3 0
3 years ago
An object is moving to the left
dybincka [34]
The object is moving with constant speed means there is no change in speed Hence the acceleration is 0

Where, u is initial speed,

v is final speed

t is time taken and

a is acceleration

a = ( v-u) /t

a =(0)/t because u=v

a=0

Therefore, Acceleration is 0
4 0
2 years ago
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