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Arte-miy333 [17]
2 years ago
13

Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a

force of 16 N. What is the net force on them?
Physics
2 answers:
sammy [17]2 years ago
6 0

Answer:

Net force acting on them is 16 N and it is acting to the right side.

Explanation:

It is given that,

Force acting by the dog, F_1 = 32\ N (right side)

Force acting by Simone , F_2 = -16\ N (backward)

Let backward direction is taken to be negative while right side is taken to be positive.

The net force will act in the direction where the magnitude of force is maximum. Net force is given by :

F=F_1+F_2

F=32+(-16)    

F = 16 N

So, the net force is 16 N and it is acting to the right side.

belka [17]2 years ago
5 0

Answer:

16 N (towards right)

Explanation:

Pulling force towards right, F1 = 32 N

Pulling force towards left, F2 = 16 N

Net force = F1 - F2

F = 32 - 16 = 16 N (towards right)

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Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a
Naily [24]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

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8 0
1 year ago
A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %
olasank [31]

Answer:

height is 69.68 m

Explanation:

given data

before it hits the ground =  46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

d = (0.5)gt²     ..............1

here d is distance and g is  acceleration and t is time

so when object falling it will be

h = 4.9 t²   ....................2

and in 1st part of question

we have (100% - 46% ) = 54 %

so falling objects will be there

0.54 h = 4.9 (t-1)²       ...................3

so

now we have 2 equation with unknown

we equate both equation

1st equation already solve for h

substitute h in the second equation and find t

0.54 × 4.9 t² = 4.9 (t-1)²  

t = 0.576 s and  3.771 s

we use here 3.771 s because  0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

so take t = 3.771 s

then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h =  69.68 m

so height is 69.68 m

6 0
2 years ago
A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm^3 of liquid. The mater
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Area for side:
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Cost for side:
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We know that the volume of the can is:
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0 = 0.08πr - 10/r²
10/r² = 0.08πr
r³ = 125/π

r = 3.41 cm
4 0
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