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wolverine [178]
3 years ago
11

I will mark brainliest if you answer please im givingalot of points also

Physics
1 answer:
Tamiku [17]3 years ago
6 0

A. two forces

B. P-f

Hope this helps!

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A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box  = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0
3 years ago
Colonel John P. Stapp, USAF, participated in studying whether a jet pilot could survive emergency ejection. On March 19, 1954, h
borishaifa [10]

we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi ​ =632mi/h=632mi/h( 1mi 1609m ​ )( 3600s 1h ​ )=282m/s (a) taking v xf ​ =v xi ​ +a x ​ t with v xf ​ =0 a x ​ = t v xf ​ −v xf ​ ​ = 1.40s 0−282m/s ​ =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f ​ −x i ​ = 2 1 ​ (v xi ​ +v xf ​ )t= 2 1 ​ (282m/s+0)(1.40s)=198m

7 0
3 years ago
A spring balance used to weigh candy is built with a spring the spring stretches 3.00 cm when a 15 n weight is placed in the pan
ira [324]

Answer: 5.9cm

Explanation:

15N/3cm=5

Convert 3kg to N (29.42N)

Use the 5 from earlier to divide 29.42/5=5.884

Round to 5.9

5.9cm is the answer

(I don’t know if I solved this the correct way but I got it right. 5.9 is the correct answer)

6 0
3 years ago
A student weighs a sample of an unkown metal and finds the mass to be 39.27g what is the metal?
xxTIMURxx [149]

The unknown metal with a mass of 39.27 g is potassium with the symbol K.

The mass of an atom is due to protons and neutrons present in it.

The protons and neutrons are together present in the nucleus of an atom.

The mass of any element is given by its mass number whose formula is

Mass Number = Number of protons + Number of neutrons

In the given case 39 = Number of protons + Number of neutrons

The only element of the periodic table which satisfies the above equation is potassium because it has 19 protons and 20 neutrons.

Thus, the unknown metal with a mass of 39.27 g is potassium with the symbol K.

To know more about "mass number", refer to the following link:

brainly.com/question/4408975?referrer=searchResults

#SPJ4

7 0
2 years ago
An automobile is traveling on a long, straight highway at a steady 80.0 mi/h when the driver sees a wreck 190 m ahead. At that i
kozerog [31]

Answer:

a) The car’s speed just after leaving the icy portion of the road is the first part

Explanation:

6 0
3 years ago
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