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goblinko [34]
3 years ago
13

A secret agent is locked in a room. He pushes against the door but cannot open it. Finally, he falls to the floor exhausted. Has

he done any work in the door? Why or why not?
Physics
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer: There is not work done at the door because the door did not move.

Explanation:  Work is defined as the movement done by a force.

So if you move to apply a force F in an object and you move it a distance D, the work applied on the object is  

W = F*D

In this case, the secret agent pushes against the door, so there is a force, but the agent does not move the door, so D = 0, so there is no motion of the door, which implies that there is no work done at the door.

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Is a cow a producer? Or a consumer or decomposer?
klemol [59]

Is a cow a producer? Or a consumer or decomposer? Cow is a consumer

7 0
2 years ago
Read 2 more answers
An airplane starts from rest and reaches a takeoff velocity of 60.0m/s due north in 4.0 s. How far has it traveled before takeof
Blizzard [7]
240 meters that's it
7 0
3 years ago
A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall
kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

F = q(\vec v \times \vec B)

so here we have

F = qvB sin90

since force is perpendicular to the velocity so here it must be centripetal force

here we have

\frac{mv^2}{R} = qvB

so we have

R = \frac{mv}{qB}

R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}

R = 4.06 cm

so here it will move in circle with radius 4.06 cm

8 0
3 years ago
The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur
Nimfa-mama [501]

The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

           P₂ is the pressure in the hose

           ρ is the density of the fluid

           h₁ is the initial height

           h₂ is the final height

           v₁ is the initial velocity of the fluid

           v₂ is the final velocity of the fluid  and

           g is the acceleration due to gravity

Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

    = (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

    = 5.92 × 10⁵ Pa

    = 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

<u>brainly.com/question/9506577</u>

#SPJ4

6 0
2 years ago
A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back
elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity=9.8 m/s^2

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates=\mu mg=0.202\times 9.8\times 35

Force by which box accelerates=62.286 N

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

ma=\mu mg

a=\mu g=9.8\times 0.202=1.9796 m/s^2

Hence, the truck can have maximum acceleration before the box slide=1.9796 m/s^2

3 0
3 years ago
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