Are moons 1-4 waxing are waning ?

Jane has a mass of 55 kg and his body covers 700 nails with a surface area of 1.00 mm,

Oksana_A [137]

We have,

Jane mass is 55 kg
His body covered with 700 nails all of them having a surface area of 1.00 mm² each = 700 × 1 = 700 mm² = 700/1000000 = 7/10000

We know that,

Pressure = Force/Area

Let's calculate force as we already have area;

F = ma
F = 55 × 9.8 { Acceleration due to gravity }
F = 539 N

Now, if should she would be on 700 nails then pressure will be;

P = F/A
P = 539/7 × 10000
P = 5390000/7
P = 770,000 Pascal

And if should would be on a 1 nail only,

P = F/A
P = 539/1 × 1000000
P = 539000000 Pascal

As, you can notice yourself that the pressure will be so high with only 1 nail and because of this, nail will pass through jane's body.

6 0

2 years ago

A 3.0 kg box is suspended by a series of ropes as shown below. The tension force in the horizontal rope is -40 N. What is the te

Nitella [24]

Answer:

50 N

Explanation:

Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:

The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.

The horizontal component of the forces:

F₁ + F₂ = -40N + F₂ = 0

F₂ = 40N

The vertical component of the forces:

F₁ + F₂ - mg = 0 + F₂ - mg = 0

F₂ = mg

If I assume the gravitational constant g = 10 m/s²:

F₂ = (3 kg) * (10 m/s²) = 30N

Adding the horizontal and vertical components of the force F₂:

F₂ = √((40N)² + (30N)²) = 50N

6 0

2 years ago

1. A truck with a mass of 8, 000 kg is traveling at 26.8 m/s when it hits the brakes. A.)What is the momentum of the truck befor

NikAS [45]

Answer:

1. A.) The moment of the truck before it hits the brakes is 214,400 kg·m/s

B.) The force it takes to stop the truck is approximately 17,290.4 N

Explanation:

1. A.) The given parameters are;

The mass of the truck, m = 8,000 kg

The velocity of the truck when it hits the brakes, u = 26.8 m/s

Momentum = Mass × Velocity

The moment of the truck = The mass of the truck × The velocity of the truck

Therefore;

The moment of the truck before it hits the brakes = 8,000 kg × 26.8 m/s = 214,400 kg·m/s

B.) The amount of momentum lost when the truck comes to a stop = The initial momentum of the truck

The time it takes the truck to come to a complete stop, t = 12.4 s

The deceleration, "a" of the truck is given by the following kinematic equation of motion

v = u - a·t

Where;

v = The final velocity of the truck = 0 m/s

u = The initial velocity = 26.8 m/s

a = the deceleration of the truck

t = The time of deceleration of the truck = 12.4 s

Substituting the known values gives;

0 = 26.8 - a × 12.4

Therefore;

26.8 = a × 12.4

a = 26.8/12.4 ≈ 2.1613

The deceleration (negative acceleration) of the truck, a ≈ 2.1613 m/s²

Force = Mass × Acceleration

The force required to stop the truck = The mass pf the truck × The deceleration (negative acceleration) given to the truck

∴ The force it takes to stop the truck = 8,000 kg × 2.1613 m/s² ≈ 17,290.4 N.

8 0

3 years ago

The stoplights on a street are designed to keep traﬃc moving at 26 mi/h. the average length of a street block between traﬃc ligh

bonufazy [111]

We use the formula,

$v = \frac{d}{t}$ .

Here, v is velocity and its value given 26 mi/h ( in m/s, $\frac{26 \times 1.609 \times 10^{3} m}{60 \times 60 s} =11.62 \ m/s$ ) and d is distance and its value is given 80 m.

Substituting these values in above formula we get,

$t = \frac{80 m}{11.62m/s } = 6.88 \ s$

Thus, the time delay between green lights on successive blocks to keep the traﬃc moving continuously is 6.88 s

3 0

2 years ago

Nuclei of u-238 atoms are

Elis [28]

Nuclei of u-238 atoms are unstable and spontaneously emit alpha particles.

5 0

3 years ago