Why can people float effortlessly on the dead sea?

On a trip to Italy, Cheyne rented a car to tour the country. As he turned onto the highway, he noticed that the speed limit was

KATRIN_1 [288]

1 mile = 1.609344 km
x miles = 95 km
x = 95 : 1.609344 = 59.03
Answer: 53.09 mph

6 0

3 years ago

A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglec

Novosadov [1.4K]

Answer:

The free-body diagram of the cannonball is found in the attachment below

Note The question is incomplete. The complete question is as follows:

A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglected.

Draw the free-body diagram of the cannonball.

Explanation:

Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.

In order to construct free-body diagrams, it is important to know the various types of forces acting on the object in that situation. Then, the direction in which each of the forces is acting is determined. Finally the given object is drawn using any given representation, usually a box, and the direction of action of the forces are represented using arrows.

In the given situation of a cannonball which has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction., the forces acting on it are:

F = force exerted by the cannon acting in the direction of angle of projection

Fdrag = drag force. The drag force acts in a direction opposite to the force exerted by the cannon

Fw = weight of the cannonball acting in a downward direction

The free body diagram is as shown in the attachment below.

6 0

3 years ago

When a small object is launched from the surface of a fictitious planet with a speed of 52.9 m/s, its final speed when it is ver

NISA [10]

Answer:

The escape speed of the planet is 41.29 m/s.

Explanation:

Given that,

Speed = 52.9 m/s

Final speed = 32.3 m/s

We need to calculate the launched with excess kinetic energy

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

$K.E=\dfrac{1}{2}\times m\times(32.3)^2$

We need to calculate the escape speed of the planet

Using formula of kinetic energy

$\text{escape kinetic energy}=\text{launch kinetic energy}-\text{excess kinetic energy}$

$\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2$

$\dfrac{1}{2}\times v^2=\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2$

$v=\sqrt{2\times(\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2)}$

$v=41.29\ m/s$

Hence, The escape speed of the planet is 41.29 m/s.

4 0

3 years ago

Which one of the following is the pigment found in redheads?

coldgirl [10]

The pigment found in redheads in pheomelanin

4 0

3 years ago

What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingr

Alexxx [7]

Answer:

E = 9.4 10⁶ N / C , The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

Ф = E. dA = qint / ε₀

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of the cylinder is the length of the circle along the length of the cable

dA = 2π dr L

A = 2π r L

They indicate that the distance at which we must calculate the field is

r = 5 R₁

r = 5 1.3

r = 6.5 mm

The radius of the outer shell is

r₂ = 10 R₁

r₂ = 10 1.3

r₂ = 13 mm

r₂ > r

When comparing these two values we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

λ = q / L

Qint = λ L

Let's replace

E 2π r L = λ L /ε₀

E = 1 / 2piε₀ λ / r

Let's calculate

E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3

E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

5 0

3 years ago