All categories

4 years ago

Nearly all physics problems will use the unit m/s squared. Why are the seconds squared?

1 answer:

7 0

Seconds squared is the time unit of acceleration. It represents the change in distance units per second per second. For example, 3 m/sec² means a distance covering 3 meters in the first second, then 9 meters in the 2nd second, and 37 meters in the third second. (3^1, 3^2, 3^3).

Acceleration is part of Newton's 2nd law: force = mass x acceleration. Units of work: joule = kg·m²/s², and power: watts = kg·m²/s³ all contain accelerations.

You might be interested in

Please help!!

Harlamova29_29 [7]

Answer:

This could represent something like sliding a small rock across an icy lake.

Explanation:

A 20N force of gravity (weight), and 20N normal force exerted back onto the object imply it is on the ground and has no vertical motion. There is a net force of 0N

An 80N force to the left and a 5N force to the right create a net force of 75N to the left. This means that there is a force acting on the object that makes it accelerate to the left. 80N represents a push or pull force and 5N represents a relatively small frictional force due to the object being slid on a surface such as steel or in this case ice.

8 0

4 years ago

A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a

Bezzdna [24]

Answer:

25 psi

Explanation:

The weight of the car is:

W = mg

W = 1550 kg * 9.8 m/s²

W = 15,190 N

Divided by 4 tires, each tire supports:

F = W/4

F = 15,190 N / 4

F = 3797.5 N

Pressure is force divided by area, so:

P = F / A

P = (3797.5 N) / (0.16 m × 0.14 m)

P ≈ 170,000 Pa

101,325 Pa is the same as 14.7 psi, so:

P ≈ 170,000 Pa × (14.7 psi / 101,325 Pa)

P ≈ 25 psi

8 0

3 years ago

During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.

Phoenix [80]

To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

$Q = \frac{\Delta P \pi r^4}{8\eta l}$

Where:

$\eta_i =$ are the viscosities of the concrete before and after the increase

l = Length of the vessel

$r_1, R_2$ = Radio of the vessel before and after the increase

$\Delta P$ = Change in the pressure

$Q_{1,2} =$ The rates of flow before and after he increase

Our values are given as:

$Q_2 = 10Q_1 \rightarrow$ 10 times her resting rate

$\eta_2 = 0.95\eta_1$ 95% of its normal value

$\Delta P_2 = 1.5\Delta P_1$ Increase of 50%

Plugging known information to get

$Q_1 = \frac{\Delta P \pi r^4}{8\eta l}$

$Q_1 8\eta_1 l = \Delta P_1 \pi r_1^4$

$r_1^4 = \frac{Q_1 8\eta_1 l}{\Delta P_1 \pi}$

$r_1 = (\frac{Q_1 8\eta_1 l}{\Delta P_1 \pi})^{1/4}$

$r_2 = (\frac{Q_2 8\eta_2 l}{\Delta P_2 \pi})^{1/4}$

$r_2 = (\frac{10Q_18 \times 0.95\eta_1 l}{1.5\Delta P_1 \pi})^{1/4}$

$r_2 = 1.586r_1$

Therefore the factor of average radio of her blood vessels increased is 1.589 the initial factor after the increase.

7 0

4 years ago

Why plate tectonics is a scientific theory and not a scientific law.

ser-zykov [4K]

Answer:

We have not drilled to the center of the earth.

Explanation:

7 0

3 years ago

Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher

Paha777 [63]

Answer:

The right sphere is negatively charged, the left sphere is charged positively.

Explanation:

When a negatively charged rod is held above the top of left sphere, the rod will attract positive charges and repel negative charges. As the sphere are initially touching each other so positive charges from the both spheres will moves toward the rod. When we separate the spheres positive charges from right sphere have already moved toward the rod i.e. left sphere, creating a deficiency of positive charges in the right sphere and excessiveness of positive charges in left sphere , hence the right sphere will remain negatively charged and left sphere will remain positively charged.

4 0

4 years ago