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3 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.

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A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re

valentinak56 [21]

Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19

E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

Final speed of the proton can be gotten by using

v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

5 0

2 years ago

Current is produced when charges are accelerated by an electric field to move to a position of lower

Vaselesa [24]

Current is created when charges are quickened by an electric field to move where the position of lower temperature. An electric current is a stream of electric charge. In electric circuits, this charge is regularly conveyed by moving electrons in a wire.

8 0

2 years ago

A 1750-kilogram cars travels at a constant speed of 15.0 meters per second around a horizontal, circular track with a radius of

bulgar [2K]

m = mass of the car moving in horizontal circle = 1750 kg

v = Constant speed of the car moving in the horizontal circle = 15 m/s

r = radius of the horizontal circular track traced by the car = 45.0 m

F = magnitude of the centripetal force acting on the car

To move in a circle . centripetal force is required which is given as

F = m v²/r

inserting the above values in the formula

F = (1750) (15)²/(45)

F = (1750) (225)/(45)

F = 1750 x 5

F = 8750 N

6 0

2 years ago

Tony has a mass of 50 kg, and his friend Sam has a mass of 45 kg. Assume that both friends push off on their rollerblades with t

Citrus2011 [14]

Answer:

Sam

Explanation:

Hes lighter

3 0

3 years ago

A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside

neonofarm [45]

Answer:

$\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$ , level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

$\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}$

$\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt} = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}$

By replacing all known variables:

$\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}$

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0

3 years ago