Friction helps your vehicle stop quickly. A. TRUE B. FALSE

What method would you use to extract the salt from the salt water mixture?

disa [49]

Answer:

maybe a strainer possibly but once mixed its rather hard

Explanation:

4 0

3 years ago

A 200g particle oscillating in SHM travels 66cm between the two extreme points in its motion with an average speed of 110cm/s. F

IgorC [24]

For a:v = d / Δt
110 = 0.66 / Δt
Δt = 0.66 / 110
Δt = 0.006 s
the period is:
T = 2Δt
T = 2*0.006
T = 0.012 s
the frequency is the inverse of the period. so: f = 1 / T
f = 83.3333333 Hz (about; Hz = 1/s)
b. T = 2π√(m/k)
being the mass m = 200g = 0.2 kg = 2*10^-1 kg, π = 3.14 (about) and T = 0.012, k is equal to:
0.012 = 6.28√(2*10^-1 / k)
0.012 / 6.28 = √(2*10^-1 / k)
0.00191082803 = √(2*10^-1 / k)
2*10^-1/ k = 0.000003
2*10^-1 / k = 3*10^-6
k = 2*10^-1 / 3*10^-6
k = 6.67*10^-5

now using hooke's law:
F = -kx
F = - 6.67*10^-5* 3.3*10^-1
F = -2.20x10^-5m
F = -0.22 *10^4 N

4 0

3 years ago

What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?

IRISSAK [1]

Answer:

The magnitude of electric field is 22.58 N/C

Solution:

Given:

Force exerted in upward direction, $\vec{F_{up}} = 3.50\times 10^{- 5} N$

Charge, Q = $- 1.55\micro C = - 1.55\times 10^{- 6} C$

Now, we know by Coulomb's law,

$F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}$

Also,

Electric field, $E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}$

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

$\vec{E} = \frac{\vec F_{up}}{Q}$

$\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}$

$\vec{E} = - 22.58 N/C$

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

$|\vec{E}| = 22.58\ N/C$

6 0

3 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

The national high magnetic field laboratory holds the world record for creating the strongest magnetic field. for brief periods

max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

F = m a ---> 1

While in magnetic forces, force can also be expressed as:

F = q v B ---> 2

where,

q = total charge

v = velocity = 45 cm / s = 0.45 m / s

B = the magnetic field = 85 T

First we solve for the total charge, q:

q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 × 10^-19 C / electron)

q = 1.594 × 10^-19 C

We equate equations 1 and 2 then solve for acceleration a:

m a = q v B

a = q v B / m

a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg

a = 160,437,862.2 m/s^2

Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.

5 0

3 years ago

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