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3 years ago

What is the force required accelerate a 5.0 kg object at 3.0m/s^ 2?

1 answer:

6 0

The equation to find force is f=ma. So, if you plug in the information that you have you'll get F=5x3 and that'll equal F=15N

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Which items in this image are electrically conductive?

9966 [12]

all except the rubber boots.

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3 years ago

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The volume of a gas can be converted to moles by

Sonja [21]

Multiplying the ideal gas law constant

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3 years ago

A car is traveling at a speed of 50 km/hr. The positive y-axis is north and the positive x-axis is east. Resolve the car's veloc

hjlf

We need to draw a coordinates. the east and south should be the north vector for horizontal line 50 KM in distance is zero. the south is a negative. the south east and north west we should draw the 45 degrees angle in the approprate quadrant and then use the 1-1-sqrt(2) have a relationship in 45-45-90 triangles to resolve the N/S, E/W components.

hope this help

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3 years ago

Simplified. The absorption spectra of ions have been used to identify the presence of the elements in the atmospheres of the sun

Nookie1986 [14]

Answer:

The answer is " $3.83 \times 10^9 \ m$ "

Explanation:

Z=2, so the equation is $E= \frac{-4B}{n^2}$

Calculate the value for E when:

n=2 and n=9

The energy is the difference in transformation, name the energy delta E Deduct these two energies

In this transition, the wavelength of the photon emitted is:

$\Delta E=2.18 \times 10^{-18} ( \frac{1}{4}- \frac{1}{81})$

$\lambda = \frac{h c}{\Delta E}$

$h ( Planck's\ constant) = 6.62 \times 10^{(-34)} \ Js \\\\ speed \ of \ light = 3 \times 10^{8} \ \frac{m}{s}\\\\= \frac{6.62 \times 10^{(-34)} \times 3 \times 10^ {8}}{2.18 \times 10^{-18}} (\frac{1}{4}- \frac{1}{81}) \\\\=3.83 \times 10^9 \ m\\\\$

6 0

3 years ago

A charged particle moving along the +x-axis enters a uniform magnetic field pointing along the +z-axis. A uniform electric field

Brums [2.3K]

Answer:

the electric field direction should be in positive y axis

Explanation:

Lets assume that charge on particle is positive and it isequal to +q

First calculate the magnetic force on it

$F_B = q V\times B =qVBsin\theta$

for direction

use Right Hand Rule which will give the direction and by using his the direction will come towards negative y axis.

As given in the question that charge particle does not change their velocity so we need to apply electric field in such a way that electric force direction should be opposite to the magnectic field.

and magnitude should be same as magnectic force and also direcion of electric force depend on the direction of elecric field when charge is positive because electric force $F_E = qE$

Hence the electric field direction should be in positive y axis

3 0

4 years ago