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Natali [406]
3 years ago
13

which of the following locations has the largest electric field? A)0.4 cm from a +3.0 uC point charge B)0.2 cm frm a +1.5 uC poi

nt charge C)0.4 cm from a +6.0 uC point charge D)0.2 cm from a +3.0 uC point charge
Physics
1 answer:
JulsSmile [24]3 years ago
7 0
<span>Electric field is proportional to q/d^2, where q is the magnitude of the charge and d is the distance. Since all the given units are identical, we can just compare their relative magnitudes without calculating for the exact values.
A) 3/(0.4)^2 = 18.75
B) 1.5/(0.2)^2 = 37.5
C) 6/(0.4)^2 = 37.5
D) 3/(0.2)^2 = 75
Therefore, choice D has the largest electric field of all.
</span>
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A sound wave takes 0.2 seconds to travel 306 meters.
chubhunter [2.5K]

Answer:

1530 m/s

Explanation:

Given that :

Speed = distance / time

Travel time take = 0.2 s

Distance covered = 306 metres

The speed of sound in the material :

Speed = distance / time = 306 m / 0.2

Speed = 1530 m/s

6 0
3 years ago
A + 8 μC charge lies 4 m to the left of a - 10 μC charge. A - 12 μC charge lies 8 m to its right. What is the resultant force on
MaRussiya [10]

Answer:

both side charges will exert force on central charge towards left so the two forces add up. Thus resultant force on centre charge is

F = 9×10^9 x 8×10^-6×10×10^-6/16

+

9× 10^9 × 10×10^-6×12×10^-6/64

= 9× 10^9 × 10^ -12/16 (80+30)

=6.18 × 10^-2 N

or

= 0.06 N

3 0
3 years ago
Each gas sample has the same temperature and pressure. which sample occupies the greatest volume? each gas sample has the same t
Anit [1.1K]

Answer:

20 gram of H as the greastest volume

Explanation:

Each gas sample has the same temperature and pressure. which sample occupies the greatest volume? each gas sample has the same temperature and pressure. which sample occupies the greatest volume? 40.0 gar 20.0 gne 20.0 gh2 4.0 ghe

Given that,

40.0 g of Ar

20.0 g of Ne

20.0 g of h

24.0 g of he

The ideal gas equation ; Pv = nRT

where P is the pressure, V is the volume, R is the gas constant, T is the temperature, n is the number of mole

which is equal to

n = m / M

where m is the given mass

M is the molar mass

The given gas sample are at the same temperature and pressure,

So, the temperature, pressure and gas constant are fixed.

Therefore the volume is proportional to the given mass and molar mass

V ∝ m/M

Substitute the given mass and molar mass of given masses for comparison

No of mole of Ar = mass / molar mass

= 40 / 40 = 1 mole

No of mole of Ne = mass / molar mass

= 20 / 20 = 1 mole

No of mole of h = mass / molar mass

= 20 / 1 = 20 mole

No of mole of he = mass / molar mass

= 24 / 4 = 6 mole

Hence, 20 gram of H as the greastest volume

3 0
3 years ago
A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for
Lorico [155]

The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N

Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>

Have a nice day!

8 0
4 years ago
A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is
lbvjy [14]

Answer:

a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\

b)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}

8 0
3 years ago
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