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3 years ago

The force experienced by a unit test charge is a measure of the strength of an electric:

Physics

2 answers:

SCORPION-xisa [38]3 years ago

8 0

<h2>Answer:</h2>

<u>The force experienced by a unit test charge is a measure of the strength of an </u><u>electric field</u>

<h2>Explanation:</h2>

Electric field strength is the intensity of an electric field at a particular location. The standard unit is the volt per meter (v/m or v. m -1). A field strength of 1 v/m represents a potential difference of one volt between points separated by one meter. A test charge shows the direction of the field. But the "test" charge should be small, since charges generate their own fields, hence they interfere with the external field.

Flauer [41]3 years ago

3 0

an electric field is the answer

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denis-greek [22]

Explanation:

A bearing if an angle is measured clockwise from north direction.

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2 years ago

Which one is having lesser ressistance. a 60w bulb or a 40w bulb?

Ivan

Power dissipation = (voltage across the component)² / (resistance of the component)

Since the resistance is in the denominator of the fraction in this formula for the
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So the <u>60w bulb</u> has lower resistance than the 40w bulb.

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3 years ago

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round

Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

6 0

3 years ago

Read 2 more answers

Star A and Star B have measured stellar parallax of 1.0 arc second and 0.75 arc second, respectively. Which star is closer? How

zhuklara [117]

Answer:

Star A is closer than Star B

Explanation:

As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle

Here we can say

$angle = \frac{1 Parsec}{distance}$

so we have

$distance = \frac{1 Parsec}{angle}$

so here we have

angle subtended by Star A = 1 arc sec

angle subtended by star B = 0.75 arc sec

now we have

distance for star A is given as

$d_a = \frac{1 Parsec}{1} = 1 Parsec$

distance of star B is given as

$d_b = \frac{1 Parsec}{0.75} = 1.33 Parsec$

So star A is closer than star B

7 0

3 years ago

A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i

Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0

3 years ago