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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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Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals

daser333 [38]

Answer:

15448

Explanation:

Compounded Quarterly:

A=P\left(1+\frac{r}{n}\right)^{nt}

A=P(1+

)

Compound interest formula

P=11000\hspace{35px}r=0.057\hspace{35px}t=6\hspace{35px}n=4

P=11000r=0.057t=6n=4

Given values

A=11000\left(1+\frac{0.057}{4}\right)^{4(6)}

A=11000(1+

0.057

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4(6)

Plug in values

A=11000(1.01425)^{24}

A=11000(1.01425)

Simplify

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2 years ago

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Answer:

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During science class while studying mixtures you mix together iron fillings and sand. You're a teacher challenges you to separat

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2 years ago

Ships A and B leave port together. For the next two hours, ship A travels at 40.0 mph in a direction 35.0° west of north while t

kompoz [17]

Answer:

Explanation:

Given

Ship A velocity is 40 mph and is traveling 35 west of north

Therefore in 2 hours it will travel $40\times 2=80 miles$

thus its position vector after two hours is

$r_A=-80sin35\hat{i}+80cos35\hat{j}$

similarly B travels with 20 mph and in 2 hours

$=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}$

Thus distance between A and B is

$r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}$

$|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}$

$|r_{AB}|=103.45 miles$

Velocity of A

$v_A=-40sin35\hat{i}+40cos35\hat{j}$

Velocity of B

$v_B=20sin80\hat{i}+20cos80\hat{j}$

Velocity of A w.r.t B

$v_{AB}=v_A-v_B$

$v_{AB}=\left ( -20sin80-40sin35\right )\hat{i}+\left ( 40cos35-20cos80\right )\hat{j}$

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3 years ago

A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.

pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

$h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t = 6.55 \ s$

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

$X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m$

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

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3 years ago