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3 years ago

What is 300+-50+-75+225

2 answers:

4 0

In general it's not "good math" to write "+-". One way to rewrite 300+-50+-75+225 more clearly would be 300 - 50 - 75 + 225.

Working from left to right, 250 - 75 + 225

175 + 225 = 400 (answer)

olasank [31]3 years ago

4 0

The answer will be 400

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Theoretical Probability (Pink Band) - 19/37
Their are 95 pink rubber bands (numerator) and when adding 95+90 = 185 (denominator) to get the theoretical probability you would get 95/185 which when simplified by the factor of 5 would be the same thing as 19/37.

Theoretical Probability (Brown Band) - 18/37

Their are 90 brown rubber bands (numerator) and we already know that are denominator will be 185 so we would get 90/185. This fraction is also divisible by a factor of 5 so when simplified you would get 18/37.

Experimental Probability (Pink/Brown Band) - 12/23 and 11/23

Pink Band: Their is a 36 (numerator) out of (36+33) 69 (denominator) chance of pulling a pink band so it would be written as 36/69 which is divisible by 3 and when simplified is written as 12/23.

Brown Band: Their is a 33 (numerator) out of our already known denominator 69 chance of pulling a brown band. When written as a fraction this would be 33/69 which is also divisible by 3 and when simplified 11/23.

We can conclude with this data that their is a slightly higher chance of pulling a pink rubber band compared to the experimental probability of picking a brown rubber band because we know that 12/23 > 11/23.

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3 years ago

Part A What is the electric field at the position (x1,y1)=(5.0 cm , 0 cm) in component form? Express your answer in terms of the

erik [133]

The complete Question is

A −12 nC charge is located at (x, y)=(1.0 cm, 0 cm).

Part A) What is the electric field at the position (x1, y1)=(5.0 cm, 0 cm) in component form?

Part B) What is the electric field at the position (x2, y2)=(−5.0 cm, 0 cm) in component form?

Part C) What is the electric field at the position (x3, y3)=(0 cm, 5.0 cm) in component form?

Express your answer in terms of the unit vectors i^ and j^. Use the 'unit vector' button to denote unit vectors in your answer.

Answer:

A) E = - 67500i + 0j N/C

B) E = 30000i + 0j N/C

C) E = 8143.18i - -40715.66j N/C

Step-by-step explanation:

Part A)

Magnitude of the charge = Q = -12 nC = $-12 \times 10^{-9}$ C

Since, its the negative charge, the direction of Electric Field lines will be directed toward the charge

Location of the charge = (1, 0)

We need to find the electric field at point (5, 0)

The formula for the magnitude of electric field due to a point charge is:

$E=\frac{kQ}{r^{2}}$

Here,

k = Coulomb's Law Constant = $9 \times 10^{9}$

Q = Magnitude of the charge

r = Distance between the charge and the point where we need to find the value of E

We can find r by using the Distance Formula.

So,

$r=\sqrt{(5-1)^{2}+(0-0)^{2}}=4$ cm = 0.04 m

Using these values in the formula, we get:

$E=\frac{9\times 10^{9} \times 12 \times 10^{-9}}{(0.04)^{2}}= 67500$ N/C

Since, two point (5, 0) is to the right of the given charge (as shown in the first image) i.e. in horizontal direction, all of the electric field experienced by it will be in horizontal direction and the vertical component would be zero. Also the direction of Electric field will be towards the charge i.e. in left direction so the x-component of Electric field will be negative.

Thus, we can write the value of E in vector form to be:

E = - 67500i + 0j N/C

Part B)

We need to find the Electric Field at point (-5, 0)

Using the similar procedure as used in the previous step, first we find r:

$r=\sqrt{(-5-1)^{2}+(0-0)^{2}}=6$ cm = 0.06

Using the values in the formula of Electric field, we get:

$E = \frac{9 \times 10^{9} \times 12 \times 10^{-9}}{(0.06)^2}=30000$ N/C

In this case again, the point is located in a horizontal direction to the given charge, so all the Electric Field experienced by it will be in horizontal direction and the vertical component will be zero. The direction of Electric field will be towards the charge i.e towards Right, so the x-component will be positive in this case.

So, value of the electric field in component form would be:

E = 30000i + 0j N/C

Part C)

We need to find the value of electric field at the point (0, 5). First we find the value of r:

$r=\sqrt{(1-0)^2+(0-5)^2}=\sqrt{26}=5.1$ cm = 0.051 m

Using the values in the formula of E, we get:

$E=\frac{kQ}{r^{2}}=\frac{9 \times 10^{9} \times 12 \times 10^{-9}}{(0.051)^{2}}=41522$ N/C

The point (0, 5) is neither exactly to the left or exactly up. So, for this point we need to find both the horizontal and vertical components as shown in the 2nd figure below.

From the triangle, we have the opposite and adjacent side to the angle, so using the tangent we can find the value of angle theta.

$tan(\theta)=\frac{5}{1}\\\theta=tan^{-1}(5)=78.69$

The two angles shown in the figure will be equal as there are alternate interior angles. Now the angle which E will make with positive x-axis will lie in the 4th quadrant as it lies below the horizontal line. So, the angle with positive x-axis would be:

360 - 78.69 = 281.31 degrees

Ex = E cos(θ) = 41522 cos(281.31) = 8143.18 N/C

Ey = E sin(θ) = 41522 sin(281.31) = -40715.66 N/C

So, in component form the Electric field will be:

E = 8143.18<em>i</em> - -40715.66<em>j</em>

3 0

3 years ago

Please help I do not understand this at all and I need help

KengaRu [80]

Answer:

Step-by-step explanation:

28+35+21+12=96

7 0

3 years ago