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3 years ago

A position vector in the first quadrant has an x-component of 6 m and a magnitude of 10 m. What is the value of its y-component

Physics

1 answer:

LUCKY_DIMON [66]3 years ago

7 0

Answer:

Explanation:

The magnitude m of a vector (x, y) is given by

m = $\sqrt{x^2 + y^2}$ -------------------------(i)

where;

x and y are the x- and y- components of the vector.

From the question;

m = 10m

x = 6m

Substitute these values into equation (i) as follows;

10 = $\sqrt{6^2 + y^2}$

Solve for y;

<em>Find the square of both sides</em>

10² = 6² + y²

100 = 36 + y²

y² = 100 - 36

y² = 64

y = √64

y = 8

Therefore, the y-component of the position vector is 8m

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Two point charges, A and B, are separated by a distance of 22.0 cm . The magnitude of the charge on A is twice that of the charg

REY [17]

Answer:

1.12×10⁻⁵ C and 2.24×10⁻⁵ C.

Explanation:

From coulomb's law,

F = kAB/r².............................. Equation 1

Where F = Force exerted by each charge, A = charge at point A, B = charge at point B, r = distance of separation between the points, k = constant of proportionality.

Given: F = 47 N, r = 22 cm = 0.22 m.

Constant: k = 9.0×10⁹ Nm²/C²

Let: B = q, the A = 2q.

Substituting these values into equation 1,

47 = 9.0×10⁹(q×2q)/0.22²

47 = 18×10⁹(q²)/0.0484

q² = (47×0.0484)/(18×10⁹)

q² = 0.126×10⁻⁹

q² = 1.26×10⁻¹⁰

q = √( 1.26×10⁻¹⁰)

q = 1.12×10⁻⁵ C

The charge at point A = 2q = 2× 1.12×10⁻⁵ = 2.24×10⁻⁵ C.

Hence the charges are 1.12×10⁻⁵ C and 2.24×10⁻⁵ C.

3 0

3 years ago

Which of the following is true about conduction? Question 2 options: It is heat transfer through direct contact It can only occu

kramer

Answer:

through direct contact

Explanation:

took the test!

3 0

3 years ago

The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be

bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

$F=\frac{Gm_{1} m_{2} }{d^{2} }$ ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

$1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }$

$d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}$

$d=\sqrt{1.48\times10^{17}}$

d = 3.85 x 10⁸ m

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3 years ago

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of wor

joja [24]

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is $\Delta U_1=222 J$

Work done on the system $W_1=-150 J$

For second step

change in internal Energy of the system is $\Delta U_2=123 J$

Work done on the system $W_2=-195 J$

Work done on the system is considered as Positive and vice-versa.

and from first law of thermodynamics

$Q=\Delta U+W$

for first step

$Q_1=222-150=72 J$

$Q_2=123-195=-72 J$

overall heat added $=Q_1+Q_2$

$Q_{net}=72-72 =0$

For overall Process Heat added is 0 J

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3 years ago

Which of the following statements is/are true for ultrasonic test?a. Equipment used for ultrasonic testing is portableb. Complic

Pavel [41]

Answer:

Equipment used for ultrasonic testing is portable

Explanation:

An ultrasonic tester is used to detect flaws such as cracks and leaks. Hence, an ultrasonic tester is an integral tool used in material testing and engineering, manufacturing quality control, equipment condition monitoring, building and bridge inspection and much more.

Ultrasonic testing can be performed on steel and other metals and alloys, though it can also be used on concrete, wood and composites, although with less resolution.

4 0

4 years ago