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2 years ago

Which vector has a y-component with a length of 4?

Physics

1 answer:

mihalych1998 [28]2 years ago

6 0

ANSWER

The correct answer is D.

EXPLANATION

The y-component of the vector, is the same as the vertical component of the vector.

We can see from the graph that each box in the grid is one unit.

The component of vector a is

$\binom {2}{4}$

We count 4 boxes along the positive y- axis to get the y-component of vector a.

See graph in attachment.

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Suppose that over a certain region of space the electrical potential V is given by the following equation. V(x, y, z) = 2x2 − 3x

liq [111]

Answer:

Rate of change of potential V at the point P (2,6,5) in the direction (u = i + j - k) is

(û.∇V) = 6.928

Explanation:

V(x,y,z) = (2x² − 3xy + xyz)

Rate of change of a potential, V, in a direction, u, at a point P, is given as (û.∇V)

where û = unit vector in the direction of u = (vector u)/(magnitude of u)

u = v = i + j − k

Magnitude of u = √[(1²) + (1²) + (-1)²] = √3

û = (i + j − k)/(√3)

∇V = [(∂/∂x)î + (∂/∂y)j + (∂/∂z)k](V) at point (2,6,5)

V(x,y,z) = (2x² − 3xy + xyz)

(∂V/∂x) = 4x - 3y + yz = 4(2) - 3(6) + (6)(5) = 20

(∂V/∂y) = - 3x + xz = -3(2) + (2)(5) = 4

(∂V/∂z) = xy = (2)(6) = 12

∇V = (20î + 4j + 12k)

Rate of change of potential V = (û.∇V)

û = (1/√3) (i + j − k)

(û.∇V) = (1/√3) [(i + j − k).(20î + 4j + 12k)]

(û.∇V) = (1/√3) [20 + 4 - 12) = 12/√3 = 6.928

3 0

2 years ago

A tough kid on a tricycle has combined mass 32 kg. She starts from rest and travels 3.5 m in 2.5 s. What is the net force used t

Darina [25.2K]

Answer:

Explanation:

Force = mass * acceleration (Newton's second law of motion)

Given

Mass = 32kg

Get the acceleration using the expression d = 1/2at^2

3.5 = 1/2 (a)2.5^2

3.5 = 3.125a

a = 3.5/3.125

a = 1.12m/s^2

Get the net force;

F = 3.2 * 1.12

F = 35.84N

Hence the net force used to accelerate is 35.84N

8 0

2 years ago

If a train (travelling from Albuquerque to Atlanta) is moving 33.9 meters per second when it accelerates at -2.02 m/s^2 for 1.01

maxonik [38]

Answer:

The final velocity is 31.86 m/s.

Explanation:

The final velocity can be found using the following equation:

$V_{f} = V_{0} + at$

Where:

$V_{f}$ is the final velocity =?

$V_{0}$ is the initial velocity = 33.9 m/s

a is the acceleration = -2.02 m/s²

t is the time = 1.01 s

$V_{f} = V_{0} + at = 33.9 m/s + (-2.02 m/s^{2})*1.01 s = 31.86 m/s$

Therefore, the final velocity is 31.86 m/s.

I hope it helps you!

5 0

2 years ago

When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin

mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

$f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

where n=1,2,3 or ...n

L=Length of string

T=Tension

$\mu =$ Mass per unit length

When string is clamped at mid-point

Effecting length becomes $L'=0.5 L$

Thus new Frequency becomes

$f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}$

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz

4 0

3 years ago

Newton’s second law states that the acceleration of an object is dependent on 1 variable which is the net force

Darya [45]

I think it’s false. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and then mass of the object.

7 0

2 years ago