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3 years ago

Displacement is which of the following types of quantities?A. vectorB. scalarC. magnitudeD. dimensional

Physics

1 answer:

agasfer [191]3 years ago

6 0

Answer: A. vector

Explanation:

Movement is the change of position of a body at a certain time. The body that experiences this change of position is called mobile.

In this sense, the displacement is the distance in a straight line between the initial and final position of this mobile, which differentiates it from the trajectory, which is the path followed by the body in its movement and a scalar magnitude.

Hence, the displacement is a vector.

You might be interested in

Find the displacement in meters a runner would travel in 5 hours at an average velocity of 12km/h to the southwest

vova2212 [387]

Answer:

60,000m

Explanation:

Convert km/h to m/s by multiplying with 1000/3600.

Convert hours to seconds by multiplying with 3600.

Because displacement is a vector quantity and deals with the shortest distance between points, simply plug it into the equation s=vt.

8 0

3 years ago

Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half-way between the two plates th

fgiga [73]

Answer:

Explanation:

Let the potential difference between the plate is V . Then in the first case

Electric field E between plate

E₁ = V / d

where d is separation between plate

When the plate separation becomes d / 2

Electric field E between plate

E₂ = V / d /2

= 2 V / d =2E₁

Or twice the earlier field

5 0

4 years ago

Scientists treat the number of stars in a given volume of space as a Poisson random variable. The density of our galaxy in the v

vladimir1956 [14]

Answer:

$P(X\ge 1) = 0.9502$

Explanation:

Given

Density = 3 starts in 10 cubic light years.

Required

Determine the probability of 1 or more in 10 cubic light years

Since the number of stars follow a Poisson distribution, we make use of:

$P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}$

$\lambda = density$

$\lambda = \frac{3}{10}$

$\lambda = 0.3$

T = the light years

$T = 10$

Calculating $P(X \ge 1)$

In probability:

$P(X \ge 1) = 1 - P(X = 0)$

Calculating P(X=0)

Substitute 0 for k and the values for $\lambda$ and T in

$P(X=k) = f(x) = (\lambda T)^k\frac{ e^{-\lambda T}}{k!}$

$P(X=0) = (0.3* 10)^0 * \frac{ e^{-0.3 * 10}}{0!}$

$P(X=0) = (3)^0 * \frac{ e^{-0.3 * 10}}{1}$

$P(X=0) = (3)^0 * e^{-0.3 * 10}$

$P(X=0) = 1 * e^{-0.3 * 10}$

$P(X=0) = 1 * e^{-3}$

$P(X=0) = e^{-3}$

$P(X=0) = 0.04979$

Substitute 0.04979 for P(X=0) in $P(X \ge 1) = 1 - P(X = 0)$

$P(X\ge 1) = 1 - 0.04979$

$P(X\ge 1) = 0.95021$

$P(X\ge 1) = 0.9502$ --- approximated

Hence, the required probability is 0.9502

3 0

3 years ago

a cepheid variable star is a star whose brightness alternately increases and decreases. suppose that cephei joe is a star for wh

MrRissso [65]

After one day, the rate of increase in Delta Cephei's brightness is;0.46

We are informed that the function has been used to model the brightness of the star known as Delta Cephei at time t, where t is expressed in days;

B(t)=4.0+3.5 sin(2πt/5.4)

Simply said, in order to determine the rate of increase, we must determine the derivative of the function that provides

B'(t)=(2π/5.4)×0.35 cos(2πt/5.4)

Currently, at t = 1, we have;

B'(1)=(2π/5.4)×0.35 cos(2π*1/5.4)

Now that the angle in the bracket is expressed in radians, we can use a radians calculator to determine its cosine, giving us the following results:

B'(1)=(2π/5.4)×0.3961

B'(1)≈0.46

To know more about:

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7 0

1 year ago

A box of books weighing 325 N moves at a constant velocity across the floor when the box is pushed with a force of 425 N exerted

Mazyrski [523]

Answer:

Mu,k between the floor and the box is 0.609

Explanation:

given information:

weight of the book, W = 325 N

applied force, F = 425 N

angle, θ = 35.2°

to find the coefficient kinetic between the floor and the box, wee need to calculate the horizontal and vertical force.

first we calculate the vertical force, there are weight, normal force(N) and applied vertical force.

according to Newton's first law

ΣF = 0

Σ $F_{y}$ = 0

W - N + F sin θ = 0

N = W + F sin θ

Next, we calculate the force in horizontal direction. applied force and friction force( $F_{friction}$ )

Σ $F_{x}$ = 0

F cos θ - $F_{friction}$ = 0

F cos θ - μk (W - F sin θ) = 0

μk = (F cos θ) / (W + F sin θ)

= (425 cos 35.2°)/(325+425 sin 35.2°)

= 0.609

3 0

3 years ago