An object is said to move from a position of 10m East to a position of 5m west. Determine the object's distance travelled.

A block of mass m=16.8 kg is sliding on a surface with initial velocity v=23.2 m/s. The block has a coefficient of kinetic frict

Firdavs [7]

Answer:

t = 23.92 s

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction, downward

FN : Normal force : perpendicular to the floor, upward

fk : Kinetic friction force: parallel to the floor and opposite to the movement

F = 86.4 N , in the direction of the motion

Calculated of the W

W= m*g

W= 16.8 kg* 9.8 m/s² = 164.64 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay ay = 0

FN - W = 0

FN = W

FN = 164.64 N

Calculated of the fk

fk = μk*FN

fk = 0.426* 164.64 N

fk = 70.13 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax , ax= a : acceleration of the block

F - fk = m*a

86.4 -70.13 = (16.8)*(-a)

16.26 = (16.8)*(-a)

a = -(16.26 )/ (16.8)

a = - 0.97 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula :

vf = v₀ + a*t Formula (2)

Where:

t: time interval (m)

v₀: initial speed (m/s)

vf: final speed (m/s)

Data:

v₀ = 23.2 m/s

vf = 0

a = -0.97 m/s²

Time it takes for the block to stop

We replace data in the formula (2) to calculate the time

vf= v₀+a*t

0 = 23.2+( -0.97)*t

(0.97)*t = 23.2

t = 23.2 / (0.97)

t = 23.92 s

7 0

3 years ago

relationship beetween image distance,focal length ,power magnification, for both concave and convex lens.

scoray [572]

Answer:

Object distance means what is the distance between pole and object. Image distance means when image is formed then the distance between pole and image is called image distance. Focal length is the distance between pole and the principal focus of the mirror.

A lens is a clear object, usually made of glass or plastic, which is used to refract, or bend light. Lenses can concentrate light rays (bring them together) or spread them out. Common examples of lenses include camera lenses, telescope lenses, eyeglasses, and magnifying glasses. Lenses are often double lenses, meaning they have two curved sides. A convex lens is rounded outward, while a concave lens curves inward. (A great way to remember this is that a concave lens creates an indent like a cave!)

The image distance can be calculated with the knowledge of object distance and focal length with the help of lens formula. In optics, the relationship between the distance of an image (i), the distance of an object (o), and the focal length (f) of the lens are given by the formula known as Lens formula. Lens formula is applicable for convex as well as concave lenses. These lenses have negligible thickness. It is an equation that relates the focal length, image distance, and object distance for a spherical mirror. It is given as,

1/i + 1/o = 1/f

i= distance of the image from the lens

o= distance of the object from the lens

f= focal length of the lens

Explanation:

Hope it is helpful....

7 0

3 years ago

Read 2 more answers

Consider two antennas separated by 9.00 m that radiate in phase at 120 MHz, as described in Exercise 35.3. A receiver placed 150

alexgriva [62]

Answer:

$\phi=4.52 rad$

Explanation:

From the question we are told that

Distance b/e antenna's $d=9.00m$

Frequency of antenna Radiation $F_r=120 MHz \approx 120*10^6Hz$

Distance from receiver $d_r=150m$

Intensity of Receiver $i= 10$

Distance difference of the receiver b/w antenna's $(r^2-r^1)=1.8m$

Generally the equation for Phase difference $\phi$ is mathematically given by

$\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)$

$\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8$

$\phi=\frac{4\pi}{5} *1.8$

Therefore phase difference f between the two radio waves produced by this path difference is given as

$\phi=4.52 rad$

7 0

3 years ago

The components of vector A are:

Korvikt [17]

here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.

So angle must be more than 270 degree and less than 360 degree

Now in order to find the value we can say that

$tan\theta = \frac{opposite\: side}{adjacent\: side}$

$tan\theta = \frac{8.6}{6.1}$

$\theta = tan^{-1}1.41$

$\theta = 54.65^0$

so it is inclined at above angle with X axis in fourth quadrant

Now if angle is to be measured counterclockwise then its magnitude will be

$\theta = 360 - 54.65 = 305.3^0$

so the correct answer will be 305 degree

3 0

3 years ago

A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),

labwork [276]

Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

FOR FORCE (P):

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

Therefore, work done by force (P) is Positive.

FOR NORMAL FORCE (Fn) AND WEIGHT (W):

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.

FOR KINETIC FRICTIONAL FORCE (fk):

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

Therefore, work done by Kinetic Frictional Force (fk) is Negative.

8 0

3 years ago