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3 years ago

6

An object moving at 38m/s takes 4s to come to a stop. What is the object’s acceleration?

1 answer:

Rudik [331]3 years ago

6 0

- 9.5 m/s^2

use the SUVAT method

0 = 38 + 4a

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An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's a

devlian [24]

Answer:

a

$D = 1162.7 \ m$

b

$\beta =- 65.55^o$

Explanation:

From the question we are told that

The speed of the airplane is $u = 92.3 \ m/s$

The angle is $\theta = 51.1^o$

The altitude of the plane is $d = 532 \ m$

Generally the y-component of the airplanes velocity is

$u_y = v * sin (\theta )$

=> $u_y = 92.3 * sin ( 51.1 )$

=> $u_y = 71.83 \ m/s$

Generally the displacement traveled by the package in the vertical direction is

$d = (u_y)t + \frac{1}{2}(-g)t^2$

=> $-532 = 71.83 t + \frac{1}{2}(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

=> $4.9t^2 - 71.83t - 532 = 0$

Solving this using quadratic formula we obtain that

$t = 20.06 \ s$

Generally the x-component of the velocity is

$u_x = u * cos (\theta)$

=> $u_x = 92.3 * cos (51.1)$

=> $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal direction is

$D = u_x * t$

=> $D = 57.96 * 20.06$

=> $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

$\beta = tan ^{-1}[\frac{v_y}{v_x } ]$

Here $v_y$ is the final velocity of the package along the vertical axis and this is mathematically represented as

$v_y = u_y - gt$

=> $v_y = 71.83 - 9.8 * 20.06$

=> $v_y = -130.05 \ m/s$

and v_x is the final velocity of the package which is equivalent to the initial velocity $u_x$

So

$\beta = tan ^{-1}[-130.05}{57.96 } ]$

$\beta =- 65.55^o$

The negative direction show that it is moving towards the south east direction

6 0

2 years ago

______ Is the distance and direction of an object's change in position from its starting point.

Vsevolod [243]

Answer:

Distance Is the distance and direction of an object's change in position from its starting point.

5 0

2 years ago

Read 2 more answers

A small lightbulb is 1.06 m from a screen. A) If you have a convex lens with 20 cm focal length, where are the two lens location

MatroZZZ [7]

Answer:

A)Explanation:

Let object distance be u.

Image distance v = 1.06 - u

Focal length = 20 cm = .2 m

Applying lens formula

1 / v - 1 / u = 1 / f

1 /( 1.06-u) - 1 / u = 1 / .2

5 u² -3.3u - 1.06 =0

It will have two roots

u₁ = .1636 m, u₂ = .8964 m

B ) Magnification

= u = .1636m , v = .8964m

m = .8964 / .1636 = 5.48

C ) When u = .8964m ,v = .1634m

m = .1634 / .8964

=.182

7 0

2 years ago

Turning on and listening your favorite TV show is an example of

tensa zangetsu [6.8K]

Answer:

Electrical Energy and Sound Energy

Explanation:

4 0

2 years ago

Light of wavelength 510 nm is used to illuminate normally two glass plates 24.3 cm in length that touch at one end and are separ

bulgar [2K]

Answer:

$m = 180 \ bright \ fringes$

Explanation:

From the question we are told that

The wavelength is $\lambda = 510 \ nm = 510 *10^{-9} \ m$

The length of the of the plate is $l = 24.3 \ cm = 0.243 \ m$

The radius of the wire is $r= 0.023 \ mm = 2.3 *10^{-5} \ m$

Generally the diameter of the wire which is the distance between the glass plates is mathematically represented as

$d = 2 * r$

=> $d = 2 * 2.3 *10^{-5}$

=> $d = 4.6 *10^{-5} \ m$

Generally the condition for constructive interference is mathematically represented as

$2 *d = m \lambda$

=> $m = \frac{2d}{\lambda}$

=> $m = \frac{2 * (4.6* 10^{-5})}{ 510 *10^{-9} }$

=> $m = 180 \ bright \ fringes$

3 0

2 years ago