Darw a velocity time graph for body moving with uniform acceleration

You're sitting in the back of a taxi and place your bag on the seat next to you. The taxi enters a traffic circle and travels th

svp [43]

Answer:

(d) III only

Explanation:

We have to observe the motion of the bag with respect to taxi , considering taxi as stationary or inertial frame . Since bag is not moving with respect to taxi , the inertial frame that means , net force on it is zero .So option i and ii are ruled out .

Now how to explain motion of the bag ie why it is stationary ie what are the balancing force acting on it. We know that on a body on circular path , a force called centripetal force is acting on it . So that force must be acting on it . The balancing force is the frictional force which is keeping it stationary with respect to taxi . Hence the third option is correct.

5 0

3 years ago

A spaceship from a friendly, extragalactic planet flies toward Earth at 0.201 times the speed of light and shines a powerful las

bagirrra123 [75]

Answer:

The wavelength of observed light on earth is 568.5 nm

Explanation:

Given that,

Velocity of spaceship $v= 0.201c$

Wavelength of laser $\lambda= 697\ nm$

We need to calculate the wavelength of observed light on earth

Using formula of wavelength

$\lambda_{0}=\lambda_{e}\times\sqrt{\dfrac{1-\dfrac{v}{c}}{1+\dfrac{v}{c}}}$

$\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-\dfrac{0.201 c}{c}}{1+\dfrac{0.201c}{c}}}$

$\lambda_{0}=697\times10^{-9}\times\sqrt{\dfrac{1-0.201}{1+0.201}}$

$\lambda=5.685\times10^{-7}\ m$

$\lambda=568.5\times10^{-9}\ m$

$\lambda=568.5\ nm$

Hence, The wavelength of observed light on earth is 568.5 nm

8 0

4 years ago

The brakes of an automobile can decelerate it at -4.7m/s^2. If the automobile is traveling at the rate of 31.0m/s How far will i

OleMash [197]

Answer:

204.5 m

Explanation:

a = v \ t

t = v \ a

t = 31 \ 4.7 = 6.6s

v = d \ t

d = v * t

d = 31 * 6.6

d= 204.5 meters

6 0

3 years ago

A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the

Aleksandr-060686 [28]

Answer:

(a): The magnitude of the electric force on the small sphere = $\dfrac{q\sigma}{2\epsilon_o}.$

(b): Shown below.

Explanation:

<u>Given:</u>

m = mass of the small sphere.
q = charge on the small sphere.
L = length of the silk fiber.
$\sigma$ = surface charge density of the large vertical insulating sheet.

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

$\rm E = \dfrac{\sigma}{2\epsilon_o}.$

<em>where,</em>

$\epsilon_o$ is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

$\rm F_e=qE.$

Thus, the magnitude of the electric force on the small sphere is given by

$\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.$

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

$\rm \tan \theta = \dfrac{F_e}{W}.$

<em>where,</em>

$\rm F_e$ = electric force on the sphere, acting along left.
$\rm W$ = weight of the sphere, acting vertically downwards.

<em />

$\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .$

g is the acceleration due to gravity.

6 0

4 years ago

You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving

BabaBlast [244]

Y₀ = initial position of the balloon at the top of the building = 44 m

Y = final position of the balloon at halfway down the building = 44/2 = 22 m

a = acceleration of the balloon = - 9.8 m/s²

v₀ = initial velocity of the balloon = 0 m/s

v = final velocity of the balloon = ?

using the kinematics equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 0² + 2 (- 9.8) (22 - 44)

v = 20.78 m/s

4 0

4 years ago