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2 years ago

12

A Martian rover is descending a hill sloped at 30° with the horizontal. It travels with a constant velocity of 2 meters/second.

Calculate its horizontal velocity.

2 answers:

kodGreya [7K]2 years ago

8 0

Vx = 2*cos30 = 1.73m's
Other words:
D = Vo*t = 1.4 * 21 = 31.5m. @ 30 Deg.
Dy = 31.5*sin30 = 15.75 m.

uysha [10]2 years ago

7 0

<u>Answer:</u> The horizontal velocity of rover is 1.73 m/s

<u>Explanation:</u>

Horizontal velocity is defined as the velocity of body proceeding in the horizontal direction. This velocity proceeds in the 'x' direction.

Mathematically,

$V_x=V_o\cos \theta$

where,

$V_o$ = constant velocity of rover = 2 m/s

$\cos \theta$ = cosine function of angle 30°

Putting values in above equation, we get:

$V_x=2\times \cos (30^o)\\\\V_x=1.73m/s$

Hence, the horizontal velocity of rover is 1.73 m/s

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Block 1 of mass m1 slides along a frictionless floor and into a one-dimensional elastic collision with stationary block 2 of mas

Elan Coil [88]

<span>Answers: (a) 2.0 m/s (b) 4 m/s

Method:

(a) By conservation of momentum, the velocity of the center of mass is unchanged, i.e., 2.0 m/s.

(b) The velocity of the center of mass = (m1v1+m2v2) / (m1+m2)

Since the second mass is initially at rest, vcom = m1v1 / (m1+m2)

Therefore, the initial v1 = vcom (m1+m2) / m1 = 2.0 m/s x 6 = 12 m/s

Since the second mass is initially at rest, v2f = v1i (2m1 /m1+m2 ) = 12 m/s (2/6) = 4 m/s </span>

7 0

2 years ago

An object is located 13.5 cm in front of a convex mirror, the image being 7.05 cm behind the mirror. A second object, twice as t

goldfiish [28.3K]

Answer:

Second object is located at 42.03 cm in front of mirror

Explanation:

In this question we have given,

object distance from convex mirror ,u=-13.5cm

Image distance from convex mirror,v=7.05cm

let focal length of convex mirror be f

we have to find the distance of second object from convex mirror

we know that u, v and f are related by following formula

$\frac{1}{f} =\frac{1}{v}+ \frac{1}{u}$ .............(1)

put values of u and v in equation (1)

we got,

$\frac{1}{f} =\frac{1}{7.05}+ \frac{1}{-13.5}$

$\frac{1}{f}=\frac{13.5-7.05}{13.5\times 7.05}$

$\frac{1}{f}=\frac{6.45}{13.5\times 7.05}\\f=13.5\times 1.09\\f=14.75$

we have given that

second object is twice as tall as the first object

and image height of both objects are same

it means

$o_{2}=2o_{1}\\i_{1}=i_{2}$ .............(2)

we know that

$\frac{v}{u}=\frac{i}{o}\\i=\frac{o\times v}{u}$

therefore,

$i_{1}=\frac{o_{1}\times v}{u}$ .................(3)

put values of v and u in equation 3

$i_{1}=-\frac{o_{1}\times 7.05}{13.5}$

$i_{1}=-0.52o_{1}$

therefore from equation 2

$i_{2}=-0.52o_{1}$

we know that

$i_{2}=\frac{o_{2}\times V}{U}$ .................(4)

put value of $i_{2}$ and $o_{2}$ in equation 4

$-.52o_{1}=\frac{2o_{1}\times V}{U}$

$U=\frac{2o_{1}\times V}{-.52o_{1}} \\U=-3.85V$

we know that U,V and f are related by following formula

$\frac{1}{f} =\frac{1}{V}+ \frac{1}{U}$ .............(5)

put values of f and U in equation 5

we got

$\frac{1}{14.75} =\frac{1}{V}- \frac{1}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}$

$\frac{1}{14.75} =\frac{2.85}{3.85V}\\V=\frac{2.85\times 14.75}{3.85}\\V=10.91 cm$

Therefore,

U=-10.91\times 3.85

U=-42.03 cm

Second object is located at 42.03 cm in front of mirror

4 0

2 years ago

A 0.20 kg mass on a horizontal spring is pulled back 2.0 cm and released. If, instead, a 0.40 kg mass were used in this same exp

KIM [24]

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

$E=U=\frac{1}{2}kA^2$ (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

8 0

2 years ago

Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward rad

mote1985 [20]

Answer:

468449163762.0812 W

Explanation:

m = Mass = $\rhoV$

V = Volume = $\dfrac{4}{3}\pi r^3$

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

$\rho$ = Density = 19 g/cm³

c = Speed of light = $3\times 10^8\ m/s$

A = Area = $\pi R^2$

I = Intensity = $\dfrac{P}{4\pi r^2}$

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

$F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}$

According to the question

$F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W$

The power required of the light source is 468449163762.0812 W

4 0

2 years ago

A particle of mass M is moving in the positive x direction with speed v. It spontaneously decays into 2 photons, with the origin

anygoal [31]

Solution :

Mass of the particle = M

Speed of travel = v

Energy of one photon after the decay which moves in the positive x direction = 233 MeV

Energy of second photon after the decay which moves in the negative x direction = 21 MeV

Therefore, the total energy after the decay is = 233 + 21

= 254 MeV

So by the law of conservation of energy, we have :

Total energy before the decay = total energy after decay

So, the total relativistic energy of the particle before its decay = 254 MeV

7 0

2 years ago