PHYSICS COLLEGE

How much longer will it taketo travel a distance of 6ookm at
a speed of 50kmh than it
would at a
speed of 6okmh?
​

Answers

Answer 1
Answer:

Answer:

2hr much longer

Explanation:

Given parameters

  Distance  = 600km

   Speed 1  = 50km/h

   Speed 2 = 60km/h

Unknown:

How much longer will it take to travel a distance  = ?

Solution:

  Speed is the distance divided by time;

          Speed  = (distance)/(time)  

 Now;

           Time taken  = (Distance)/(Speed)  

Time 1;

                        = (600)/(50)

                       = 12hr

Time 2;

                       = (600)/(60)

                       = 10hr

To find how much more time;

            Time 1 will take 12hr - 10hr, 2hr much longer to travel the distance at that rate.

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1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of θ = 37◦ with the horizontal as shown in figure-1. The block is sliding at a constant velocity over a frictionless floor.(a) Find the value of the normal force on the block.

(b) Find the magnitude of force F~2 that is acting on the block

(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .

Answers

Answer:

Normal force=7.48 N

Explanation:

N+F~1 sinθ-mg=0

=>N=1.5*9.8-12 sin37◦

=>N=14.7-7.22=7.48 N

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?

Answers

Explanation:

Given

Acceleration of the pebble is

At t=0, velocity is

considering horizontal motion

\Rightarrow x=ut+0.5at^2 \n\Rightarrow 11=4.3t+0.5(4.6)t^2\n\Rightarrow 2.3t^2+4.3t-11=0\n\Rightarrow (t-1.4435)(t+3.3131)=0\n\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\n

Velocity acquired during this time

\Rightarrow v_x=4.3+4.6* 1.44\n\Rightarrow v_x=4.3+6.624\n\Rightarrow v_x=10.92\ s

Consider vertical motion

\Rightarrow v_y=0+7(1.44)\n\Rightarrow v_y=10.08\ m/s

Net velocity is

\Rightarrow v=√(10.92^2+10.08^2)\n\Rightarrow v=√(220.85)\n\Rightarrow v=14.86\ m/s

Angle made is

\Rightarrow \tan \theta =(10.08)/(10.92)\n\n\Rightarrow \tan \theta =0.92307\n\n\Rightarrow \theta =42.7^(\circ)

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters, t in seconds, and c and b are positive constants. What are the units of (a) constant c and (b) constant b? Find a formula in terms of c, b, and t of the (c) velocity v and (d) acceleration a. (e) At what time t ≥ 0 does the particle reach its maximum x value?

Answers

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=(2c)/(3b).

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=(dx)/(dt)\n=(d)/(dx)(ct^2-bt^3)\n=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = (dv)/(dt)\n=(d)/(dt)(2ct-3bt^2)\n=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left ((dx)/(dt)\right )_(t=t_o)=0.
  2. \rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.

Applying both these conditions,

\rm \left ( (dx)/(dt)\right )_(t=t_o)=0\n2ct_o-3bt_o^2=0\nt_o(2c-3bt_o)=0\nt_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = (2c)/(3b).

For \rm t_o = 0,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = (2c)/(3b),

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6b\cdot (2c)/(3b)=2c-4c=-2c.

Here,

\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.

Thus, the particle reach its maximum x value at time \rm t_o = (2c)/(3b).

A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?

Answers

Answer:

55.3

Explanation:

The computation of the number of bright-dark-bright fringe shifts observed is shown below:

\triangle m = (2d)/(\lambda) (n - 1)

where

d = 3.95 * 10^(-2)m

\lambda = 400 * 10^(-9)m

n = 1.00028

Now placing these values to the above formula

So, the  number of bright-dark-bright fringe shifts observed is

= (2 *3.95 * 10^(-2)m)/(400 * 10^(-9)m) (1.00028 - 1)

= 55.3

We simply applied the above formula so that the number of bright dark bright fringe shifts could come

Which is true about inelastic collisions: a. An inelastic collision does not obey conservation of energy. b. An inelastic collision conserves kinetic energy. c. Objects will stick together upon collision. d. Momentum is not conserved in inelastic collisions..

Answers

Answer:

Option c is correct

Explanation:

There are two types of collisions-elastic collision and inelastic collision.

In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.

Total energy is always conserved in both types. Thus, option a is incorrect.

In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.

Answer:

i believe its a?

Explanation:

In an inelastic collision, momentum is conserved

A straight trail with a uniform inclination of 16° leads from a lodge at an elevation of 600 feet to a mountain lake at an elevation of 7,000 feet. What is the length of thetrail (to the nearest foot)?
O A. 6,658 ft
OB. 25,396 ft
OC. 7,282 ft
OD. 23,219 ft​

Answers

I believed the answer is d
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