For the Doppler effect to occur ?

A tiger travels 3m/s^2 for 4.1s,what was its initial speed if it's final speed was 55k/h?

zaharov [31]

Acceleration a=3m/s^2
time t= 4.1seconds
Final velocity V= 55km/h
initial velocity U= ?
First convert V to m/s
36km/h=10m/s
55km/h= 55*10/36=15.28m/s
Using the formula V= U+at
U= V-at
U= 15.28-3*4.1=15.28-12.3=2.98m/s
Initial velocity U= 2.98m/s or 10.73km/h (Using the conversion rate 36km/h=10m/s)

8 0

2 years ago

How is it technically correct to say that a car making a u-turn can have a constant speed but cannot have a constant velocity?

saw5 [17]

During the "U" part of the turn, the car would follow an approximately circular path, and if it's moving at a constant speed, it would have to accelerate toward the center of the circle in order to change its direction.

5 0

2 years ago

During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla

Butoxors [25]

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of $u=-40\ m/s$

ball returned with speed of $v=40\ m/s$

average Force imparted by racquet on the ball is given by

$F_{racquet}=\frac{m(v-u)}{\Delta t}$

where $m=mass\ of\ ball$

$\Delta t=$ time of contact of ball with racquet

$F_{racquet}=\frac{m(40-(-40))}{\Delta t}$

$F_{racquet}=\frac{80m}{\Delta t}-----1$

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

$F_{player}=\frac{m(0-40)}{\Delta t}$

$F_{player}=\frac{-40m}{\Delta t}-----2$

From 1 and 2 we get

$F_{racquet}=-2F_{player}$

Hence the magnitude of Force by racquet is twice the Force by player

5 0

3 years ago

The temperature of a room is 77°F what would be the temperature in Celsius scale

anygoal [31]

Formulas change from F to degree C : C = ( F - 32 )/1.8

so we have (77-32)/1.8 = 25 oC

ok done. Thank to me :>

5 0

2 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution decrease and the vapor pressure of the solution decrease .

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

2 years ago