The new deal helped fix the economy but was not the perfect solution

Salmon often jump waterfalls to reach their

natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

$v_x = u cos \theta$

where u is the initial speed and $\theta=37.7^{\circ}$ . The horizontal distance travelled by the salmon is

$d=v_x t = (ucos \theta)t$

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

$t=\frac{d}{v_x}=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$y=h+u_y t -\frac{1}{2}gt^2$

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

$u_y = u sin \theta$ is the vertical component of the initial velocity of the salmon

$g=9.81 m/s^2$ is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

$y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}$

and we can solve this formula for u, the initial speed of the salmon:

$y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s$

5 0

4 years ago

If the period of a wave is 0.05 s, then what is its frequency?

yKpoI14uk [10]

Frequency = 1/time period = 1/0.05 = 20s^-1.

8 0

3 years ago

Read 2 more answers

An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pip

OverLord2011 [107]

Answer:

The value is $v_2 = 5.53 \ m /s$

Explanation:

From the question we are told

The pipe diameter at location 1 is $d = 8.8 \ cm = \frac{8.8 }{10} = 0.88 \ m$

The velocity at location 1 is $v_1 = 2.4 \ m /s$

The diameter at location 2 is $d_2 = 5.80 \ cm = 0.58 \ m$

Generally the area at location 1 is

$A_1 = \pi * \frac{d^2}{ 2}$

=> $A_1 = \pi * \frac{0.88^2}{ 2}$

=> $A_1 = 3.142 * \frac{0.88^2}{ 2}$

=> $A_1 = 1.2166 \ m^2$

Generally the area at location 1 is

$A_2 = \pi * \frac{d_1^2}{ 2}$

=> $A_2 = \pi * \frac{0.58^2}{ 2}$

=> $A_2 = 0.528 \ m^2$

Generally from continuity equation we have that

$A_1 * v_1 = A_2 * v_2$

=> $1.2166 * 2.4 = 0.528 * v_2$

=> $v_2 = 5.53 \ m /s$

3 0

3 years ago

Two blocks of ice, one four times as heavy as the other, are at rest on a frozen lake. A person pushes each block the same dista

qaws [65]

Answer:b

Explanation:

Given

mass of heavy object is 4m

mass of lighter object is m

A person pushes each block with same force F

According to Work Energy theorem Change in kinetic energy of object is equal to Work done by all the object

As launching velocity is same for both the object so heavier mass must possess greater kinetic energy . For same force heavier mass must be pushed 4 times farther than the light block .

$\Delta (K.E.)_H=\frac{1}{2}(4m)v^2$