A ball is thrown vertically up ward with a velocity of 25m/s. a/how fast was it moving after 2sec?

Due Sun 06/09/2019 11:59 pm Skip Navigation Questions correct Q 1 [1/1] correct Q 2 [1/1] correct Q 3 [1/1] untried Q 4 (0/1) un

astraxan [27]

Due Sun 06/09/2019 11:59 pm (you're already more than a week late) Skip Navigation Questions, correct Q 1 [1/1], correct Q 2 [1/1], correct Q 3 [1/1], untried Q 4 (0/1), untried Q 5 (0/1), untried Q 6 (0/1), untried Q 7 (0/1), untried Q 8 (0/1), untried Q 9 (0/1), untried Q 10 (0/1), Grade: 3/10, Print Version, Start of Questions:

Questions Two cyclists, 42 miles apart, start riding toward each other at the same time. One cycles 2 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?

Start of Answer:

-- When they started, they were 42 miles apart. When they met, 2 hours later, they were no miles apart. The distance between them shrank at the rate of 21 miles per hour. Since they rode directly toward each other, the sum of their individual speeds must have been 21 miles per hour.

-- One biker was two times the speed of the other. Slower biker: 1 time. Faster biker: 2 times. Sum of their speeds: 3 times = 21 miles per hour. Each 'time' = 7 miles per hour.

-- Slower cycler = 1 time = 7 mi/hr

Faster cycler = 2 times = 14 mi/hr

5 0

3 years ago

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

NeX [460]

Answer:

The transverse displacement is $y(1.51 , 0.150) = 0.055 m$

Explanation:

From the question we are told that

The generally equation for the mechanical wave is

$y(x,t) = Acos (kx -wt)$

The speed of the transverse wave is $v = 8.25 \ m/s$

The amplitude of the transverse wave is $A = 5.50 *10^{-2} m$

The wavelength of the transverse wave is $\lambda = 0540 m$

At t= 0.150s , x = 1.51 m

The angular frequency of the wave is mathematically represented as

$w = vk$

Substituting values

$w = 8.25 * 11.64$

$w = 96.03 \ rad/s$

The propagation constant k is mathematically represented as

$k = \frac{2 \pi}{\lambda}$

Substituting values

$k = \frac{2 * 3.142}{0. 540}$

$k =11.64 m^{-1}$

Substituting values into the equation for mechanical waves

$y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03 * 0.150))$

$y(1.51 , 0.150) = 0.055 m$

4 0

3 years ago

The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and

viktelen [127]

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

$1\ m=\frac{1}{1609.344}\ miles$

1 hour = 60×60 seconds

$1\ s=\frac{1}{3600}\ hours$

$28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h$

Top speed of the cheetah is 64.2 mi/h

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2$

Acceleration of the cheetah is 8.68 m/s²

2)

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s$

It takes a cheetah 3.31 seconds to reach its top speed.

3)

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m$

It travels 47.5 m in that time

4) When s = 120 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s$