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2 years ago

Cam Newton can sprint 40 meters in 5.79 seconds! How fast can he run? Show your work

Physics

2 answers:

Setler79 [48]2 years ago

6 0

Speed=Distance/Time

Distance=40m,time=5.79seconds

S=40/5.79

=6.908m/s

vivado [14]2 years ago

4 0

Answer:

speed=v=6.9m/s

Explanation:

we know the formula

s=vt...........(1)

we are given

Distance=s=40m

and

time=t=5.79 s

we have to find

speed =v=?

from (1)

v=s/t

speed=v=40/5.79

speed=v=6.9m/s

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3 years ago

A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at

storchak [24]

Answer

given,

wavelength = λ = 18.7 cm

= 0.187 m

amplitude , A = 2.34 cm

v = 0.38 m/s

A) angular frequency = ?

$f = \dfrac{v}{\lambda}$

$f = \dfrac{0.38}{0.187}$

$f =2.03\ Hz$

angular frequency ,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) the wave number ,

$K = \dfrac{2\pi}{\lambda}$

$K= \dfrac{2\pi}{0.187}$

$K =33.59\ m^{-1}$

as the wave is propagating in -x direction, the sign is positive between x and t

y ( x ,t) = A sin(k x - ω t)

y ( x ,t) = 2.34 x sin(33.59 x - 12.75 t)

4 0

3 years ago

Can an object be accelerated while traveling at constant velocity? Why or why not?

AysviL [449]

Answer:

An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. If an object's velocity is constant, however, its acceleration will be zero.

3 0

3 years ago

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa

tester [92]

Answer:

Part a)

$t = 2.83 s$

Part b)

Ball thrown downwards = $v_f = 23.8 m/s$

Ball thrown upwards = $v_f = 23.8 m/s$

Part c)

$d = 22.24 m$

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

$\Delta y = 0 = v_y t + \frac{1}{2}at^2$

$0 = 13.9 t - \frac{1}{2}(9.81) t^2$

$t = 2.83 s$

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

$v_f^2 - v_i^2 = 2 a \Delta y$