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I am Lyosha [343]

2 years ago

12

Hideki is having trouble completing long workouts. What does he need to improve?

1 answer:

KonstantinChe [14]2 years ago

3 0

He needs to improve his endurance time.

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An airplane accelerates from rest down a runway at 3.20 m/s2 for 32.8 seconds until it lifts off the ground. Determine the dista

vagabundo [1.1K]

Answer:

s=1721.344m ,v=104.96m/s.

Explanation:

using thr equation of motion;

$s=ut+\frac{1}{2}at^{2}$

u=0, plane starts from rest,

$s=\frac{1}{2}at^{2}$

$a=3.2m/s^{2}, t=32.8s \\ s=\frac{1}{2}*3.2*32.8^{2}$

s=1721.344m

v=u+at

v=0 +3.2*32.8

v=104.96m/s

8 0

2 years ago

Statistically, consumers tell more people about negative experiences in business encounters than about positive experiences. A)

elena-s [515]

true hope this helps

8 0

3 years ago

Read 2 more answers

A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o

Ad libitum [116K]

Answer:

The thickness of the oil slick is $1.95\times10^{-7}\ m$

Explanation:

Given that,

Index of refraction = 1.28

Wave length = 500 nm

Order m = 1

We need to calculate the thickness of oil slick

Using formula of thickness

$2nt= m\lambda$

Where, n = Index of refraction

t = thickness

$\lambda$ = wavelength

Put the value into the formula

$2\times1.28 \times t=1\times\times500\times10^{-9}$

$t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }$

$t=1.95\times10^{-7}\ m$

Hence, The thickness of the oil slick is $1.95\times10^{-7}\ m$

4 0

2 years ago

A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a

Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

$v_x = \frac{x}{t}$

$v_x = \frac{67}{4.5}$

$v_x = 14.89m/s$

The vertical component of velocity is

$-h = v_y t -\frac{1}{2} gt^2$

Here,

h = Height

g = Gravitational acceleration

t = Time

$v_y$ = Vertical component of velocity

$-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2$

$-1.23= 4.5v_y - 99.225$

$v_y = 21.77m/s$

The direction of the velocity will be given by the tangent of the components, then

$tan\theta = \frac{v_y}{v_x}$

$\theta = tan^{-1} (\frac{21.77}{14.89})$

$\theta = 55.59\°$

The magnitude is given vectorially as,

$|V| = \sqrt{v_x^2+v_y^2}$

$|V| = \sqrt{14.89^2 +21.77^2}$

$|V| = 26.37m/s$

Therefore the angle is 55.59° and the velocity is 26.37m/s

6 0

2 years ago

A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a

Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial Velocity of Each object

$V_f$ = Final Velocity

Our values are given as

$m_1 = 2190Kg$

$v_1 =10.5m/s$

$m_2 = 3220kg$

$V_f = 4.74m/s$

Replacing we have that

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

$(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)$

$v_2 = 0.8224m/s$

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0

3 years ago