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3 years ago

If Mia ran 15 meters at a speed of 5 m/s, how long did it take her?

Physics

2 answers:

satela [25.4K]3 years ago

7 0

It took her 3 seconds. That’s 15/5.

photoshop1234 [79]3 years ago

5 0

It took her 3 seconds

Explanation: 15 Divided by 5 is 3.

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A soccer ball was kicked off the roof of a building going

WINSTONCH [101]

Answer:

Final vertical velocity = -29m/s

Horizontal distance = 100m

Height = 20.41m

Explanation:

1. The vertical final velocity can be calculated thus: vy = vyo - gt

Where;

vy = vertical velocity (m/s)

vyo = vertical initial velocity (20m/s)

g = acceleration due to gravity (9.8m/s²)

t = time (5s)

Hence, vy = vyo - gt

vy = 20 - (9.8 × 5)

vy = 20 - 49

vy = -29m/s

2. x = V0 x t

Where;

x = horizontal distance (m)

Vo = initial velocity

t = time (s)

x = 20 × 5

x = 100m

3. Maximum height = (voy)²/2g

= 20²/ 2 × 9.8

= 400/19.6

= 20.41m

4 0

3 years ago

An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in

asambeis [7]

Answer:

$E=(-17.08 i +7.2 j -7.6 k )N/C$

Explanation:

$v= (19.0j+18.k)km/s$

$a=3.0x10^{13}m/s^2$

$\beta= 400x10{-6}T$

Electron information needed to solve the question:

$m_e=9.11x10^{-31}kg$

$q=-1.6x10{-19}C$

$F=F_E+F_B=q*(E+Vx \beta)$

$F=m*a$

$m*a=q*(E+Vx\beta)$

$E=\frac{m*a}{q}-(Vx\beta )$

$E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]$

$E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C$

$E=(-17.08 i +7.2 j -7.6 k )N/C$

6 0

3 years ago

Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat

kvasek [131]

Answer:

a) $a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

b) $k = \frac{9.8}{9.74}=1.006$

Explanation:

Part a

For this case we can begin finding the period like this:

$T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s$

Then we know that the centripetal acceleration is given by:

$a= \frac{v^2}{r}$

And the velocity is given by:

$v=\frac{2\pi r}{T}$

If we replace this into the acceleration we got:

$a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}$

And we can replace the values and we got:

$a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

Part b

For this case we want to find a value of k such that:

$a= k 9.8$

Where a = 9.74, so then we can solve for k like this:

$k = \frac{9.8}{9.74}=1.006$

8 0

3 years ago

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st

irina [24]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, $31 {\rm {N}/{mm}}$ . What is the difference in maximum stored energy between the sprinters and the nonathlethes?