PLEASE PLEASE HELP!!!Answer the following questions

Calculate the gravitational potential energy of a body of mass 40 kg at a vertical height of 10 m. ( g = 9.8 m/s2)

olganol [36]

Ep= mgh
Ep = 40 x 9.8 x 10
Ep = 3920J
Ep = 3900J (2sf)

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

A spring stretches 3.5 cm when a 9 g object is hung from it. The object is replaced with a block of mass 26 g which oscillates i

evablogger [386]

Answer:

The period of motion of new mass T = 0.637 sec

Explanation:

Given data

Mass of object (m) = 9 gm = 0.009 kg

Δx = 3.5 cm = 0.035 m

We know that spring force is given by

F = m g

F = 0.009 × 9.81 = 0.08829 N

Spring constant

$k = \frac{F}{x}$

$k = \frac{0.08829}{0.035}$

k = 2.522 $\frac{N}{m}$

New mass $(m_1)$ = 26 gm = 0.026 kg

Now the period of motion is given by

$T = 2 \pi \sqrt{\frac{m}{k} }$

$T = 2 \pi \sqrt{\frac{0.026}{2.522} }$

T = 0.637 sec

This is the period of motion of new mass.

3 0

2 years ago

What is the angular displacement of a minute hand of a clock after 3 minutes?

Inessa [10]

Answer:

π/10 rads

Explanation:

It takes an hour (60 minutes) for the minute's hand to turn a full circle or achieve an angular rotation of

2πl rad.

Now, number of periods of 3 minutes in an hour is;

Number of periods = 60/3 = 20 periods

Thus, 3 minutes rotation accounts for 1/20 of 2π the rotation of the minute's hand in an hour.

Thus;

Angular displacement = (1/20) * 2π = π/10 rads

6 0

2 years ago

Explain the working and performance of a centrifugal clutch with a sketch

timama [110]

a centrifugal clutch works, as the name suggests, through centrifugal force. ... The rotation of the hub forces the shoes or flyweights outwards until they come into contact with the clutch drum, the friction material transmits the torque from the flyweights to the drum. The drive is then connected

7 0

3 years ago