Ask question

Ask question

All categories

MariettaO [177]

3 years ago

8

How do you identify the zero reference position for calculating mechanical potential energy? What is the most common zero refere

nce position for system that has gravitational potential energy?

1 answer:

ANTONII [103]3 years ago

8 0

<span>The zero reference position should be the position the system has the least potential energy. As an object falls, its potential energy is converted into kinetic energy. Gravity's ability to do work (potential energy) reaches zero when it hits the ground and stops falling, hence the ground is the most common zero reference position.</span>

You might be interested in

find the velocity of the object for all relevent times find the position of the object for all relevent times a softball is popp

KATRIN_1 [288]

Answer:

whats the formula

Explanation:

4 0

2 years ago

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes th

Sonbull [250]

Answer:

V = 331.59m/s

Explanation:

First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.

S = ut + 1/2at²

Given height of the cliff S = 80m

initial velocity u = 0m/s²

a = g = 9.81m/s²

Substitute

80 = 0+1/2(9.81)t²

80 = 4.905t²

t² = 80/4.905

t² = 16.31

t = √16.31

t = 4.04s

Next is to get the vertical velocity

Vy = u + gt

Vy = 0+(9.81)(4.04)

Vy = 39.6324

Also calculate the horizontal velocity

Vx = 1330/4.04

Vx = 329.21m/s

Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.

V² = Vx²+Vy²

V² = 329.21²+39.63²

V² = 329.21²+39.63²

V² = 108,379.2241+1,570.5369

V² = 109,949.761

V = √ 109,949.761

V = 331.59m/s

Hence the speed of the shell as it hits the ground is 331.59m/s

7 0

3 years ago

Tides are a result of

Virty [35]

The answer should be D but if its not then im sorrry, idk

6 0

3 years ago

Read 2 more answers

A laser beam is incident at an angle of 33.0° to the vertical onto a solution of cornsyrup in water.(a) If the beam is refracted

IrinaK [193]

Answer:

1.29649

488.08706 nm

$6.14644\times 10^{14}\ Hz$

231715700.28346 m/s

Explanation:

n denotes refractive index

1 denotes air

2 denotes solution

$\lambda_0$ = 632.8 nm

From Snell's law we have the relation

$n_1sin\theta_1=n_2sin\theta_2\\\Rightarrow n_2=\dfrac{n_1sin\theta_1}{sin\theta_2}\\\Rightarrow n_2=\dfrac{1\times sin33}{sin24.84}\\\Rightarrow n_2=1.29649$

Refractive index of the solution is 1.29649

Wavelength is given by

$\lambda=\dfrac{\lambda_0}{n_2}\\\Rightarrow \lambda=\dfrac{632.8}{1.29649}\\\Rightarrow \lambda=488.08706\ nm$

The wavelength of the solution is 488.08706 nm

Frequency is given by

$f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{488.08706\times 10^{-9}}\\\Rightarrow f=6.14644\times 10^{14}\ Hz$

The frequency is $6.14644\times 10^{14}\ Hz$

$v=\dfrac{c}{n_2}\\\Rightarrow v=\dfrac{3\times 10^8}{1.29469}\\\Rightarrow v=231715700.28346\ m/s$

The speed in the solution is 231715700.28346 m/s

8 0

3 years ago

Can someone please help me with these physics problems? I just don’t even know where to start.

KIM [24]

#1

for the block of mass 5 kg normal force is given as

$F_n = mg$

$F_n = 5*9.8 = 49 N$

friction force is given as

$F_f = \mu F_n$

$F_f = 0.1*49 = 4.9 N$

Net force is given as

$F_{net} = ma$

$F_{net} = 5*2 = 10 N$

now we know that

$F_{net} = F_{app} - F_f$

$10 = F_{app} - 4.9$

$F_{app} = 14.9 N$

#2

Normal force is given as

$F_n = mg$

$F_n = 6*9.8$

$F_n = 58.8 N$

now we know that

$F_{net} = F_{app} - F_f$

$F_{net} = 0$

as object moves with constant velocity

$F_{app} = F_f = 15 N$

now for coefficient of friction we can use

$F_f = \mu F_n$

$15 = \mu * 58.8$

$\mu = 0.255$

#3

net force upwards is given as

$F = 1.2 * 10^{-4} N$

mass is given as

$m = 7 * 10^{-5} kg$

now as per newton's law we can say

$F = ma$

$1.2 * 10^{-4} = 7 * 10^{-5} * a$

$a = 1.71 m/s^2$

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

$F_{net} = F_{app} - F_f$

$ma = F_{app} - F_f$

$5*6 = 40 - F_{f}$

$F_f = 40 - 30 = 10 N$

part b)

for coefficient of friction we can use

$F_f = \mu F_n$

$10 = \mu * F_n$

here normal force is given as

$F_n = mg = 5*9.8 = 49 N$

now we have

$\mu = \frac{10}{49} = 0.204$

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$20 = 0 + \frac{1}{2}a(5^2)$

$a = 1.6 m/s^2$

now net force is given as

$F_{net} = ma$

$F_{net} = 10*1.6 = 16 N$

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 25}{1.5} = -16.67 m/s^2$

now the force is gievn as

$F = ma$

$F = 5*16.67 = 83.3 N$

3 0

3 years ago