Ask question

Ask question

All categories

3 years ago

14

A 40 kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hillside? (Formula:

PE = mgh)
3.9 m
4.0 m
39.2 m
40.0 m

2 answers:

igomit [66]3 years ago

5 0

Answer:

B) 4.0

Explanation:

Jus took the test on E D G E

Vika [28.1K]3 years ago

5 0

Answer:

it's b

Explanation:

i got it right

You might be interested in

In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal

kramer

Explanation:

The gravitational force equation is the following:

$F_G = G * \frac{m_1 m_2}{r^2} \\$

Where:

G = Gravitational constant = $6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}$

m1 & m2 = the mass of two related objects

r = distance between the two related objects

The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.

8 0

3 years ago

Correct shape will reduce fluid friction.<br> True <br> False

AURORKA [14]

I believe that is true.

hope this helps!

8 0

3 years ago

Read 2 more answers

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

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

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

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

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

A train goes up a hill with a 15º incline. If the train has constant speed of 22 m/s, what are the vertical and horizontal compo

Kobotan [32]

Any two-dimensional vector in cartesian (x,y) coordinates can be broken down into individual horizontal and vertical components using trigonometry. If a train goes up a hill with 15 degree incline at a speed of 22 m/s, the horizontal component is 22cos(15)=21.3 m/s and the vertical component is 22sin(15)=5.5 m/s.

8 0

3 years ago

Can anyone plz help me

Alekssandra [29.7K]

Answer:

1.) volume = 8 cm^3

2.) density = 2.6 g/cm^3

Explanation:

1.) volume =

$side {}^{3}$

=》

${2}^{3}$

=》

$8 \: cm {}^{3}$

2.) density =

$\frac{mass}{volume}$

=》

$\frac{20.8}{8}$

=》

$2.6 \: g \: cm {}^{ - 3}$

density = 2.6

3 0

3 years ago