Ask question

Ask question

All categories

3 years ago

13

A book that weighs 5 newtons is sitting on a table 1.5 meters above the floor. How much gravitational potential energy does the

book have?6.5 joules
4.5 joules
7.5 joules
3.33 joules

1 answer:

valina [46]3 years ago

5 0

Answer:7.5J

Explanation:7.5

J

Explanation:

P

E

=

m

g

h

Since

F

=

m

g

in this case

P

E

=

F

h

P

E

=

5

N

⋅

1.5

m

=

7.5

You might be interested in

The energy that is released by cellular respiration is the form of

Y_Kistochka [10]

I believe the energy released in cellular respiration is in the form of ATP.

6 0

4 years ago

Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea

steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at P = 100 k Pa and T = 500 K

mass is constant

$m_1 = m_2$

$\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}$

Pressure is same because it's not changing due to process

$\dfrac{V}{500} = \dfrac{2 V}{T_2}$

$T_2 = 1000\ K$

$\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}$

$\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})$

$m = \dfrac{P_1V_1}{RT_1}$

$m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}$

m = 1.39 x 10⁻³ Kg

$\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)$

$\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K$

5 0

3 years ago

9. A Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table. How long does it take to hit the floor if no one sneezes? Wh

Paha777 [63]

By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s

Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.

Height h = 1.0 m

As the ball rolls off the table, it will be fallen under gravity. Where

g = 9.8 m/ $s^{2}$

Initial vertical velocity $u_{y}$ = 0

Initial horizontal velocity $u_{x}$ = 5 m/s

Considering the vertical distance, the formula to use to calculate the time will be;

h = ut + 1/2g $t^{2}$

1 = 0 + 1/2 x 9.8 $t^{2}$

1 = 4.9 $t^{2}$

$t^{2}$ = 1/4.9

t = $\sqrt{0.204}$

t = 0.45 seconds

It takes 0.45 seconds to hit the floor if no one sneezes.

To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.

Vertical component

$V_{y}$ = $U_{y}$ + gt

$V_{y}$ = 0 + 9.8(0.45)

$V_{y}$ = 4.41 m/s

Horizontal component

$V_{x}$ = $u_{x}$ + at

but a = 0

$V_{x}$ = 5 m/s

Final velocity V = $\sqrt{5^{2} + 4.41^{2} }$

V = 6.67 m/s

Therefore, it will hit the floor at a velocity of 6.7 m/s

Learn more here: brainly.com/question/5063616

8 0

3 years ago

Consider a thin, spherical shell of radius 14.5 cm with a total charge of 29.5 µC distributed uniformly on its surface. (a) Find

ra1l [238]

Answer:

10.0 zero, by Gauss' Law the symmetrical distribution will produce no internal electric fields

21.5 E = k Q / R^2 behaves as if all charge were at center

E = 9 E9 * 29.5 E-6 / .215^2 = 5.74E6 N/C

8 0

3 years ago

In a "Rotor-ride" at a carnival, people rotate in a vertical cylindrically walled "room." If the room radius is 5.5 m, and the r

Crank

Answer:

0.181

Explanation:

We can convert the 0.5 rps into standard angular velocity unit rad/s knowing that each revolution is 2π:

ω = 0.5 rps = 0.5*2π = 3.14 rad/s

From here we can calculate the centripetal acceleration

$a_c = \omega^2r = 3.14^2*5.5 = 54.3 m/s^2$

Using Newton 2nd law we can calculate the centripetal force that pressing on the rider, as well as the reactive normal force:

$F = N = a_cm = 54.3 m$

Also the friction force and friction acceleration

$F_f = N\mu = 54.3 m \mu N$

$a_f = F_f / m = 54.3 \mu$

For the rider to not slide down, friction acceleration must win over gravitational acceleration g = 9.81 m/s2:

$g = a_f = 54.3 \mu$

$9.81 = 54.3 \mu$

$\mu = 9.81 / 54.3 = 0.181$

6 0

3 years ago