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LiRa [457]
2 years ago
14

If a rock weighing 3.5 newtons is resting on cliff's ledge 12 meters from the canyon floor, what is its potential energy?

Physics
1 answer:
podryga [215]2 years ago
6 0

Answer:

42 Joules

Explanation:

Potential energy

= weight × height

= (3.5)(12)

= <u>42 joules</u>

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When a vapor condenses to a liquid, an amount of thermal energy (Q
mina [271]

Answer:

true

Explanation:

I did this unit for science

6 0
2 years ago
Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s
Zielflug [23.3K]

Answer:

there will be collision

Explanation:

v_{s} =  speed of sue = 34 m/s

v_{v} = speed of van = 5.20 m/s

v_{sv} = speed of sue relative to van  = v_{s} - v_{v} = 34 - 5.20 = 28.8 m/s

d_{s} = stopping distance after brakes are applied

D = distance between sue and van = 160 m

v_{f} = final speed of sue = 0 m/s

a = acceleration = - 1.80 m/s²

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a d_{s}

0^{2} = 28.8^{2} + 2 (1.80) d_{s}

d_{s} = 230.4 m

Since  d_{s} < D

hence there will be collision

7 0
3 years ago
In 1939 or 1940, Emanuel Zacchini took his human-cannonball act to an extreme. After being shot from a cannon, he soared over th
Arlecino [84]
 We first determine the vertex by using the formula,<span>-b/2a = vertex, in order to get the values for the t-coordinate. That is why we got 
</span>
v_y=26.5 sin(53)=21.163v_x=26.5 cos(53)=15.948 
then

let x=0since you are going to land on a 3m tally=-.5(9.8)t^2+ 21.163*t
y=0=-4.9t+21.163t=4.31
vx*4.31= total distance travelled=68.88m
Then for the first wheel, you have 15.948m=vxdetermine the time when he reaches 23 meters, that is

23/15.948=1.44218 sec

substitute t with1.44218 sec, then determine the height.
h(1.44218)=20.329

determine vertex by using a graphing calculatort=2.1594s h=22.85m
using the time value of the vertex, determine horizontal distance travelled
34.438m away from cannon
8 0
3 years ago
Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

c) wavelength

        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

  T = \dfrac{v^2\ m}{L}

  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

g) Power transmitted by the wave

  P = \dfrac{1}{2}m\ v \omega^2\ A^2

  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

      P = 0.438 W

5 0
3 years ago
Star A and Star B have different apparent brightnesses but identical luminosities. Star A is 10 light years away from earth and
STALIN [3.7K]

intensity of a star is inversely depends on the square of the distance from the star

we can say it is given as

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

here we know that

\frac{I_1}{I_2} = 36 times

also we know that

r_1 = 10 Ly

now we will have

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

36 = \frac{r_2^2}{10^2}

r_2 = 60 Ly

so other star is at distance 60 Light years

8 0
3 years ago
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