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olga nikolaevna [1]
3 years ago
8

Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the p

uck begins to move to the moment it hits the spring. Use 0.253 m for the displacement of the puck along the ramp and 9.80 m/s2 for the acceleration due to gravity. Assume that the mass of the puck is 0.180 kg. Express your answer using SI units to three significant figures.
Physics
1 answer:
jekas [21]3 years ago
7 0

Answer:

0.16joules

Explanation:

Using the relation for The gravitational potential energy

E= Mgh

Where,

E= Potential energy

h = Vertical Height

M = mass

g = Gravitational Field Strength

To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

So E = mghsintheta

= 0.18 x 0.98 x 0.253 sin22

=0.16joules

Explanation:

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On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
3 years ago
Help me, please with this queastion
sergejj [24]

Answer:

Kinetic energy is the energy due to motion. Potential energy is energy stored in matter. The joule (J) is the SI unit of energy and equals (kg×m2s2) ( kg × m 2 s 2 ) .

please mark me brainliest and follow me my friend.

7 0
3 years ago
when a tuning fork vibrates over an open pipe and the air in the pipe starts to vibrate, the vibrations in the tube are caused b
malfutka [58]
Sound waves....................................
7 0
4 years ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
Read 2 more answers
When you drive through deep water, you should dry out your brakes by _____________. pumping your brakes and driving in low gear
Mandarinka [93]

Answer:

By Applying pressure to the brakes

Explanation:

Driving cars through deep water that is more than 10cm can make the cars to float. Most modern cars are usually water- tight so they can start to float through water that is about 30cm deep, fast moving water is very powerful so one needs to be very careful when driving.

If the brakes are wet test them by pressing or tapping on them gently.

You can as well dry brakes by driving in low gear and applying pressure to the brakes.

7 0
3 years ago
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