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3 years ago

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A force of 6.00 newtons on a car's bumper makes an angle of 28.5 degrees with the ground. What is the vertical component of this

force vector?

1 answer:

Zielflug [23.3K]3 years ago

7 0

Your answer by using the following calculation"

6sin(28.5) = 2.862

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A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system,

NARA [144]

As per the question the initial speed of the car [ u] is 42 m/s.

The car applied its brake and comes to rest after 5.5 second.

The final velocity [v] of the car will be zero.

From the equation of kinematics we know that

$v=u+at$ [ here a stands for acceleration]

$0=42 +5.5a$

$a =\frac{-42}{5.5} m/s^2$

$a= -7.64 m/s^2$

Here a is taken negative as it the car is decelerating uniformly.

We are asked to calculate the stopping distance .

From equation of kinematics we know that

$S=ut+\frac{1}{2} at^2$ [here S is the distance]

$= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m$

$=115.445 m$ [ans]

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3 years ago

Why does the sun appear to rise in the east and set in the west every day?

anygoal [31]

Answer: even heating

Explanation:

the earth rotates

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3 years ago

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You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.

valentinak56 [21]

Answer:

$F=6\times 10^{-7}\ N$

Explanation:

Given:

quantity of point charge, $q=-4\times 10^{-9}\ C$

radial distance from the linear charge, $r=0.09\ m$

linear charge density, $\lambda=3\times 10^{-9}\ C.m^{-1}$

<u>We know that the electric field by the linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.\epsilon_0.r}$

$E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}$

$E=150\ N.C^{-1}$

<u>Now the force on the given charge can be given as:</u>

$F=E.q$

$F=150\times 4\times 10^{-9}$

$F=6\times 10^{-7}\ N$

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4 years ago

What is the kinetic energy of a baseball moving at a speed of 40 m/s if the baseball has mass of 0.15 kg?

bonufazy [111]

0.5mv^2 > 0.5*0.15*40*40=120J

4 0

3 years ago

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An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

x = initial distance of the electron from the proton = 6 cm = 0.06 m

y = initial distance of the electron from the proton = 3 cm = 0.03 m
u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

m = mass of an electron = $9.1\times 10^{-31}\ kg$
v = final velocity of the electron
e = magnitude of charge on an electron = $1.6\times 10^{-19}\ C$

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

$\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\$

$\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\$

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

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3 years ago