A 1200 kg car accelerates from 0 m/s to 25 m/s in 10 seconds. how much work was done on the car by the net force?

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m

son4ous [18]

Answer:

A. 30.7cm

B. $1.7*10^{-10}C$

C. The electric field is directed away from the point of charge

Explanation:

A.

$because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm$

B.

Considering Gauss's law

$EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C$

C. The electric field directed away from the point of charge when the charge is positive.

3 0

3 years ago

Hey guys how to prevent from shock

Pie

U need an insulator to prevent the shock passing through , that's why there is a plastic sheet covering the copper wires so that it can prevent the electricity passing through as plastic is an insulator and does not conduct electricity

8 0

3 years ago

Which of the following is the correct formula for sulfur hexaflouride? SF 6 SF 5 SF 8 SF

alexira [117]

Answer: SF6

Explanation: Hexa is a prefix means 6. Sulfur Hexafluoride is expressed in SF6

6 0

3 years ago

What is the temperature of a star (in Kelvin) if its peak wavelength is 150 nm (that is, 150 x 10^-9 m)?

liq [111]

Answer:

19320 K

Explanation:

The temperature of a star is related to its peak wavelength by Wien's displacement law:

$T=\frac{b}{\lambda}$

where

T is the absolute temperature at the star's surface

$b=2.898 \cdot 10^{-3}m\cdot K$ is Wien's displacement constant

$\lambda$ is the peak wavelength

Here we have

$\lambda=150 nm = 150\cdot 10^{-9}nm$

Substituting into the equation, we find

$T=\frac{2.898\cdot 10^{-3}}{150\cdot 10^{-9}}=19320 K$

7 0

3 years ago

A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and

il63 [147K]

Answer:

He becomes a croco-dile food

Explanation:

From the question we are told that

The height is h = 2.0 m

The angle is $\theta = 20^o$

The distance is $w = 10m$

The speed is $u = 11 m/s$

The coefficient of static friction is $\mu = 0.02$

At equilibrium the forces acting on the motorcycle are mathematically represented as

$ma = mgsin \theta + F_f$

where $F_f$ is the frictional force mathematically represented as

$F_f =\mu F_x =\mu mgcos \theta$

where $F_x$ is the horizontal component of the force

substituting into the equation

$ma = mgsin \theta + \mu mg cos \theta$

$ma =mg (sin \theta + \mu cos \theta )$

making a the subject of the formula

$a = g(sin \theta = \mu cos \theta )$

substituting values

$a = 9.8 (sin(20) + (0.02 ) cos (20 ))$

$= 3.54 m/s^2$

Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as

$sin \theta = \frac{h}{l}$

making $l$ the subject

$l = \frac{h}{sin \theta }$

substituting values

$l = \frac{2}{sin (20)}$

$l = 5.85m$

Apply Newton equation of motion we can mathematically evaluate the final velocity at the end of the ramp as

$v^2 =u^2 + 2 (-a)l$

The negative a means it is moving against gravity

substituting values

$v^2 = (11)^2 - 2(3.54) (5.85)$

$v= \sqrt{79.582}$

$= 8.92m/s$

The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is

Initial velocity along the x-axis which is mathematically evaluated as

$v_x = vcos 20^o$

substituting values

$v_x = 8.92 * cos (20)$

$= 8.38 m/s$

Initial velocity along the y-axis which is mathematically evaluated as

$v_y = vsin\theta$

substituting values

$v_y = 8.90 sin (20)$

$= 3.05 m/s$

Now the motion through the pool in the vertical direction can mathematically modeled as

$y = y_o + u_yt + \frac{1}{2} a_y t^2$

where $y_o$ is the initial height,

$u_y$ is the initial velocity in the y-axis

$a_y$ is the initial acceleration in the y axis with a constant value of ( $g = 9.8 m/s^2$ )

at the y= 0 which is when the height above ground is zero

Substituting values

$0 = 2 + (3.05)t - 0.5 (9.8)t^2$

The negative sign is because the acceleration is moving against the motion

$-(4.9)t^2 + (2.79)t + 2m = 0$

Solving using quadratic formula

$\frac{-b \pm \sqrt{b^2 -4ac} }{2a}$

substituting values

$\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}$

$t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}$

$t = -0.39s \ or \ t = 1.02s$

since in this case time cannot be negative

$t = 1.02s$

At this time the position the motorcycle along the x-axis is mathematically evaluated as

$x = u_x t$

x $=8.38 *1.02$

$x =8.54m$

So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool

7 0

3 years ago