THE PROBLEM!

Which type of radiation occur during nuclear processes? (Select all that apply)

Rainbow [258]

Answer:

Particle radiation

Explanation:

nuclear process is in atom nucleas maybe emit the electron ,neutrino or anti-neutrino

7 0

2 years ago

Newtons law of gravitation is called a universal law why

mihalych1998 [28]

Answer:

it is applicable on all bodies having mass and the bodies Will be governed by the same law

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

Anyone know how to do this?

Nadya [2.5K]

nope not really sorry

8 0

3 years ago

A man pushes a lawn mower on a level lawn with a force of 250N. If 40% of this force is directed downward, how much work is done

Dvinal [7]

When considering work, we always take the force directed along the axis of motion (in this case, the horizontal axis). If 40% of the force is directed downward, then 60% of the force is being directed horizontally, so the horizontal force is 250*0.6 = 150N. Work = Force * distance = 150N * 6.2m = 930J

4 0

3 years ago