Which of the following is a derived unit? ampere newton second kilogram

You use 8x binoculars were used on a warbler (14cm long) in a tree 18cm away. What angle (in degrees) does the image of the warb

mafiozo [28]

Answer:

The angle it subtend on the retina is $\theta_z = 0.44586^o$

Explanation:

From the question we are told that

The length of the warbler is $L = 14cm = \frac{14}{100} = 0.14m$

The distance from the binoculars is $d = 18cm = \frac{18}{100} = 0.18m$

The magnification of the binoculars is $M =8$

Without the 8 X binoculars the angle made with the angular size of the object is mathematically represented as

$\theta = \frac{L}{d}$

$\theta = \frac{0.14}{0.18}$

$= 0.007778 rad$

Now magnification can be represented mathematically as

$M = \frac{\theta _z}{\theta}$

Where $\theta_z$ is the angle the image of the warbler subtend on your retina when the binoculars i.e the binoculars zoom.

So

$\theta_z = M * \theta$

=> $\theta_z =8 * 0.007778$

$= 0.0622222224$

Generally the conversion to degrees can be mathematically evaluated as

$\theta_z = 0.062222224 * (\frac{360 }{2 \pi rad} )$

$\theta_z = 0.44586^o$

7 0

3 years ago

After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. when the box reaches position

KiRa [710]

As we know by work energy theorem

total work done = change in kinetic energy

so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

$W_p = KE_f - KE_i$

initial the box is at rest at position x = x1

so initial kinetic energy will be ZERO

at final position x = x2 final kinetic energy is given as

$KE_f = \frac{1}{2}mv_1^2$

now work done is given as

$W_p = \frac{1}{2}mv_1^2 - 0$

so we can say

$W_p = \frac{1}{2}mv_1^2$

so above is the work done on the box to slide it from x1 to x2

3 0

3 years ago

What is the distance that a car travels if it was brought to stop in 5 seconds and if it was traveling at 110 Km/h

Triss [41]

Answer:

Suppose that the acceleration is a constant, a.

a(t) = a.

To write the velocity equation, we must integrate over time, and the constant of integration will be equal to the initial velocity, in this case is 110km/h.

v(t) = a*t + 110km/h

And we know that at t = 5s, the car was brought to stop, so the velocity must be zero.

v(5s) = 0 = a*5s + 110km/h.

a = (110km/h)*(1/5s)

now we have that:

1 hour = 3600 seconds.

1km = 1000m

then:

110km/h = (110*1000/3600)m/s = 30.56 m/s

Then we have:

a = (-30.55 m/s)/5s = -6.11 m/s^2

Now the velocity equation is:

v(t) = -6.11m/s^2*t + 30.56m/s

To write the positon equation we must integrate over time again, we can get:

p(t) = (1/2)*(-6.11m/s^2)*t^2 + (30.56m/s)*t + p0

Where p0 is the initial position, here i will assume that is zero, because it does no really mater.

The total displacement of the car will be equal to p(5s)

p(5s) = (1/2)*(-6.11m/s^2)*(5s)^2 + (30.56m/s)*(5s) = 76.425 meters.

6 0

3 years ago

Consider the hydrogen atom as described by the Bohr model. The nucleus of the hydrogen atom is a single proton. The electron rot

riadik2000 [5.3K]

To solve this problem we will apply the concept of centripetal acceleration. This type of acceleration is described as the product between the square of the angular velocity and the turning radius. Mathematically the expression can be expressed as

$a_c = \omega^2 r$

Here,

$\omega =$ Angular velocity

r = Radius

Our values are given as,

$\omega = 4.12*10^{16}rad/s$

$r = 5.29*10^{-11}$

Replacing,

$a_c = (4.12*10^{16})^2( 5.29*10^{-11})$

$a_c = 8.979*10^{22}m/s^2$

Therefore the electron's centripetal acceleration is $8.979*10^{22}m/s^2$

7 0

3 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At t = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive x-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

4 0

2 years ago