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3 years ago

Birdman is flying horizontally at a

Physics

1 answer:

den301095 [7]3 years ago

5 0

Answer:

68 m

Explanation:

Given that the horizontal velocity of the birdman = 17 m/s and

the height, h= 78 m.

The gravitational force is acting in the downward direction, so it will not change the horizontal speed.

The horizontal speed will remains be constant and will be equal to the initial horizontal speed of the turd.

Initially, the turd was also flying horizontally with the birdman, so the initial velocity of the turd is the same as the horizontal velocity of the birdman, i.e In the horizontal direction, $u_0=17$ m/s.

In the vertical direction, u = 0,

The distance to be traveled, in the direction of application of force, is equals to the height of the turd, i.e

s= 78 m

Let t be the time taken to cover a distance of 78 m.

Now, applying the equation of motion in the vertical direction,

$s=ut+\frac 12 at^2$

where u is the initial velocity and a is the acceleration due to gravity in the direction of displacement,s.

Here, $a=g=9.81 m/s^2$ , so

$78=0\times t +\frac 12 (9.81)t^2$

$\Rightarrow t^2=(78\times2)/9.81$

$\Rightarrow t = 4$ seconds.

Hence, the time taken to reach the ground is 4 seconds.

As the horizontal speed, $u_0=17$ m/s, is constant throughout the journey, so

the horizontal distance covered by turd

$= u\times t$

$= 17 \times 4 = 68$ m.

So, the distance of landing from the start of the field is 68 m as the birdman releases a turd directly above the start of the field.

Hence, the robot must hold the bucket at a distance of 68 m from the start of the field.

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Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the

BlackZzzverrR [31]

relation between potential difference and electric field is given as

$E . d = \Delta V$

so here we know that

d = 3 cm

$\Delta V = 30 V$

$E \times 0.03 = 30$

$E = 1000 N/C$

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0

3 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

and

$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

where:

x = 1322 ft is the final position of the shell when it strikes the hill

$u=156 ft/s$ is the initial velocity of the shell

$\theta=49.0^{\circ}$ is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

$v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s$

Instead, the vertical component of the velocity is given by

$v_y=usin \theta -gt$

And substituting at t = 12.9 s (time at which the shell strikes the hill),

$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

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5 0

3 years ago

What is the velocity of a wave that has a wavelength of 20 meters and frequency of 0.5<br> Hz?

k0ka [10]

Answer:

v = 10 m/s

Explanation:

recall that velocity is related to wavelength and frequency by the formula

v = fλ

where v = velocity, f = frequency and λ= wavelength

Simply substitute these into the formula:

v = fλ

v = (0.5)(20)

v = 10 m/s

3 0

3 years ago

Read 2 more answers

The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p

aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

$F_{p}$ = $q_{p}$ E

and, we know that, F = ma

So,

$m_{p}$ a = $q_{p}$ E

a = $\frac{q_{p}.E }{m_{p} }$ Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2 $at^{2}$

Here, u = 0.

S = 1/2 $at^{2}$

Put equation 1 into the above equation:

S = 1/2 x ( $\frac{q_{p}.E }{m_{p} }$ ) $t^{2}$ Equation 2

So,

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ( $\frac{q_{e}.E }{m_{e} }$ ) $t^{2}$ Equation 3

We know that the charge of electron is equal to the charge of proton so,

$q_{p}$ = $q_{e}$ = q

By dividing the equation 2 by equation 3, we get:

$\frac{S}{D-S}$ = $\frac{m_{e} }{m_{p} }$

Solve the above equation for S,

S $m_{p}$ = $m_{e}$ D - $m_{e}$ S

So,

S = $\frac{m_{e}.D }{(m_{e} + m_{p}) }$

Plugging in the values,

As we know the mass of electron is 9.1 x $10^{-31}$ and the mass of proton is 1.67 x $10^{-27}$

S = $\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }$

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = $\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }$

As we know the mass of chlorine is 35.5 and of sodium is 23

S = $\frac{35.5 . 4.22}{(35.5 + 23)}$

S = 2.56 cm

7 0

2 years ago

One particle has a charge of 2.15 x 10^ -9 while another particle has a charge of 3.22 x 10^ -9 If the two particles are separat

marin [14]

Answer:

B. 2.77 x $10^{-4}$ N

Explanation:

The required force can be calculated by:

F = $\frac{Kq_{1}q_{2} }{d^{2} }$

Where F is the force between the particles, K is the coulomb's constant (9 x $10^{9}$ $Nm^{2}/C^{2}$ ), $q_{1}$ is the charge on the first particle, $q_{2}$ is the charge on the second particle and $d^{2}$ is the distance between the particles.

So that:

F = $\frac{9*10^{9}*2.15*10^{-9} *3.22*10^{-9} }{(0.015)^{2} }$

= $\frac{6.2307*10^{-8} }{(2.25*10^{-4} }$

= 2.7692 x $10^{-4}$

The force between the particles is 2.77 x $10^{-4}$ N.

3 0

2 years ago