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3 years ago

PLEASE HELP! I will give brainliest and ten points!

Physics

2 answers:

Dmitry_Shevchenko [17]3 years ago

4 0

I need help with this too! I also do Connexus

Andrews [41]3 years ago

3 0

Answer:

t = 3.64 s

Explanation:

As we know that the speed of the rock in vertical direction is ZERO

so here we will have

$\Delta Y = \frac{1}{2}gt^2$

$64.8 = \frac{1}{2}(9.8)t^2$

$64.8 = 4.9 t^2$

$t = 3.64 s$

Answer:

The package appears to fall straight downwards

Explanation:

Since the package is dropped from the plane so the horizontal speed of the plane and package will be same

So here the package will always appear vertically down the plane

So here it will appear to move vertically down

Answer:

8 cm to the right and 16 cm below

Explanation:

After one tringlet the position of the object is

$x = 4 cm$

$y = 4 cm$

Now we know that

$x = v_x t$

$y = \frac{1}{2}gt^2$

so here x coordinate is proportional to the time while y coordinate is square proportional to the time

so here in x direction it will move double distance in double time while in y direction its distance becomes 4 times

so we have

$x = 8 cm$

$y = 16 cm$

Answer:

y = 0.945 m

Explanation:

distance of the home plate

$d = 60 ft 6 in$

$d = 60.5\times 0.3048m$

$d = 18.44 m$

now the time taken by ball to reach the home plate

$t = \frac{d}{v}$

$t =\frac{18.44}{42} = 0.44 s$

Now the distance dropped by the ball in same time

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.8)(0.44^2)$

$y = 0.945 m$

Answer:

d = 282 m

Explanation:

Time taken by the packet to reach the waves is given as

$y = \frac{1}{2}gt^2$

$55 = \frac{1}{2}(6.91)t^2$

$t = 3.99 s$

now in the same time distance moved by the packet

$d = vt$

$d = 70.6 \times 3.99$

$d = 282 m$

Answer:

Between B to C

Explanation:

While moving downwards under gravity the attraction force of Earth will increase the speed

So its speed will increase during its return journey from B to C

Answer:

y = 2040 m

Explanation:

given that

$v_x = 50 m/s$

$\Delta x = 1000 m$

now the time taken by the ball to reach the given distance is

$t = \frac{d}{v}$

$t = \frac{1000}{50} = 20 s$

now in the same time the vertical distance moved by it

$y = v_y t - \frac{1}{2}gt^2$

$y = 200(20) - \frac{1}{2}(9.8)(20^2)$

$y = 2040 m$

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3 years ago

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vlabodo [156]

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3 years ago

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0

3 years ago

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

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3 years ago

What is it called when there’s an atom that has fewer neutrons than protons and more electrons than protons??l

HACTEHA [7]

Answer: The answer is an atom.

Explanation: This is because an atom has fewer neutrons than protons and more electrons than protons

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3 years ago