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Rina8888 [55]
3 years ago
5

A squirrel jumps into the air with a velocity of 4 m/s at an angel of 50 degrees. What is the maximum height reached by the squi

rrel?
Physics
1 answer:
Debora [2.8K]3 years ago
7 0

Answer:

Explanation:

Assuming the squirrel is jumping off the ground, here's what we know but don't really know...

v₀ = 4.0 at 50.0°

So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:

v_{0y}=4.0sin(50.0) which gives us that the upward velocity is

v₀ = 3.1 m/s

Moving on here's what we also know:

a = -9.8 m/s/s and

v = 0

Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:

v = v₀ + at and filling in:

0 = 3.1 - 9.8t and

-3.1 = -9.8t so

t = .32 seconds.

Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:

Δx = v_0t+\frac{1}{2}at^2 and filling in:

Δx = 3.1(.32)+\frac{1}{2}(-9.8)(.32)^2 and using the rules for adding and subtracting sig fig's correctly, we can begin to simplify this:

Δx = .99 - .50 so

Δx = .49 meters

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Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving
Iteru [2.4K]

Explanation:

Let m_p is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

\dfrac{mv^2}{r}=qvB

r is the radius of path,

r=\dfrac{mv}{qB}

Time period is given by :

T=\dfrac{2\pi r}{v}

T=\dfrac{2\pi m_p}{qB}

Frequency of proton is given by :

f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}

The wavelength of radiation is given by :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{2\pi m_pc}{qB}

So, the wavelength of radiation produced by a proton is \dfrac{2\pi m_pc}{qB}. Hence, this is the required solution.

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3 years ago
Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

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Answer:

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Explanation:

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Your friend tells you “I no the moon does not rotate because we always see the same side.” do you agree or disagree with your fr
nydimaria [60]

Answer:

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3 years ago
An airplane maintains a speed of 585 km/h relative to the air it is flying through as it makes a trip to a city 815 km away to t
navik [9.2K]

Answer:

a)   t = 1.47 h    b) t = 1.32 h

Explanation:

a)  In this problem the plane and the wind are in the same North-South direction, whereby the vector sum is reduced to the scalar sum (ordinary). Let's calculate the total speed

     v = v_{f}f - v_{w}

     v = 585 -32.1

     v = 552.9 km / h

We use the speed ratio in uniform motion

     v = x / t

     t = x / v

     t = 815 /552.9

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b)  We repeat the calculation, but this time the wind is going in the direction of the plane

      v=  v_{f}f - v_{w}

      v 585 + 32.1

      v = 617.1 km / h

      t = 815 /617.1

      t = 1.32 h

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3 years ago
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