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2 years ago

A water skier is pulled behind a boat by a rope. The rope has a tension of 450

Physics

2 answers:

nataly862011 [7]2 years ago

7 0

Answer:

161 N like the other person said

Explanation:

WARRIOR [948]2 years ago

4 0

Answer:

161 N

Explanation:

Fy = F sin θ

Fy = 450 sin 21°

Fy ≈ 161

The y component of the tension is approximately 161 N.

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A special triangular frame is being made for a piece of artwork. The the base of the triangular frame must be 90 cm. If the area

Mashutka [201]

Answer:

Explanation:

base of triangular frame, b = 90 cm

Area, A = 765 cm²

Let the height is h.

Area of a triangular frame = 1/2 x base x height

765 = 0.5 x 90 x h

h = 17 cm

Thus, the height of triangular frame is 17 cm.

5 0

3 years ago

Dmitry [639]

Answer: 176.4 J

Explanation:

5 0

2 years ago

Calculate the wavelength of an orange light wave with a frequency of 5.085 x 10^14 Hz. The speed of light is 3.0 x 10^8 m/s.

Margarita [4]

Answer:

5.9 x 10⁻⁷m

Explanation:

Given parameters:

Frequency = 5.085 x 10¹⁴Hz

Speed of light = 3.0 x 10⁸m/s

Unknown:

Wavelength of the orange light = ?

Solution:

The wavelength can be derived using the expression below;

wavelength = $\frac{v}{f}$

v is the speed of light

f is the frequency

wavelength = $\frac{3 x 10^{8} }{5.085 x 10^{14} }$ = 5.9 x 10⁻⁷m

3 0

3 years ago

Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the

Assoli18 [71]

Answer:

Part a)

$\rho = 3\epsilon_0 k r^2$

Part b)

$Q = 4\pi \epsilon_0kR^5$

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

$E = kr^3 \hat r$

now electric flux through a spherical surface of radius r is given as

$\phi = E. A$

$\phi = kr^3(4\pi r^2)$

now by Guass law we know that

$E.A = \frac{q}{\epsilon_0}$

$q = 4\pi \epsilon_0kr^5$

now volume charge density is given as

$\rho = \frac{q}{\frac{4}{3}\pi r^3}$

$\rho = 3\epsilon_0 k r^2$

Part b)

Total charge inside the radius R is given as

$Q = 4\pi \epsilon_0kR^5$

7 0

2 years ago

There are two identical, positively charged conducting spheres fixed in space. The spheres are 40.4 cm apart (center to center)

Nadya [2.5K]

Answer:

$q_1=5.64\times 10^{-7}\ \text{C}$ and $q_2=2.32\times 10^{-6}\ \text{C}$

Explanation:

$F_1=0.072\ \text{N}$

$F_2=0.115\ \text{N}$

r = Distance between shells = 40.4 cm

$q_1$ and $q_2$ are the charges

$k$ = Coulomb constant = $8.99\times10^{9}\ \text{Nm}^2/\text{C}^2$

Force is given by

$F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q_1q_2=\dfrac{F_1r^2}{k}\\\Rightarrow q_1q_2=\dfrac{0.072\times 0.404^2}{8.99\times 10^{9}}\\\Rightarrow q_1q_2=1.307\times 10^{-12}\\\Rightarrow q_1=\dfrac{1.307\times 10^{-12}}{q_2}$

$F_2=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{F_2r^2}{k}}\\\Rightarrow q=\sqrt{\dfrac{0.115\times 0.404^2}{8.99\times 10^{9}}}\\\Rightarrow q=1.44\times 10^{-6}\ \text{C}$

$q=\dfrac{q_1+q_2}{2}\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2\times1.44\times 10^{-6}\\\Rightarrow q_1+q_2=2.88\times 10^{-6}$

Substituting the above value of $q_1$ we get

$\dfrac{1.307\times 10^{-12}}{q_2}+q_2=2.88\times 10^{-6}\\\Rightarrow q_2^2-2.88\times 10^{-6}q_2+1.307\times 10^{-12}=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4\times \:1\times \:1.307\times 10^{-12}}}{2\times \:1}\\\Rightarrow q_2=2.32\times 10^{-6}, 5.64\times 10^{-7}$

$q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{2.32\times 10^{-6}}\\\Rightarrow q_1=5.63\times 10^{-7}$

$q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{5.64\times 10^{-7}}\\\Rightarrow q_1=2.32\times 10^{-6}$

Since we know $q_1$

$q_1=5.64\times 10^{-7}\ \text{C}$ and $q_2=2.32\times 10^{-6}\ \text{C}$ .

5 0

2 years ago