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Vlad1618 [11]
3 years ago
11

Find the surface area of the prism 14yd 6yd 10yd

Physics
1 answer:
castortr0y [4]3 years ago
5 0

the answer is 568 yd^2

.......

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A fish swimming in a horizontal plane has velocity i = (4.00 + 1.00 ) m/s at a point in the ocean where the position relative to
ryzh [129]

Answer:

a. The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. -9.19° = 350.81° from the the positive x-axis

Explanation:

The initial velocity v₁ of the fish is v₁ = 4.00i + 1.00j m/s. Its final velocity after accelerating for t = 19.0 s is v₂ =  17.0i - 1.00j m/s

a. The acceleration a = (v₂ - v₁)/t = [17.0i - 1.00j - (4.00i + 1.00j)]/19 = [(17.0 -4.0)i - (-1.0 -1.0)j]/19 = (13.0i - 2.0j)/19 = 0.68i - 0.11j m/s²

The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. The direction of the acceleration relative to the unit vector i,

tanθ = a₂/a₁ = -0.11/0.68 = -0.1618

θ = tan⁻¹(-0.1618) = -9.19° ⇒ 360 + (-9.19) = 350.81° from the the positive x-axis

6 0
3 years ago
What is the rotational kinetic energy of the Earth about the Sun? Assume the earth is a uniform sphere, mass of the Earth is 5.9
Thepotemich [5.8K]

Answer:

2.66x10^33 J

Explanation:

In order to do this, we first need to know the expression for kinetic energy:

E = 1/2 I*w²  (1)

I is moment of innertia

w is angular speed.

Moment of Innertia can be calculated using the following expression:

I = 2/5 M*R²  (2)

M is mass of earth, R is radius of earth

Replacing the data in expression (2) we have:

I = 2/5 * 5.97x10^24 * (6.37x10^6)²

I = 9.69x10^37 kg m²

Next, we need to calculate the angular speed of Earth over it's axis, this is easy, because we know the Earth rotates over it's own axis once a day, 24 hours (86400 s), and assuming Earth is a perfect sphere, we can calculate the speed:

w = 2π / 86400 = 7.27x10^-5 rad/s

next thing we need to do is calculate the rotational kinetic energy of earth on it's axis, using equation (1) so:

E = 1/2 * 9.69x10^37 * (7.27x10^-5)²

E = 2.56x10^29 J

Now that we have this value, we can finally calculate the rotational kinetic energy of earth about the sun. For that, we need to calculate again the angular speed of earth about the sun. The Earth rotates around the sun once a year, or 365 days, which is 3.1536x10^7 s, so the angular speed would be:

w = 2π/3.1536x10^7 = 1.99x10^-7 rad/s

finally the energy is the combination of the sun and earth so:

K = 1/2 (Ie + Me*Rorb²)wo²

I is innertia for earth

Me mass of earth

Rorb RAdius of orbit around the sun

wo is angular speed around the sun

Replacing the data we finally have:

K = 1/2 [9.69x10^37 + 5.97x10^24 * (1.5x10^11)²]*(1.99x10^-7)²

K = 2.66x10^33 J

4 0
3 years ago
A.) Determine the work done by Zach on the bull.
tresset_1 [31]

Explanation:

The unit of work done is in Joules

Work done is a physical quantity that is defined as the force applied to move a body through a particular distance.

Work is only done when the force applied moves a body through a distance.

    Work done  = Force x distance

The maximum work is done when the force is parallel to the distance direction.

The minimum work is done when the force is at an angle of 90° to the distance direction.

 So to solve this problem;

 multiply the force applied by Zack and distance through which the bull was pulled.

8 0
3 years ago
The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t
slamgirl [31]

Answer:

The charge is 2.75\times10^{-13}\ C

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length L=5.0\times5.0\times10^{-4}\ m

We need to calculate the linear charge density

Using formula of linear charge density

E=\dfrac{2k\lambda}{r}

\lambda=\dfrac{Er}{2k}

Put the value into the formula

\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}

\lambda=1.1\times10^{-10}\ C/m

We need to calculate the charge

Using formula of charge

Q=\lambda\timesL

Put the value into the formula

Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})

Q=2.75\times10^{-13}\ C

Hence, The charge is 2.75\times10^{-13}\ C

5 0
3 years ago
Which have the longest wavelengths--light waves, x-rays, or radio waves?
Tresset [83]

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