How does the tide cycle affect erosion along a sea coast?

The volume of a balloon an be approximated by V = 4 3 π r 3 V=43πr3. If air is leaking from the balloon at a rate of 56 cubic ce

11Alexandr11 [23.1K]

Answer:

The rate the radius of the balloon shrinking at the moment the radius is 5 centimeters is 0.1783 cm/s

Explanation:

Here we have

dV/dt = 56 cm³/s

$\frac{dV}{dt} = \frac{d}{dt}(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi\cdot3r^2 \frac{dr}{dt}$

When the radius is 5 cm we have

$56 \hspace {0.09cm}cm^3/s= \frac{4}{3}\pi\cdot3\cdot 5^2 \cdot \frac{dr}{dt} = 314.16 \times \frac{dr}{dt}$

Therefore,

$56 \hspace {0.09cm}cm^3/s= 314.16 \hspace {0.09cm}cm^2\times \frac{dr}{dt}$

From which,

$\frac{dr}{dt} = 56 \hspace {0.09cm}cm^3/s \div314.16 \hspace {0.09cm}cm^2$

$\frac{dr}{dt} = 0.1783 \hspace {0.09cm}cm/s$

The rate the radius of the balloon shrinking at the moment the radius is 5 centimeters = 0.1783 cm/s.

6 0

3 years ago

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Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and plac

mylen [45]

To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.

Mathematically the conservation of these two energies can be given through

$W = U_f - U_i$

Where,

W = Work

$U_f =$ Final gravitational Potential energy

$U_i =$ Initial gravitational Potential energy

When the spacecraft of mass m is on the surface of the earth then the energy possessed by it

$U_i = \frac{-GMm}{R}$

Where

M = mass of earth

m = Mass of spacecraft

R = Radius of earth

Let the spacecraft is now in an orbit whose attitude is $R_{orbit} \approx R$ then the energy possessed by the spacecraft is

$U_f = \frac{-GMm}{2R}$

Work needed to put it in orbit is the difference between the above two

$W = U_f - U_i$

$W = -GMm (\frac{1}{2R}-\frac{1}{R})$

Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is

$W = \frac{GMm}{2R}$

3 0

3 years ago

What is the length of the x-component of the vector shown below?

ollegr [7]

Answer:

Option B. 8.1

Explanation:

From the question given above, the following data were obtained:

Angle θ = 71°

Hypothenus = 25

Adjacent = x

Thus, we can obtain the x component of the vector by using the cosine ratio as illustrated below:

Cos θ = Adjacent /Hypothenus

Cos 71 = x/25

Cross multiply

x = 25 × Cos 71

x = 25 × 0.3256

x = 8.1

Therefore, the x component of the vector is 8.1

4 0

3 years ago

The mass of car is 1200kg. If the car is moving with tue velocity of 30m/s. How much force should be applied to stop the car in

faust18 [17]

Answer:

momentum of car=1200kg *10m/s =12000kgm/s

momentum =force * time

12000 kgm/s =force * 20s

force = 600 N

7 0

3 years ago

Albert's laboratory is filled with a constant uniform magnetic field pointing straight up. Albert throws some charges into this

guajiro [1.7K]

Answer:

$\vec{F}=qB(v_y \hat{i} - v_x\hat{j})$

Explanation:

The force excerted by a magnetic field on a charged particle is given by the Lorentz force:

$\vec{F} = q \vec{v} \times \vec{B}$

Lets consider the z-direction of our coordinate system the same direction of the magnetic field, that is:

$\vec{B} = B \hat{k}$

Let us consider that the velocity of a given particle is:

$\vec{v} = v_x\hat{i} + v_y \hat{j} + v_z \hat{k}$

Therefore, since k×k = 0

$\vec{v} \times \vec{B} = (v_x\hat{i} + v_y \hat{j} + v_z \hat{k}) \times B\hat{k}\\\vec{v} \times \vec{B} = (Bv_x \hat{i} \times\hat{k}) + (Bv_y \hat{j}\times\hat{k})$

And since i, j , k are a rigth hand system:

i × j = k

j × k = i

k × i = j --> i × k= -j

$\vec{v} \times \vec{B} = Bv_x (-\hat{j}) + Bv_y \hat{i} = Bv_y \hat{i} - Bv_x\hat{j}$

Threfore, if the particle has charge q and velocity v = (vx,vy,vz), the magnetic force it will feel will be

$\vec{F}=qB(v_y \hat{i} - v_x\hat{j})$

4 0

4 years ago