Which statement is true about the image produced by a plane mirror?

lilavasa [31]

Answer:

it ends when clouds above start to break apart. Some tornadoes only last seconds. Others can last much longer. They come in many shapes and sizes.

8 0

3 years ago

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A 0.01 kg spring toy is compressed 0.02 m and released vertically. The toy is measured to reach 0.25 m in the air. Determine the

Scorpion4ik [409]

Answer:

122.5 N/m

Explanation:

According to the law of conservation of energy, if there is no air resistance or frictional forces, the initial elastic potential energy of the spring toy is entirely converted into gravitational potential energy when the toy reaches the highest point.

Therefore, we can write:

$\frac{1}{2}kx^2=mgh$

where the term on the left is the initial elastic potential energy while the term on the right is the gravitational potential energy, and where

k is the spring constant

x = 0.02 m is the compression of the spring

m = 0.01 kg is the mass of the toy

h = 0.25 m is the height reached by the toy

$g=9.8 m/s^2$ is the acceleration due to gravity

Solving for k,

$k=\frac{2mgh}{x^2}=\frac{2(0.01)(9.8)(0.25)}{(0.02)^2}=122.5 N/m$

8 0

2 years ago

Weak magnetic fields can be measured at the surface of the brain. Although the currents causing these fields are quite complicat

STALIN [3.7K]

To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.

By definition we know that the magnetic field in a sphere can be described as

$B = \frac{\mu_0}{2}\frac{Ia^2}{(z^2+a^2)^{3/2}}$

Where,

a = Radius

z = Distance to the magnetic field

I = Current

$\mu_0 =$ Permeability constant in free space

Our values are given as

$D=2a = 16cm \rightarrow$ diameter of the sphere then,

$a = 0.08m$

Thus z = a

$B = \frac{\mu_0}{2}\frac{Ia^2}{(a^2+a^2)^{3/2}}$

$B = \frac{\mu_0I}{2(2^{3/2})a}$

$B = \frac{\mu_0 I}{2^{5/2}a}$

Re-arrange to find I,

$I = \frac{2^{5/2}Ba}{\mu_0}$

$I = \frac{2^{5/2}(3*10^{-12})(8*10^{-2})}{4\pi*10^{-7}}$

$I = 1.08*10^{-6}A$

Therefore the current at the pole of this sphere is $1.08*10^{-6}A$

5 0

2 years ago

To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit

DiKsa [7]

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

$R = 350ft$

$a_t = 1.1ft/s^2$

PART A )

$a_c = \frac{V^2}{R}$

$a_c = \frac{V^2}{350}$

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $5.25ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$27.5625 = 1.21 + \frac{v^4}{122500}$

$v=42.3877ft/s$

Now calculate the angular velocity of the motorcycle

$v = r\omega$

$42.3877 = 350\omega$

$\omega = 0.1211rad/s$

Calculate the angular acceleration of the motorcycle

$a_t = r\alpha$

$1.1 = 350\alpha$

$\alpha = 3.1428*10^{-3}rad/s^2$

Calculate the time needed by the motorcycle to reach an acceleration of

$5.25ft/s^2$

$\omega = \alpha t$

$0.1211 = 3.1428*10^{-3}t$

$t = 38.53s$

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

$v=48.2796ft/s$

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

$a_r = \frac{v^2}{R}$

$a_r = \frac{21.5^2}{350}$

$a_r =1.3207ft/s^2$

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

$a = \sqrt{a_t^2+a_r^2}$

$a = \sqrt{(1.1)^2+(1.3207)^2}$

$a = 1.7187ft/s^2$

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

$v=48.2796ft/s$

3 0

2 years ago

Why do surface winds tend to cross the isobars?

scZoUnD [109]

Geostrophic winds blows parallel to the isobars. That is because the Coriolis force and pressure gradient force ( PGF ) are in balance. But near the surface the friction can act to change the direction of the wind and to slow it down. Coriolis force decreases at the surface and PGF stays the same. The difference in terrain conditions affects how much friction is exerted. Hills and forests force the wind to change direction more than flat areas. Answer: Friction reduces the speed so Coriolis is weakened.

3 0

3 years ago

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