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MariettaO [177]

3 years ago

13

swelling magma chambers below the surface can cause the crust above to uplift and form a large flat regions higher than the surr

ounding land known as

1 answer:

klemol [59]3 years ago

3 0

Answer: I belive the answer is 666

Explanation:

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Papessa [141]

) 5 -5 1 2 3 4 5 Other than at t = 0, when is the velocity of the object equal to zero? 1. 5.0 s 2. 4.0 s 3. 3.5 s 4. At no other time on this graph. correct 5. During the interval from 1.0 s to 3.0 s. Explanation: Since vt = Z t 0 a dt, vt is the area between the acceleration curve and the t axis during the time period from 0 to t. If the area is above the horizontal axis, it is positive; otherwise, it is negative. In order for the velocity to be zero at any given time t, there would have to be equal amounts of positive and negative area between 0 and t. According to the graph, this condition is never satisfied. 005 (part 1 of 1) 0 points Identify all of those graphs that represent motion at constant speed (note the axes carefully). a) t x b) t v c) t a d) t v e) t a

6 0

3 years ago

A mass m is attached to an ideal massless spring. When this system is set in motion with amplitude a, it has a period t. What is

Luden [163]

The period will be the same if the amplitude of the motion is increased to 2a

What is an Amplitude?

Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.

Here,

mass m is attached to the spring.

mass attached = m

time period = t

We know that,

The time period for the spring is calculated with the equation:

$T = 2\pi \sqrt{\frac{m}{k} }$

Where k is the spring constant

Now if the amplitude is doubled, it means that the distance from the equilibrium position to the displacement is doubled.

From the equation, we can say,

Time period of the spring is independent of the amplitude.

Hence,

Increasing the amplitude does not affect the period of the mass and spring system.

Learn more about time period here:

<u>brainly.com/question/13834772</u>

#SPJ4

7 0

2 years ago

A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t

kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

$\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W$

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

$emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV$

hope this helps!!

5 0

3 years ago

PLEASE PLEASE PLEASE A mad scientist places massive amounts of charge on basketball sized aluminum balls. The charge on the ball

bazaltina [42]

Answer:

4.2 x 10⁷N

Explanation:

Given parameters:

Charge on ball:

q₁ = 3C

q₂ = 14C

Distance between balls = 9000m

Unknown:

Force acting on the two balls

Solution:

The force experienced by the two charges is given by coulombs law. It is mathematically expressed as;

F = $\frac{k q_{1} q_{2} }{r^{2} }$

where k = 9 x 10⁹Nm²/C²

q is the charges

r is the distance

Input the variables and solve;

F = $\frac{9 x 10^{9} x 3 x 14 }{9000}$ = 4.2 x 10⁷N

8 0

3 years ago

What is the pendulum length whose period is 2.0s ?

Mashutka [201]

$Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m$ $T=2 \pi \sqrt{\frac{L}{g}} \\
 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi } = \frac{L}{g} \\
 \frac{T^2}{2 \pi }*g=L\\
L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m
$

7 0

3 years ago