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tresset_1 [31]
3 years ago
6

I need this ASAP please!!!

Physics
2 answers:
WINSTONCH [101]3 years ago
4 0
It’s E we just had a test in this and I got it right
Mademuasel [1]3 years ago
3 0

Answer:

C. the help our understanding of the formation of planetary bodies.

Explanation:

this is the answer of PLATO

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The water-ice particles forming Saturn's rings are frozen together into a thin sheet that rotates around Saturn like a solid bod
vaieri [72.5K]

Answer:

B. False

Explanation:

According to research by several scientists, Saturn's rings aren't solid, as they appear from Earth.  They are actually made up of floating chunks of water ice, rocks and dust that range in diferent sizes from specks to enormous, even house-sized pieces that orbit Saturn in a ring pattern.

4 0
3 years ago
Name two factors that can affect the function of an enzyme
myrzilka [38]

Answer:

1. pH

2. Temperature

Hope this helps

3 0
3 years ago
If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.
AleksAgata [21]

Answer:

a) f_1=5.587Hz

b) f_{n+1}-f_n=5.587Hz

Explanation:

The frequency of the n^{th} harmonic of a vibrating string of length <em>L, </em>linear density \mu under a tension <em>T</em> is given by the formula:

f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}

a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:

f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz

3 0
3 years ago
An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car
saveliy_v [14]

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

F = ma

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

-10000 =2000\times a

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

a = -\dfrac{10000}{2000}

a=-5\ m/s^2

For the distance,

Using third equation of motion

v^2-u^2=2as

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

0-30^2=2\times(-5)\times s

s = 90\ m

Hence, The car traveled the distance before stopping is 90 m.

6 0
3 years ago
A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi
grin007 [14]

Explanation:

It is given that,

Magnitude of charge, q=15\ \mu C=15\times 10^{-6}\ C

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, B=0.08\ j

Velocity, v=(5\ cos25)i+(5\ sin25)j

v=[(4.53)i+(2.11)j]\ m/s

We need to find the magnitude of force on the charge. Magnetic force is given by :

F=q(v\times B)

F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]

<em>Since</em>, i\times j=k\ and\ j\times j=0

F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]

F=0.00000543\ kN

F=5.43\times 10^{-6}\ kN

So, the force acting on the charge is 5.43\times 10^{-6}\ kN and is moving in positive z axis. Hence, this is the required solution.

6 0
3 years ago
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