Waves do NOT carry.<br><br> A)Weight<br> B)Matter<br> C)Energy<br> D)Color<br> (Choose one)

A ring placed along y 2 + z 2 = 4, x = 0 carries a uniform charge of 5 μ C/m. ( a ) Find D at P(3,0,0) . ( b ) If two identical

GuDViN [60]

For right now idk :p

3 0

3 years ago

Why can't you feel rhe force of attraction between you and mars

NISA [10]

Because Mars is too far away for its gravitational pull to affect us, in addition Earths gravitational pull is greater than Mars anyways.

4 0

3 years ago

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G

Romashka-Z-Leto [24]

There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) = - 3.675 m

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time

So

acceleration(-g)= dv/dt

Solve it

dv = a dt

dv = -g dt

v - v₀ = -gt

v= dy/dt

dy = v dt

dy = ( v₀ - gt ) dt

y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) = - 3.675 m

The cat falls 3.675 m between time 1/2 s and 1 s.

6 0

3 years ago

5. A family leaves for an evening walk around the neighborhood. They leave at 5 pm and return at 6 pm. They traveled a total of

vlabodo [156]

This question involves the concepts of average speed, distance, and time.

The average speed of the family is "2.5 mi/hr".

<h3>AVERAGE SPEED</h3>

The average speed is defined as the ratio of the total distance covered with the total time taken to cover it.

$v=\frac{s}{t}$

where,

s = distance covered = 2.5 miles
t = total time taken = 1 hour
v = average speed = ?

Therefore,

$v=\frac{2.5\ mi}{1\ hr}$

v = 2.5 mi/hr

Learn more about average speed here:

brainly.com/question/12322912

4 0

2 years ago

A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s

Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

$T=m\frac{v^2}{r}$

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

$m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg$

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

$T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N$

5 0

3 years ago

Waves do NOT carry. A)Weight B)Matter C)Energy D)Color (Choose one)