Information is in the photo

Consider two identical wave pulses on a rope having a fixed end. Suppose the first pulse reaches the end of the rope, is reflect

jek_recluse [69]

Answer:

zero

Explanation:

Consider two identical wave pulses on a rope having a fixed end. Suppose the first pulse reaches the end of the rope, is reflected back, and then meets the second pulse, both waves will be out of phase by π radians. Therefore, they form destructive interference and hence the amplitude of the resultant pulse would be zero.

4 0

4 years ago

I have a 3 cc piece of aluminum with a density of steel with a mass of 24.0 g. I cut it into 2 equal pieces. How has the density

tresset_1 [31]

Answer:

What happens to the density of an object if the object is cut in half? ... The density remains the same because cutting the object in half will divide the mass & volume by the same amount. Also, the density of a substance remains the same no matter what size it is.

Explanation:

What happens to the density of an object if the object is cut in half? ... The density remains the same because cutting the object in half will divide the mass & volume by the same amount. Also, the density of a substance remains the same no matter what size it is.

3 0

3 years ago

if a body of mass 2kg travels with 2 m/s and other body of mass 2kg travels with 1m/s then find v1 and v2 in given statement

KIM [24]

Answer:

Your answer is here,

Explanation:

v1 = 4 kg m/s

v2 = 2 kg m/s

8 0

3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f

Verdich [7]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

$u = 0m/s$

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

$s = ut+\frac{1}{2} at^2$

Replacing,

$1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2$

$u =6.4013m/s$

The height of the acorn above the meter stick can be calculated as,

$v^2 = u^2 +2gh$

$h = \frac{v^2-u^2}{2g}$

$h = \frac{6.4013^2-0^2}{2(9.8)}$

$h = 2.0906m$

Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

$h = 2.0906+1.87$

$h = 3.9606m$

4 0

4 years ago

You drive in a straight line at 22 m/s for 12 miles, then at 30 m/s for another 12 miles. (a) How does your average speed sav co

Setler [38]

Answer:

Explanation:One method of calculating average speed is to divide the total distance by the total time spent

d1=d2= 12 miles

d1=d2= 19312.1 m

$t1=\frac{d1}{u1} \\t1=19312.1/22\\t1=877.8 secs$

$t2=\frac{d2}{u2} \\t2=19312.1/30\\t2=673.7 secs$

$u_a=\frac{d1+d2}{t1+t2} \\u_a=25.385 m/s$

There is slight difference between the average speeds due to the reason that average speed is not just the midpoint of the speed but is the total distance traveled by total time.

6 0

3 years ago