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A same-size iron ball and wooden ball are dropped simultaneously from a tower and reach the ground at the same time. The iron ba

Papessa [141]

Answer:

momentum of iron ball is greater than wooden ball

Explanation:

when metal ball (iron ball) and wooden drop are drop from same elevation and reaching the ground after same time. at this position the iron ball has greater momentum than wooden ball

we know that momentum is defined as

P=Mv

and we know also that mass of iron ball is greater than mass of wooden ball and they reached on ground at same time and same distance it mean also velocity will be same for both ball. therefore from above relation we have

Miron*V > Mwood*V i.e.

momentum of iron ball is greater than wooden ball

5 0

4 years ago

Line segment Q R , Line segment R S and Line segment S Q are midsegments of Î"WXY. Triangle R Q S is inside triangle X Y W. Poin

Whitepunk [10]

The perimeter of ΔWXY is : ( D ) 14.5 cm

Calculating the perimeter of ΔWXY

QR = WY / 2

RS = XW / 2

QS = XY / 2

Given that : QR = 2.93 cm , RS = 2.04 cm, QS = 2.28 cm

Therefore

Perimeter of ΔWXY = ∑ WY + XW + XY

= 2SR + 2QS + 2QR

= 2(2.04) + 2(2.28) + 2(2.93)

= 14.5 cm

Hence we can conclude that the perimeter of ΔWXY = 14.5 cm

learn more about perimeter calculations : brainly.com/question/24744445

8 0

3 years ago

Read 2 more answers

A radar used to detect the presence of aircraft receives a pulse that has reflected off an object 5 ✕ 10−5 s after it was transm

Sunny_sXe [5.5K]

Answer:

7500 m

Explanation:

The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.

We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so

d = 2L

In a time of

$t=5\cdot 10^{-5}s$

Electromagnetic waves travel in a vacuum at the speed of light, which is equal to

$c=3.0\cdot 10^8 m/s$

Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:

$d=vt$ (1)

where

$v=c=3.0\cdot 10^8 m/s$

$t=5\cdot 10^{-5}s$

$d=2L$ , where L is the distance between the radar and the object

Re-arranging eq(1) and substituting, we find L:

$L=\frac{vt}{2}=\frac{(3.0\cdot 10^8)(5\cdot 10^{-5})}{2}=7500 m$

7 0

4 years ago

At a distance of 41 ft , an ionizing radiation source delivers 5.0 rem of radiation. How close could you get to the source and s

mash [69]

Answer:

at distance 18.33 ft no biological effects

Explanation:

given data

distance = 41 ft

radiation = 5 rem

to find out

how close get so no biological effects

solution

we know here that intensity of radiation R is inversely proportional to (radiation)²

so here equation will be

$\frac{I1}{I2} = (\frac{R2}{R1})^{2}$ .............1

here I is intensity

so here I1 = 25

because for 0 to 25 there is no detectable effects and for 25 to 100 it will . temporary decrease so

we take 25 because dose is less than 25 so we take that highest value

so

$\frac{I1}{I2} = (\frac{R2}{R1})^{2}$

$\frac{25}{5} = (\frac{41}{R1})^{2}$

R1 = 18.33 ft

so at distance 18.33 ft no biological effects

6 0

3 years ago

There are 2 questions into 1

kiruha [24]

As we know that magnitude of the vector is 670 m

and it is inclined 25 degree with x axis

now we know that component of vector along x direction is given as

$x = L cos\theta$

$x = 670 cos25$

$x = 610 m$

Part 2)

In order to find the other component of the vector we can use

$y = L sin\theta$

$y = 670 sin25$

$y = 280 m$

3 0

4 years ago