Ask question

Ask question

All categories

3 years ago

10

a telephone pole casts a shadow that is 29 m long. find the height of the telephone pole if a statue that is 38cm tall casts a s

hadow 78cm long

1 answer:

mafiozo [28]3 years ago

6 0

You do cross multiplication so multiply 29 by 38 and then divide it by 78 which gives you answer of 14.128 cm

You might be interested in

if the kinetic energy of an aeroplane moving at a velocity of 100m/s is 200j ,what is the mass of the aeroplane

Black_prince [1.1K]

Answer:

<u>0.04kg</u>

Step-by-step explanation:

Kinetic energy Formula: eK=0.5×m×v²

to work out mass

we can rearrange the formula and make it:

~m=eK÷(0.5×v²)

now insert the missing digits into the formula:

m=200÷(0.5×100²)

work it out, and you get your answer

m=0.04kg

Hopefully this helped-have a good day)

8 0

2 years ago

Please help giving brainlist, the answers i put arent the correct ones i just misclicked.

QveST [7]

1. B 2. A 3. B that’s what you looking for

6 0

3 years ago

(y+3) solve by using the distributive property

MA_775_DIABLO [31]

Answer:

y-3

Step-by-step explanation:

4 0

3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago

In 2000 the small town

Elodia [21]

Answer:

what the question ?

4 0

4 years ago