All categories

3 years ago

A tug boat traveled -15 miles in 0.3 hours what was its velocity

Mathematics

1 answer:

GenaCL600 [577]3 years ago

4 0

Answer:

22.35 m/s

Step-by-step explanation:

The distance is always a positive unit, so distance traveled (d) = 15 miles

Time (t) = 0.3 hours

Time (t) = 0.3 * 60 = 18 minutes

(1 hour = 60 minutes)

Velocity (v) = ?

The formula of distance is: d = v*t

=> v = d/t = 15/18 = 0.83 miles/minute

we can also calculate velocity is meter/sec, for that

1 minute = 60 sec

=> 18 mins = 18*60 = 1080 sec

1 mile = 1609.34 meters

=> 15 miles = 15 * 1609.34 = 24,140 m

hence,

v = 24,140/1080 = 22.35 m/s

You might be interested in

The circumference of a circle is 127 cm. What is the area,

bogdanovich [222]

A≈ 1283.51
I hope this will help u

6 0

3 years ago

-34 -2p = 8(8 - 2p)

Lera25 [3.4K]

Answer:

p = 7

Step-by-step explanation:

Look at the picture hope this helps!

8 0

3 years ago

Please help on this asap

Lelechka [254]

I believe that would be the last choice, because it is not impossible that someone without contacts or glasses would be chosen. And it is a fairly similar chance that someone with glasses or with contacts will be chosen. Hope this helps! :)

7 0

4 years ago

Water flows at a rate of 8000 cubic inches per minute into a cylindrical tank. The tank has a diameter of 128 inches and a heigh

beks73 [17]

Answer:

The height of the water = 6.2 in. to the nearest tenth

Step-by-step explanation:

∵ The rate of flows of water into the tank = 8000 in.³/min.

∴ The volume of the water in the tank after 10 min. = 8000 × 10 = 80000 in.³

∵ The water take the shape of the tank

∴ The height of the water = volume of water ÷ Area of the base of the tank

∵ The tank is cylinder with diameter 128 in. and height 72 in.

∴ The area of the base of the tank = π(128/2)²

∴ The height of the water = $\frac{80000}{\pi(64)^{2}}=6.2169899$

∴ The height of the water = 6.2 in. to the nearest tenth

5 0

3 years ago

In ΔJKL, k = 9.6 cm, l = 2.7 cm and ∠J=43°. Find ∠L, to the nearest 10th of a degree.

Bogdan [553]

Answer:

L = 10.64°

Step-by-step explanation:

From the given information:

In triangle JKL;

line k = 9.6 cm

line l = 2.7 cm; &

angle J = 43°

we are to find angle L = ???

We can use the sine rule to determine angle L:

i.e

$\dfrac{j}{SIn \ J} = \dfrac{l}{ SIn \ L}$

Using Pythagoras rule to find j

i,e

j² = k² + l²

j² = 9.6²+ 2.7²

j² = 92.16 + 7.29

j² = 99.45

$j = \sqrt{99.45}$

j = 9.97

∴

$\dfrac{9.97}{Sin \ 43} = \dfrac{2.7}{ Sin \ L}$

${9.97 \times Sin (L ) = (2.7 \times Sin \ 43)$

$= Sin \ L = \dfrac{ (2.7 \times Sin \ 43)}{9.97 } \\ \\ = Sin \ L = \dfrac{ (2.7 \times 0.6819)}{9.97 } \\ \\ = Sin \ L = 0.18466 \\ \\ L = Sin^{-1} (0.18466) \\ \\ L = 10.64 ^0$

3 0

3 years ago