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3 years ago

What unit do we use to measure angles??

Mathematics

2 answers:

Katarina [22]3 years ago

7 0

Degrees (not like celsius or temperature though)

ArbitrLikvidat [17]3 years ago

3 0

You can measure angles in degrees

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I really need help

Assoli18 [71]

Answer: the first 1

Step-by-step explanation:

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2 years ago

PLEASE PLEASE HELP ME ASAPPP

Ivenika [448]

Answer:

a. 14:45pm (It's really easy to convert to a 24-hour clock time, all you need to do is add 12 hours onto the hour. Eg. 2 + 12 = 14 so it will be 14:45.pm

b. The length of time between 11:45am and 2:45pm is 3 hours.

I don't know what c is sorry.

Hope that helps. x

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2 years ago

Boat travel 27 miles in two hours how many miles travel in 1/2 hours

Sloan [31]

Answer:

6.75

Step-by-step explanation:

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2 years ago

What is the length of side AB as shown on the coordinate plane?

Alenkinab [10]

6 units (or whatever you’re measuring in)

Each little square represents 1 unit.

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3 years ago

33. Suppose you were on a planet where the

tatyana61 [14]

Answer:

a) 3.675 m

b) 3.67m

Explanation:

We are given acceleration due to gravity on earth = $9.8ms^-2$

And on planet given = $2.0ms^-2$

A) Since the maximum jump height is given by the formula

$\mathrm{H}=\frac{\left(\mathrm{v} 0^{2} \times \sin 2 \emptyset\right)}{2 \mathrm{g}}$

Where H = max jump height,

v0 = velocity of jump,

Ø = angle of jump and

g = acceleration due to gravity

Considering velocity and angle in both cases

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 2}{\mathrm{g} 1}$

Where H1 = jump height on given planet,

H2 = jump height on earth = 0.75m (given)

g1 = 2.0 $ms^-2$ and

g2 = 9.8 $ms^-2$

Substituting these values we get H1 = 3.675m which is the required answer

B) Formula to find height of ball thrown is given by

$\mathrm{h}=(\mathrm{v} 0 * \mathrm{t})+\frac{\mathrm{a} *\left(t^{2}\right)}{2}$

which is due to projectile motion of ball

Now h = max height,

v0 = initial velocity = 0,

t = time of motion,

a = acceleration = g = acceleration due to gravity

Considering t = same on both places we can write

$\frac{\mathrm{H} 1}{\mathrm{H} 2}=\frac{\mathrm{g} 1}{\mathrm{g} 2}$

where h1 and h2 are max heights ball reaches on planet and earth respectively and g1 and g2 are respective accelerations

substituting h2 = 18m, g1 = 2.0 $ms^-2$ and g2 = 9.8 $ms^-2$

We get h1 = 3.67m which is the required height

6 0

3 years ago