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3 years ago

A golf ball is given an initial speed of 20. meters

2 answers:

5 0

In calculus, neglecting air resistance, the formula for range is R = v(0)^2 sin 2A/g, where v(0) is the initial speed, A is the angle, and g is the acceleration of gravity. Here v(0) = 20 m/s, and g is a constant.

R = 400 sin 2A 

For maximum range, sin 2A = 1, so 2A= 90°, and A= 45

45 degrees is my answer.

andre [41]3 years ago

4 0

Neglecting air resistance, the greatest horizontal distance
always results from a launch angle of 45° above the horizon,
no matter what the initial speed is. This is a handy thing to know.
You'll use it many times, and you should carry it in your toolbox
(i.e. memorize it).

You might be interested in

Two objects of equal mass are a fixed distance apart. If the mass of each object could be tripled, the gravitational force betwe

antiseptic1488 [7]

gravitational force between two objects is given as

F = G m₁ m₂/r²

where m₁ = mass of first object , m₂ = mass of second object , r = distance between the two objects .

Initial case :

m₁ = m₂ = m

gravitational force between the objects is given as

F = G m²/r²

Final Case :

m₁ = m₂ = 3 m

new gravitational force between the objects is given as

F' = G (3m)²/r²

F' = 9 G m²/r²

F' = 9 F

hence the gravitational force between the two objects becomes 9 times.

5 0

3 years ago

a blackbody is radiating with a characteristic wavelength of 9 microns what is the blackbody temperature answer in kelvin

Daniel [21]

This question involves the concepts of Wein's displacement law and characteristic wavelength.

The blackbody temperature will be "3.22 x 10⁵ k".

<h3>WEIN'S DISPLACEMENT LAW</h3>

According to Wein's displacement law,

$\lambda_{max} T = c\\\\T=\frac{c}{\lambda_{max}}$

where,

$\lambda_{max}$ = characteristic wavelength = 9 μm = 9 x 10⁻⁹ m
T = temperature = ?
c = Wein's displacment constant = 2.897 x 10⁻³ m.k

Therefore,

$T=\frac{2.897\ x\ 10^{-3}\ m.k}{9\ x\ 10^{-9}\ m}$

T = 3.22 x 10⁵ k

Learn more about characteristic wavelength here:

brainly.com/question/14650107

7 0

3 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago

Which wave characteristic determines color?

Veseljchak [2.6K]

I would said A is the best option if i’m wrong sorry

3 0

3 years ago

If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o

tangare [24]

The correct answer is letter A. 6 millimeters. If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front of the lens, the height of the image is 6 millimeters.

Solution:
18 / x = 12 / 4
12x = 72
x = 6mm

7 0

3 years ago