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3 years ago

If a cannonball were launched with a speed of 37m/s at a 60° angle from the height of Om, what would be the cannonball's range?

Physics

1 answer:

JulijaS [17]3 years ago

8 0

Answer:

Range = 120.98 m, (or rounded to the nearest whole number =121 meters)

Explanation:

We can use the formula for range in projectile motion at ground level:

$Range = \frac{v^2\,\,sin(2\theta)}{g} \\Range= \frac{37^2\,\,sin(120)}{9.8} \\Range\approx 120.98\,\,m$

Then the cannon's range would be approximately 120.98 m, which may be rounded to the nearest whole number as 121 meters

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The force of Earths gravity keeps earth in orbit true or false

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Answer:

Explanation:

The force of gravity keeps all of the planets in orbit around the sun

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3 years ago

3 a There is a thin layer of water between the blade and the ice. Suggest how this affects friction .

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Answer:

The water acts like a lubricant therefore has a smooth motion over the ice.

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3 years ago

What is the difference between kinetic and gravitacional energy?

kondaur [170]

Answer:

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3 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

3 years ago

Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.360 mmmm wide. The diffraction pattern is obse

777dan777 [17]

Answer:

Δy= 5,075 10⁻⁶ m

Explanation:

The expression that describes the interference phenomenon is

d sin θ = (m + ½) λ

As the observation is on a distant screen

tan θ = y / x

tan θ= sin θ/cos θ

As in ethanes I will experience the separation of the vines is small and the distance to the big screen

tan θ = sin θ

Let's replace

d y / x = (m + ½) λ

The width of a bright stripe at the difference in distance

y₁ = (m + ½) λ x / d

m = 1

y₁ = 3/2 λ x / d

Let's use m = 1, we look for the following interference,

m = 2

y₂ = (2+ ½) λ x / d

The distance to the screen is constant x₁ = x₂ = x₀

The width of the bright stripe is

Δy = λ x / d (5/2 -3/2)

Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)

Δy= 5,075 10⁻⁶ m

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3 years ago