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2 years ago

9) For a horizontally launched projectile, decreasing the height of the

Physics

1 answer:

Volgvan2 years ago

4 0

Answer:

Increasing the launch height increases the downward distance, giving the horizontal component of the velocity greater time to act upon the projectile and hence increasing the range.

Explanation:

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Which image illustrates refraction

photoshop1234 [79]

Answer:

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4 years ago

Read 2 more answers

The color that an object appears to be depends on the select one:

ddd [48]

B is the right answer

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3 years ago

What grade of sprain is a completely torn ligament?

Alex Ar [27]

Grade 1: Stretching or slight tearing of the ligament with mild tenderness, swelling and stiffness. The ankle feels stable and it is usually possible to walk with minimal pain.

Grade 2: A more severe sprain, but incomplete tear with moderate pain, swelling and bruising. Although it feels somewhat stable, the damaged areas are tender to the touch and walking is painful.

Grade 3: This is a complete tear of the affected ligament(s) with severe swelling and bruising. The ankle is unstable and walking is likely not possible because the ankle gives out and there is intense pain.

source - https://www.rushcopley.com/health/physician-articles/varying-degrees-of-ankle-sprains/

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3 years ago

Star A and Star B have measured stellar parallax of 1.0 arc second and 0.75 arc second, respectively. Which star is closer? How

zhuklara [117]

Answer:

Star A is closer than Star B

Explanation:

As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle

Here we can say

$angle = \frac{1 Parsec}{distance}$

so we have

$distance = \frac{1 Parsec}{angle}$

so here we have

angle subtended by Star A = 1 arc sec

angle subtended by star B = 0.75 arc sec

now we have

distance for star A is given as

$d_a = \frac{1 Parsec}{1} = 1 Parsec$

distance of star B is given as

$d_b = \frac{1 Parsec}{0.75} = 1.33 Parsec$

So star A is closer than star B

7 0

3 years ago

A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle

Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of θ = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height y at time t would be

y = (40 m/s) t - 1/2 g t ²

Set y = 160 m, and you'll find that there is no (real) solution for t, so the ball never attains the given maximum height.

From another perspective: recall that

v ² - v₀² = 2a ∆y

where

• v₀ = initial velocity

• v = final velocity

• a = acceleration

• ∆y = displacement

At its maximum height, the ball has zero vertical velocity, and ∆y = maximum height = 160 m. The ball is in free fall once it's launched, so a = -g.

So we have

0² - (40 m/s)² = -2g (160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0

3 years ago