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10 months ago

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A ball is projected at an angle of 30° above the horizontal with a speed of 35 m/s. What will be its approximate horizontal rang

e across a level surface?

1 answer:

BaLLatris [955]10 months ago

3 0

37.5 m will be its approximate horizontal range across a level surface.

we have given in the data

u (initial velocity) = 35 m/s

projected angle = 30°

Horizontal range =?

range of horizontal distance $=\frac{u^2 Sin 2 \theta}g$

PUTTING all the values in the equation

horizontal range= $\frac{35\cdot35\cdot sin60}{9.8}$

by solving, we got

horizontal range=37.5 m

The horizontal distance traversed is the projectile's range,The thing moves in one direction until it hits the ground.
When an item is hurled at an angle of 90°, only vertical motion occurs. As a result, the projectile's range will be zero.
horizontal range variates with angle, initial velocity, and by the value of g.

To know more about horizontal range visit : brainly.com/question/23827445

#SPJ1

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Answer:

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2 years ago

Use the drop-down menus to complete each sentence about the layers of the atmosphere. If the did not exist, Earth might be destr

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Answer:

If the mesosphere did not exist, Earth might be destroyed by chunks of rock from space.

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Explanation:

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Above the stratosphere, the mesosphere extends upward to about 85 km from Earth's surface. The chunks of rock from space such as meteors burn upon entering the mesosphere and protect the Earth. In this layer, the temperature gets colder as altitude increases and so the coldest temperatures (about -90° C) in Earth's atmosphere are found at the top of this layer.

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2 years ago

The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

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$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

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Finding $\theta$ :

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2 years ago