A 38kg child is at the top of a slide that is 2. 1m tall.

The maximum range of projectle is R° if projected at angle of 45°. It'll be 300% of Range of projectile if projected at angle? a

geniusboy [140]

Answer:

The maximum range is 300% of the range of the projectile is projected at an angle of 9.74°.

None of the options are correct.

Explanation:

Normally, ignoring air resistance, for projectile motion, the range (horizontal distance travelled) of the motion is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile

θ = angle above the horizontal at which the projectile was launched

g = acceleration due to gravity = 9.8 m/s²

The range is maximum when θ = 45°

R when θ = 45° is

R = (u² sin 90°)/g = (u²/g)

We are then told that this maximum range is 300% of the value obtainable for the range at a particular angle

Maximum range = 3R

(u²/g) = 3(u² sin 2θ)/g

Sin 2θ = (1/3)

2θ = sin⁻¹ (1/3) = 19.47°

θ = (19.47°/2) = 9.74°

Hope this Helps!!!

4 0

3 years ago

A beam of monochromatic light is aimed at a slit of width and forms a diffraction pattern. In which case is the width of the cen

AysviL [449]

When the incident light is yellow the width of the central band greater. Single-wavelength light sources are known as monochromatic lights, where mono stands for one and chroma for color. Monochromatic lights are defined as visible light that falls inside a specific range of wavelengths. It has a wavelength that falls within a constrained wavelength range.

A laser beam is the ideal illustration of monochromatic light. A monochromatic light beam produced by a single atomic transition with a particular single wavelength is what makes up a laser. A color scheme that consists solely of different shades of one color is referred to as monochromatic.

To learn more about monochromatic, click here.

brainly.com/question/23624834

#SPJ4

4 0

2 years ago

What is the wavelength of a wave that travels at a speed of 3.0x10^8 m/s and has a frequency of 1.5x10^16 m/s?

Vlad [161]

Speed of wave = wavelength * frequency

Therefore, wavelength is 2.0 x 10^-8

5 0

3 years ago

A body of mass 8 kg moves in a (counterclockwise) circular path of radius 10 meters, making one revolution every 10 seconds. You

Sav [38]

Answer:

Centripetal force is equal to 31.55 N

Explanation:

We have given mass of the body m = 8 kg

Radius of the circular path r = 10 m

It is given that it makes 1 revolution in 10 seconds

Distance traveled in 10 seconds is equal to $d=2\pi r=2\times 3.14\times 10=62.8m$

Velocity is equal to $velocity=\frac{distance}{time}=\frac{62.8}{10}=6.28m/sec$

We have to find the centripetal force

Centripetal force is equal to $F=\frac{mv^2}{r}=\frac{8\times 6.28^2}{10}=31.55N$

So centripetal force will be equal to 31.55 N

8 0

3 years ago

Starting from rest, a small block of mass m slides frictionlessly down a circular wedge of mass M and radius R which is placed o

guapka [62]

Answer:

Part a)

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

$mv + MV = 0$

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR$

now we have

$\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR$

$\frac{1}{2}MV^2(\frac{M}{m} + 1) = mgR$

so we have

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

and another block has speed

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

7 0

3 years ago