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3 years ago

Select each example of a projectile

Physics

2 answers:

mafiozo [28]3 years ago

4 0

Baseball, javelin, and maybe the clock but not sure on that... Just say baseball and javelin

Ludmilka [50]3 years ago

3 0

Answer:

2. A baseball

4. A javelin thrown

Explanation:

In projectile motion all the objects will have two components of motion under gravity.

Horizontal component of motion must remains constant as there is no force in that direction while in vertical direction due to gravity the speed must change.

So whenever an object is projected at some angle with the horizontal which is less than 90 degree then in that case the motion must be projectile motion.

So here the correct options are

2. A baseball

4. A javelin thrown

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You find yourself in the middle of a frozen lake with a surface so slippery (ms = mk = 0) you cannot walk. However, you happen t

EleoNora [17]

Answer:Maybe if you would slide on your stomach you can get out of there.

Explanation:

8 0

3 years ago

The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which

emmainna [20.7K]

Answer:

Explanation:

A ) angular velocity ω = 2π / T

= 2 x 3.14 / 60

= .10467 rad / s

linear velocity v = ω R

= .10467 x 50

= 5.23 m / s

centripetal force = m v² / R

= mg v² / gR

= 834 x 5.23² / 9.8 x 50

= 46.55 N

B )

apparent weight

= mg - centripetal force

= 834 - 46.55

= 787.45 N

C ) apparent weight

= mg + centripetal force

= 834 + 46.55

= 880.55 N.

D )

For apparent weight to be zero

centripetal force = mg

mg = mv² / R

v² = gR

= 9.8 x 50

= 490

v = 22.13 m /s

time period of revolution

= 2π R /v

2 x 3.14 x 50 / 22.13

= 14.19 s

8 0

3 years ago

Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside

jek_recluse [69]

Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

From the continuity equation,

$\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}$

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

$\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}$

Thus, the flow rate of the water at the later position is 5.99 m/s

4 0

1 year ago

How many earths could fit inside jupiter

Snezhnost [94]

1,300 or more.

hope this helped :)

8 0

3 years ago

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 20 cm apart. The sound

Sophie [7]

Answer:

a. Wavelength = λ = 20 cm

b. Next distance of maximum intensity will be 40 cm

Explanation:

a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.

Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.

This we get the equation:

(nλ + λ/2) - nλ = 30-20

λ/2 = 10

λ = 20 cm

b. at what distance, sound intensity will be maximum again.

For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.

= 30 + λ/2

= 30 + 20/2

=30+10

=40 cm

7 0

3 years ago