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Arisa [49]
2 years ago
12

A stunt man projects himself horizontal from a height of 60m. He lands 150m away from where he was launched. How fast was he lau

nched?
A.) 54.87 m/s
B.) 43.98 m/s
C.) 47.46 m/s
D.) 42.87 m/s
Physics
1 answer:
koban [17]2 years ago
4 0

Answer:

D) 42.87 m/s

Explanation:

First, find the time it takes him to land.  Given in the y direction:

Δy = 60 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

60 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 3.5 s

Next, find the speed needed to travel the horizontal distance in that time.  Given in the x direction:

Δx = 60 m

a = 0 m/s²

t = 3.5 s

Find: v₀

Δy = v₀ t + ½ at²

150 m = v₀ (3.5 s) + ½ (0 m/s²) (3.5 s)²

v₀ = 42.87 m/s

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Levi observed properties of four different waves and recorded observations about each one in his chart. A 2-column table with 4
makvit [3.9K]

Answer:

Wave W is a sound wave, Waves X and Y are light waves, and it is impossible to tell what kind of wave Wave Z is.

Explanation:

W      travels fastest through metal

X       travels fastest through air,

Y       travels more slowly through water than air    

Z       travels more slowly at cool temperatures

W appears to be sound wave  as sound travels fastest through metal .

X appears to be light wave as light travels fastest in air .

Y also appears to be light wave as  speed of light is reduced when it passes from air to water .

Z  It is impossible to tell anything about the nature of Z wave .

5 0
2 years ago
Read 2 more answers
In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9
slega [8]

Answer:

1.6 secs

Explanation:

In a circus act, an acrobat upwards from the surface of a trampoline

At that same moment another acrobat perched 9.0m above him

A ball is released from rest

While still in motion the acrobat catches the ball

He left the ball with a trampoline of 5.6m/s

Since the ball is falling downwards from a distance then acceleration will be negative

a= -9.8

s= d

s= 1/2at^2

= 1/2 × (-9.8)t^2

= 0.5× (-9.8)t^2

d = -4.9t^2

Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows

9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2

9 - 4.9t^2= 5.6t + (-4.9)t^2

9 - 4.9t^2= 5.6t - 4.9t^2

9= 5.6t

t= 9/5.6

t= 1.6 secs

6 0
2 years ago
An example of potential energy is a ball sitting _____ of the stairs.
expeople1 [14]

Answer:

at the top

Explanation:

Potential energy is the stored energy, mechanical energy,

or energy possessed by by virtue of the position of an object.an example of potential energy is the energy that a ball possesses by virtue of its sitting at the top of the stairs it being about to roll down the stairs.

3 0
2 years ago
A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a
Lapatulllka [165]

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that,

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :

y = 0.049 \cos(7t) .......(1)

The general equation of a wave is given by :

y=A\cos(\omega t) .......(2)

A is amplitude of wave

On comparing equation (1) and (2) we get :

A = 0.049 meters

So, the amplitude of the wave is 0.049 meters.

3 0
2 years ago
An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0
2 years ago
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