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Ierofanga [76]
4 years ago
9

Please help me answer this

Physics
1 answer:
yanalaym [24]4 years ago
6 0

Answer:

care of it and I will be there at work and I will be there at work

Explanation:

g the day off and I will be there at work and I will be there at work and I will be there

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A simple machine does work because it applies
tigry1 [53]
Force 
or motion im not toatally sure
7 0
3 years ago
A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the tables edge in the
GREYUIT [131]

A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the edge of the table in the horizontal direction,

  • t= 0.45 seconds.
  • V=2.22m/s
  • VT=4.95 m/s

This is further explained below.

<h3>What is its speed when it hits the floor...?</h3>

Generally, the equation for motion is mathematically given as

S= ut + 0.5at²

Therefore

y = Voy t + 0.5gt^2

1 = 0.5x 98 x 6²

1=4.9t^2

t=\sqrt{0.2041 }

t= 0.45 seconds.

b) Horizontal motions are uniform.

V=Horizontal displacement/time

V=1/0.45

V=2.22m/s

C)

Vx: 2.22 m/s At bottom,

Vy² = Voy² + 2as

Vy² = 2x95x1

Vy² = 19.6

Total velocity

VT=\sqrt{( 2.22 m/)^2+19.6}

VT=4.95 m/s

Read more about  Arithmetic

brainly.com/question/22568180

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4 0
2 years ago
A uniform film of a material that has index of refraction 1.30 covers the front surface of a pane made of glass with index of re
Orlov [11]

Answer:

The energy of each photon in the light beam is 1.73\times10^{-18}\ J.

Explanation:

Given that,

Index of refraction of glass =1.55

Index of refraction of material =1.30

Wavelength = 115 nm

We need to calculate the energy of each photon in the light beam

Using formula of energy

E=\dfrac{hc}{\lambda}

Where, h = Planck constant

\lambda=wavelength

c = speed of light

Put the value into formula

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{115\times10^{-9}}

E=1.73\times10^{-18}\ J

Hence, The energy of each photon in the light beam is 1.73\times10^{-18}\ J.

3 0
4 years ago
A constant force of 53 newtons was applied to a 12-kilogram mass for 15 seconds. a.) What was the change in momentum of the mass
vazorg [7]

Explanation:

Impulse = change in momentum

FΔt = Δp

(53 N) (15 s) = Δp

Δp = 795 kg m/s

For constant mass:

Δp = m Δv

795 kg m/s = (12 kg) Δv

Δv = 66.25 m/s

Rounding to two significant figures, Δp = 800 kg m/s and Δv = 66 m/s.

8 0
3 years ago
Calculate how fast the ball would be moving at the instant it leaves the projectile launcher of the spring is compressed by 3.75
Mnenie [13.5K]

Answer:

V = 8.34m/s

Explanation:

Given that

1/2ke^2 = 1/2mv^2 ......1

Where e = 3.75cm = (3.75/100)m

e = 0.0375m

K = 500 N/m

m = 10g = 10/1000

= 0.01kg

Substitute the values into equation 1

0.5×500×(0.0375)^2 = 0.5×0.01×v^2

250×0.001395 = 0.005v^2

0.348 = 0.005v^2

v^2 = 0.348/0.005

v^2 = 69.6

V = √69.6

V = 8.34m/s

The ball launches at the speed of V = 8.34m/s

6 0
3 years ago
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