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Verizon [17]
3 years ago
9

2) How much work is required to pull a sled 15 meters if you use 30N of force?

Physics
1 answer:
geniusboy [140]3 years ago
7 0

2 people

Explanation:

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Danny Diver weighs 500 N and steps off a diving board 10 m above the water. Danny hits the water with kinetic energy of
Morgarella [4.7K]

Answer:

Danny hits the water with kinetic energy of 5000 J.

Explanation:

Given that,

The Weight of Danny Diver,

F = 500 N

m*g= 500 N

He steps off a diving board 10 m above the water.

h=10 m

when Danny diver hits water he generates the kinetic energy.

We need to find the kinetic energy of the water.

Let kinetic energy is K.

K = m*g*h

Where g is acceleration due to gravity.

that g= 9.8 m/s^2

now substituting the values in above equation

K= (500) * 10

K= 5000 J

Hence,

he hits the water with kinetic energy of 5000 J.

Learn more about Kinetic energy here:

<u>brainly.com/question/15587458</u>

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#SPJ4

7 0
2 years ago
1)Light of wavelength 588.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 55.5 cm from the slit
Talja [164]

Answer:

These are Diffraction Grating Questions.

Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:  

Given as  

y = nDλ/w                                                       Eqn 1

where  

w = width of slit  

D = distance to screen  

λ = wavelength of light  

n = order number  

Making x the subject of the formula gives,  

w = nDλ/y  

Given  

y = 0.0149 m  

D = 0.555 m  

λ = 588 x 10-9 m  

and n = 3

w = 6.6x10⁻⁵m

Hence, the width of the slit w, in micrometers (μm) = 66μm

Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen

i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx

Recall Eqn 1,     y = nDλ/w  

given, D = 27cm = 0.27m  

λ = 632 x 10-9 m  

w = 0.1mm = 1.0x10⁻⁴m

For the 9th order, n = 9,

y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m

Similarly, for n = 5,

y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m

Recall,  Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m

Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm

8 0
3 years ago
Nio conducts a science experiment. He puts yeast into a bottle with water and sugar, and then caps the bottle with a balloon. He
KATRIN_1 [288]

Answer: Option (C) is the correct answer.

Explanation:

When yeast is added into a bottle with water and sugar then after some time yeast becomes activated. This results in the formation of a chemical change as bubbles are formed on the surface of the mixture in the bottle.

This indicates that a gas has been released due to the chemical change.

Thus, we can conclude that most likely a chemical change has occurred because a gas was produced.

3 0
3 years ago
Read 2 more answers
The aim of the newton's first law experiment ​
Semmy [17]

Answer:

Application of Newton's first law of motion

A body in motion will continue in motion in a straight line unless acted upon by an outside force.

Explanation:

4 0
2 years ago
A driver drives for 30.0 minutes at 80.0 km/h, then 45.0 minutes at 100 km/h. She then stops 30 minutes for lunch. She then trav
bija089 [108]

Answer:

b) 68,9 km/h a) picture

Explanation:

In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.

Using this formula we pass time to hours:

t_{hours}=t_{min}*\frac{1 h}{60 min}\\30min*\frac{1 h}{60 min}=0,5h\\45min*\frac{1 h}{60 min}=0,75h

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.

Using the values of the speed we calculate the distance in each interval

d=v*t\\80km/h*0.5h=40km\\100km/h*0.75h=75km

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).

Finally, we compute the average speed with the distance over time:

v_{average}=\frac{155km}{2.25h}=68.9km/h

6 0
3 years ago
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