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Ira Lisetskai [31]
2 years ago
7

A box of mass 3kg is lifted 1.5m onto a shelf. Calculate the change in its gravitational potential energy. The gravitational fie

ld strength is 10N/kg.
Physics
1 answer:
Vedmedyk [2.9K]2 years ago
5 0

Answer:

The change in gravitational potential energy is 45 J.

Explanation:

Given that,

Mass = 3 kg

Distance = 1.5 m

Gravitational field strength = 10 N/kg

We need to calculate the change in gravitational potential energy

Using formula of  gravitational potential energy

Change\ in\ gravitational\ potential\ energy =gravitational\ field\ strength\times mass\times distance

Put the value into the formula

Change\ in\ gravitational\ potential\ energy =10\times3\times1.5

Change\ in\ gravitational\ potential\ energy =45\ J

Hence, The change in gravitational potential energy is 45 J.

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B 
Think of inertia of getting into a car accident without a seat belt although the car stops you will not you would likely fly out the window
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1 year ago
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In the produce section of a supermarket, five pears are placed on a spring scale. The placement of the pears stretches the sprin
tankabanditka [31]

Answer:

The displacement of the spring due to weight is 0.043 m

Explanation:

Given :

Mass m = 2 Kg

Spring constant k = 450 \frac{N}{m}

According to the hooke's law,

  F = -kx

Where F = force, x = displacement

Here,

F = mg         ( g = 9.8 \frac{m}{s^{2} } )

F = 2 \times 9.8 = 19.6 N

Now for finding displacement,

  x = \frac{F}{k}

Here minus sign only represent the direction so we take magnitude of it.

  x = \frac{19.6}{450}

  x = 0.043 m

Therefore, the displacement of the spring due to weight is 0.043 m

8 0
2 years ago
A hummingbird 3.4m above the ground flies 1.2 m along a straight line path. Upon spotting a flower below, the hummingbird drops
Anit [1.1K]

A = horizontal displacement of the humming bird = 1.2 m

B = vertical displacement of the humming bird = 1.4 m

C = net displacement of the humming bird from initial to final position = ?

In the triangle drawn , Using Pythagorean theorem

C = √(A² + B²)

inserting the values

C = √(1.2² + 1.4²)

C = √(1.44 + 1.96)

C = √(3.4)

C = 1.4 m

Hence the net displacement of hummingbird comes out to be 1.4 m

4 0
1 year ago
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The tonga trench in the pacific ocean is 36,000 feet deep. assuming that sea water has an average density of 1.04 g/cm3, calcula
MakcuM [25]
1110 atm    

Let's start by calculating how many cm deep is 36,000 feet. 

 36000 ft * 12 in/ft * 2.54 cm/in = 1097280 cm   

 Now calculate how much a column of water 1 cm square and that tall would mass. 

 1097280 cm * 1.04 g/cm^3 = 1141171.2 g/cm^2   

 We now have a number using g/cm^2 as it's unit and we desire a unit of Pascals ( kg/(m*s^2) ).  

 It's pretty obvious how to convert from g to kg. But going from cm^2 to m is problematical. Additionally, the s^2 value is also a problem since nothing in the value has seconds as an unit. This indicates that a value has been omitted. We need something with a s^2 term and an additional length term. And what pops into mind is gravitational acceleration which is m/s^2. So let's multiply that in after getting that cm^2 term into m^2 and the g term into kg.   

 1141171.2 g/cm^2 / 1000 g/kg * 100 cm/m * 100 cm/m = 11411712 kg/m^2 

 11411712 kg/m^2 * 9.8 m/s^2 = 111834777.6 kg/(m*s^2) = 111834777.6 Pascals   

 Now to convert to atm 

 111834777.6 Pa / 1.01x10^5 Pa/atm = 1107.2750 atm   

 Now we gotta add in the 1 atm that the atmosphere actually provides (but if you look closely, you'll realize that it won't affect the final result). 

 1107.274 atm + 1 atm = 1108.274 atm   

 And finally, round to 3 significant figures since that's the accuracy of our data, giving 1110 atm.
8 0
1 year ago
A boy stands at a certain distance from a large building and blows a whistle. After 2.3s he hears the echo of the sound. He move
ddd [48]

Answer:

I) 57.5 m

Il) 50 m/s

Explanation:

Given that a boy stands at a certain distance from a large building and blows a whistle.

After 2.3s he hears the echo of the sound. The speed V of the sound will be:

V = 2X / T

Where X is the distance covered

V = 2X / 2.3

V = 0.87X ...... (1)

He moves 50m towards the building and blows his whistle again; this time the echo reaches him after 2.0s.

V = (2×50) / 2

V = 50 m/s

Substitutes the V into equation 1

50 = 0.87X

X = 50 / 0.87

57.5 m

Therefore, the Boys original distance from the building is 57.5m and the

Speed of sound in air is 50m/s

6 0
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