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Naddik [55]
4 years ago
7

it takes 1.4 minutes for Jonny to get to class which is a distance of 82 meters away s. what is Johnny’s average speed in second

s?
Physics
1 answer:
Marat540 [252]4 years ago
4 0
We have to use the equation speed=distance/time.
We want the average speed in second so we have to change the minutes into seconds. We can do this by multiplying the minutes by 60 (60seconds in 1 minute). 60x1.4= 84 seconds.
Speed=distance/time, distance is 82 and time is 84 so speed=82/84
Average speed = 0.98ms^-1 (2dp)
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Please help!! I’m so confused and struggling rn
DIA [1.3K]

Explanation:

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6 0
3 years ago
a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while th
anzhelika [568]

The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

F_{1} = 250

A_{1} = Area of the small piston = πr^{2}

A_{1} = 22/7 x 0.4^{2}

A_{1} = 0.5 m^{2}

F_{2} = ?

A_{2} = Area of the large piston = πr^{2}

A_{2} = π x 1

A_{2} = 3.14 m^{2}

To calculate the force on the large piston, we will use the below formula

F_{1}/ A_{1} = F_{2} / A_{2}

Substitute all the parameters into the equation

250/0.5 =  F_{2}/3.14

F_{2} = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)​

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

F_{1}/ 0.25A_{1} = F_{2} / A_{2}h

250/0.125 = 1570/3.14h

make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

Learn more here: brainly.com/question/13596980

5 0
2 years ago
A velocity selector uses a fixed electric field of magnitude E and the magnetic field is varied to select particles of various e
defon

Answer: we need B/√2 for having twice the energy i.e 0.17B

Explanation:

This is actually a simple one, so i will guide you through it.

We know that Energy here means Kinetic Energy  (K.E)

and the expression is given thus;

Energy = 1/2 mv2

and velocity v = E/B

let us make x the energy intially, foe electric field E, magnetic field B, mass m

so, x = 1/2 m (E/B)2

we have to find magnetic field for which twice the energy so , let magnetic field be y, so

2x = 1/2m(E/y)2

y = B/√2  = 0.17B

therefore, we need B/√2 for having twice the energy

cheers i hope this helps!!!!

6 0
3 years ago
PLEASE HELPPP ILL DO ANYTHINGGGGGG Write 1 paragraph or more and use all three of Newton’s laws of motion to describe what happe
Lilit [14]

Answer:

The image shows an object at rest, then a force pushes the car into the water. Newton's First Law describes that objects at rest remain at rest, and objects at motion remain at motion, unless they are bothered by a force. This law comes into play with the image because the object is first seen at rest, on the top of another object. Then it is seen falling, and is now in motion at it descends into the water. Newton's Second Law, F = MA, is seen in the image is the object accelerates into the water from the force. Newton's Third Law, which states 'for every action there is an equal but opposite reaction', is in this image as the object is falling, then makes contact with the water.

Explanation: Newton's Laws

4 0
3 years ago
22. A force of 52 N acts upon a 4 kg block sitting on the ground. Calculate the acceleration of the obj
chubhunter [2.5K]

Explanation:

to calculate the acceleration we use this formula

force=mass × acceleration

F = M × A

we are given the value of force(F) which is 52 and mass(M) which is 4

to find the acceleration we just make A the subject of formula

A = F/M

A=52/4

A =13m/s²

please rate as brainliest it will really go a long way

5 0
1 year ago
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