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densk [106]
3 years ago
8

A 1000 kg car is rolling slowly across a level surface at 1m/s, heading toward a group of small innocent children.

Physics
2 answers:
Ne4ueva [31]3 years ago
7 0

Answer:

The force is -1620.73 N.

Explanation:

Given that,

Mass of car = 1000 kg

Velocity = 1 m/s

Distance = 2 m

Angle = 30°

We need to calculate the force

Using formula of work done

W_{net}=K_{f}-K_{i}

F\times d=K_{f}-K_{i}

Fd\cos\theta=-K_{i}

F=\dfrac{-K_{f}}{d\cos\theta}

F=-\dfrac{\dfrac{1}{2}mv^2}{d\cos\theta}

Put the value into the formula

F=-\dfrac{\dfrac{1}{2}\times1000\times(1)^2}{2\times\cos30}

F=-1620.73\ N

Hence, The force is -1620.73 N.

AnnyKZ [126]3 years ago
4 0

Answer:

Force, F = 288.67 N

Explanation:

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, \theta=30^{\circ}

Distance, d = 2 m

Let F is the force must you push to stop the car. using work energy theorem to find it. According to this theorem, the work done is equal to the change in kinetic energy as :

W=\dfrac{1}{2}m(v^2-u^2)

F\times d=\dfrac{1}{2}m(v^2-u^2)

It stops, v = 0

Fd\ cos\theta=\dfrac{1}{2}m(u^2)      

F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}

F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}

F = -288.67 N

So, the force required to push to stop the car is 288.67 N. Hence, this is the required solution.                                            

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I NEED HELP PLEASE, THANKS! :)
Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio  = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = \frac{k\ q_1\ q_2}{r^2}     .....................1

and here k is 9 × 10^9 N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = \frac{k\ q^2}{d^2}

and force between charges at A to C, at 2d distance

F1 = \frac{k\ q^2}{(2d)^2}  =  \frac{k\ q^2}{4d^2}

force between charges at A to D,  3d distance

F1 = \frac{k\ q^2}{(3d)^2}  = \frac{k\ q^2}{9d^2}  

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a)  = -F1 - F2 + F3             ....................2

put here value

F(a) = -\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

solve it

F(a) = \frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})  

F(a) = -\frac{41}{36}\ F1   = 1.13 F1  

and

Charge b It  receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2            ....................3

F(b)  =  \frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}    

F(b)  = \frac{1}{4} \ F1    =  0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1      ....................4

F(c) = \frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}      

F(c) =  \frac{9}{4} \ F1   = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1      ....................5

F(d) = -\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}  

so

F(d) = -\frac{49}{36} \ F1    = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = \frac{2.25}{0.25}

 ratio  = 9 : 1

5 0
2 years ago
Imagine a negative test charge sitting at the coordinate origin (0,0). Two bunches of positive charges are located on the x-axis
Oxana [17]

Answer:

the total force vector, on test charge is points from origin to point C( 1, 1 )

Explanation:

Given the data in the question, as illustrated in the image below;

from the Image, OA = 1, OB = AC = 1

so using Pythagoras theorem

a² = b² + c²

a = √( b² + c² )

so

OC = √( OB² + AC² )

we substitute

OC = √( OA² + AC² )

OC = √( 1² + 1² )

OC = √( 1 + 1 )

OC = √2

Coordinate of C( 1, 1 )

Hence, the total force vector, on test charge is points from origin to point C( 1, 1 )

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What is heat energy on earth escapes into space
Firdavs [7]

Answer:

Latent heat, along with birds, ride those rising columns of air. This brings up a third and the ultimate mechanism by which the Earth's heat escapes into space, which is electromagnetic radiation. Every object, including the Earth's surface, absorbs and radiates heat electromagnetically

Explanation:

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5 0
2 years ago
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maksim [4K]

Answer:

Microlensing.

Explanation:

This techniques is called Microlensing.

Microlensing is a method of gravitational lensing where light from a backdrop point of origin is curved to develop distorted, numerous and/or lightened images by the gravity field of a foreground lens.

This method is very effective in discovering planets that are far-far from earth.It is actually an astronomical effect that was predicted by Albert Einstein's general theory of relativity.

7 0
3 years ago
A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

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In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0
2 years ago
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