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givi [52]
3 years ago
13

Four boys push on the front back and sides of a shopping cart. the boys in the front on the two sides push with a force of 60N.

the boy in back pushes with a force of 50N. which way does the cart move
*ASAP

Physics
1 answer:
Art [367]3 years ago
3 0

Answer:

The cart will move backwards

Explanation:

To solve this problem we must make a free body diagram of the shopping cart and the forces acting on them. You have two boys pushing on the sides with a force of 60 [N] and the child on the front also pushes with a force of 60[N], the child behind pushes with a force of 50 [N].

 In the attached image we can see the free body diagram.

As shown in the attached image the shopping cart is displayed from above, the forces of children pushing on the sides, are canceled with each other as they are equal in magnitude and opposite in direction. While the force of the child pushing with 60 [N] is greater than the force of the child pushing with 50 [N] therefore the cart will move backwards, that is to the left as seen in the attached scheme.

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Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. The heat capacity of Object A is
Oksi-84 [34.3K]

Answer:

Explanation:

Heat capacity A = 3 x heat capacity of B

initial temperature of A = 2 x initial temperature of B

TA = 2 TB

Let T be the final temperature of the system

Heat lost by A is equal to the heat gained by B

mass of A x specific heat of A x (TA - T) = mass of B x specific heat of B x ( T - TB)

heat capacity of A x ( TA - T) = heat capacity of B x ( T - TB)

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8 0
3 years ago
When the number of atoms on the right side a chemical equation matches the number of atoms on the left side of a chemical equati
Hitman42 [59]

Answer:

Balanced.

Explanation:

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7 0
2 years ago
please help me with my question I will like and mark as brainliest NO LINKS THEY DON'T WORK AND IF U DON'T KNOW THE ANSWER PLS D
bearhunter [10]

Answer:

a)

Weight in Air = 0.3N

Weight in Water = 0.25N

Weight in Liquid = 0.24N.

Upthrust /Buoyant Force = Weight in Air – Weight in Fluid(Water in this case)

= 0.3 – 0.25

= 0.5N.

b) R.D of Body = Density of Body/Density of Standard Fluid(Water).

There's a Derived Formula for RD.

I'm gonna Apply it here.

Ask me for the derivation in the Comment section if you need it.

RD = α/ρ = (Weight in Air) / (Upthrust Force)

Where

α = density of the Body(or reference substance)

ρ = density of standard fluid (water)

= 0.3/0.05 = 6.

c) RD of Liquid = (Density of Liquid) /(Density of standard Fluid(water)

Or we just go by that formula

RD of Liquid = Weight in Air/Upthrust(In Liquid)

We'll be using the Upthrust in that Liquid now.

= 0.3 – 0.24 = 0.06

RD = 0.3/0.06 = 5.

7 0
2 years ago
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