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worty [1.4K]
3 years ago
14

the force of a bag of potatoes on a vegetable stand is 22.5 N down. what is the force of the vegetable stand on the bag of potat

oes?
Physics
1 answer:
natima [27]3 years ago
4 0

As per Newton's III law we can say that

Force applied by object 1 on object 2 is always equal in magnitude and opposite in direction of the force that object 2 apply on object 1.

So we can say it as

F_{12} = -F_{21}

now here above question is based upon the same

if a bag of vegetables applied a force F = 22.5 N of the surface stand the the same surface will apply same magnitude of force in opposite direction on the vegetables bag

So our answer will be F = 22.5 N (upwards).

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the answer would be b. it`s half as great

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We are on an asteroid with thrusters pointed in 4 directions; Go Left, GO Right, Go Up, and Go Down. We start at coordinate (3,3
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Which of the following is the correct formula for sulfur hexaflouride? SF 6 SF 5 SF 8 SF
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3 years ago
The car has a maximum acceleration of 4m/s2 . The car’s mass is 1,250 Kg what is the car max speed?
marusya05 [52]
The net force acting on the car is
3 × 10^3
Newtons.
Force is defined as the product of the mass of the body and its aaceleration,
⇒ F = ma

Substituting the above given values we get,
F =(1500kg) (2.0 m/s^2) = 3000N = 3 × 10^3N .
3 0
2 years ago
A rectangular metal tank with an open top is to hold 171.5 cubic feet of liquid. what are the dimensions of the tank that requir
Semmy [17]
To minimize the material usage we have to have the volume requested with the minimum surface area.
The volume is:
171.5 =xyz
And the surface is:
S=xy+2xz+2yz
From the first equation we get:
z=\frac{171.5}{xy} ; k=171.5\\ z=\frac{k}{xy}\\
I will use k instead of a number just for the conveince.
We plug this into the second equation and we get:
S=xy+2k\frac{1}{x}+2k\frac{1}{y}
To find the minimum of this function we have to find the zeros of its first derivative.
Sx will denote the first derivative with respect to x and Sy will denote the first derivative with respect to Sy.
S_x=y-2k\frac{1}{x^2}\\ S_y=x-2k\frac{1}{y^2}
Now let both derivatives go to zero and solve the system (this will give us the so-called critical points).
0=y-2k\frac{1}{x^2}\\
0=x-2k\frac{1}{y^2}\\
y=2k\frac{1}{x^2}\\
x=2k\frac{1}{y^2}\\
Now we plug in the first equation into the other and we get:
x=\frac{\frac{2k}{1}}{\frac{4k^2}{x^4}}\\
x^3=2k\\
x=(2\cdot171.5)^{1/3}\\
x=7

Now we can calculate y:
y=2k\frac{1}{x^2}\\
y=2\cdot 171.5\frac{1}{7^2}=7
And finaly we calculate z:
z=\frac{171.5}{xy}\\
z=\frac{171.5}{7\cdot7}=3.5
And finaly let's check our result:
V=xyz=7\cdot7\cdot3.5=171.5
4 0
3 years ago
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