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Sveta_85 [38]
3 years ago
11

A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force resisting the carts moti

on is 15.0 N, how hard does the man have to push so that the cart accelerates at 1.50m/s/s.
Physics
1 answer:
-BARSIC- [3]3 years ago
4 0

The force applied by the man is 60 N

Explanation:

We can solve this problem by applying Newton's second law, which states that:

\sum F = ma (1)

where

\sum F is the net force acting on the child+cart

m is the mass of the child+cart system

a is their acceleration

In this problem, we have:

m = 30.0 kg is the mass

a=1.50 m/s^2

And there are two forces acting on the child+cart system:

  • The forward force of pushing, F
  • The force resisting the cart motion, R = 15.0 N

Therefore we can write the net force as

\sum F = F -R

where R is negative since its direction is opposite to the motion

So eq.(1) can be rewritten as

F-R=ma

And solving for F,

F=ma+R=(30.0)(1.50)+15.0=60 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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The force required to extend a rod increases as the cross sectional area

increases.

The rod that experiences the largest force is <u>rod B</u>

Reason:

The elongation of a rod by the application of a force is given by the

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\Delta L = \dfrac{F \cdot L}{A \cdot E}

From the above equation, we have that the elongation is inversely

proportional to the cross sectional area, such that the extension of a rod by

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Therefore, the force required to extend the length of a rod by a specific

amount increases as the cross sectional area of the rod increases,

indicating that the rod with the largest cross sectional area require the

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The rod that experiences the largest force is the rod with the largest cross

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Learn more here:

brainly.com/question/12937199

4 0
2 years ago
four forces act on an object of mass 100kg, they are 12N(N), 70N(S), 33N(E), 10N(w). calculate the reseulting acceleration of th
trapecia [35]

Answer:

0.62 m/s² at 68° S of E

Explanation:

Net force north = 12 - 70 = -58 N

Net force east = 33 - 10 = 23 N

Net force = √(-58² + 23²) = 62.3939... N

acceleration = F/m = 62.3939/100 = 0.623939... ≈ 0.62 m/s²

θ = arctan(-58/23) = -68.3691... ≈ 68° S of E

6 0
3 years ago
Consider a constant density gas flowing steadily over an airfoil. Far upstream the velocity is V0. Halfway along the top surface
Mashutka [201]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

5 0
3 years ago
Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

8 0
3 years ago
Which describes one event that causes an eclipse?
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Answer:

earth's shadow covering the moon,thats lunar eclipse

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