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Sveta_85 [38]
2 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]2 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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blsea [12.9K]

Answer:

the new diameter of the third ring = 0.607 mm

Explanation:

Consider the radius of \\m ^{th}\\ bright ring when air is in between the lens and the plate ;

r = \sqrt {\frac{(2m+1)\lambda R}{2}}

Using the expression: r (n)= \sqrt {\frac{(2m+1)\lambda R}{2n}} for the radius of the  \\m ^{th}\\ bright ring since the liquid (water ) of the refractive index 'n' is now in between the lens and the plate;

where;

\\m ^{th}\\ = number of fringe

λ = wavelength

R = radius

n = refractive index of water

Now ;

r (n)=\frac {r}{n}}

the radius r of the third bright ring when the air is in between lens and plate = r= \frac{0.700 \ mm}{2} \\\\

r= 0.35 \ mm

The new radius of the third bright fringe is r(3) = \frac{0.35}{\sqrt 1.33}

r(3) = 0.3035 \ mm

Calculating the new diameter ; we have:

d(3) = 2(r(3))

d(3) = 2(0.3035)

d = 0.607 mm

Thus, the new diameter of the third ring = 0.607 mm

7 0
2 years ago
Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.
m_a_m_a [10]

Answer:

F = 10.788 N

Explanation:

Given that,

Charge 1, q_1=6\ \mu C=6\times 10^{-6}\ C

Charge 2, q_2=2\ \mu C=2\times 10^{-6}\ C

Distance between charges, d = 0.1 m

We know that there is a force between charges. It is called electrostatic force. It is given by :

F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N

So, the force applied between charges is 10.788 N.

3 0
2 years ago
Read 2 more answers
A penny is placed on a rotating turntable. Where on the turntable does the penny require the largest centripetal force to remain
Artyom0805 [142]

Answer:

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m"  and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.  

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

Explanation:

5 0
2 years ago
Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a
garik1379 [7]

Answer:

T=2.94*10^-10  N/m.

Explanation:

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m³.

To have a fundamental frequency at 150Hz , to what tension must a spider adjust a 14cm -long strand of silk?

l=length of the spider silk, 14cm

velocity of wave = √(T/μ)          

where T = tension and

μ = mass per unit length)

λ/2=l

for fundamental frequency λ/2 =14cm    

 (λ= wavelength of standing wave;  as there will be no node

   except the endpoints of silk strand)

               λ = 28 cm = 0.28 m

and since frequency * wavelength = speed of wave. we have,

                  150 * 0.28 = √(T/μ)                                        ..................(#)

now μ = mass/length = [volume * density]/length = [(length*area) * density] / length = area * density

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now putting this in equation (#) we get

    150 * 0.28 = √(T/[13π * 10^(-8)]).

thus T = [13π * 10^(-8)] * (42)²     =  

2.94*10^-10  N/m.

6 0
2 years ago
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Nookie1986 [14]
R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
L : wire length
A : cross section area of wire
by sub.
0.757= (rou*25)/ 3.5*10^-6
25*rou = 2.6495*10^-6
rou= 1.0598*10^-7 ohm.m
4 0
2 years ago
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