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worty [1.4K]
2 years ago
11

Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimu

m necessary. Find the magnitude of the frictional force between the person and the wall
Physics
1 answer:
Likurg_2 [28]2 years ago
7 0

Answer:

Explanation:

The man is moving on a circular path . The reaction of wall is providing centripetal force .so

R = mv² / r

Frictional force = μR where  μ is coefficient of friction

= μ x mv² / r

Frictional force = μ x mv² / r

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The position of a car at time t is given by the function p(t)=t2 2t−4. What is the velocity when p(t)=11? assume t≥0
AysviL [449]

The velocity when function p(t)=11 is 8 .

According to the question

The position of a car at time t  represented by function :

p(t)=t^{2} +2t-4

Now,

When  function p(t) = 11 , t will be

p(t)=t^{2} +2t-4

11 = t²+2t-4

0 = t² + 2t - 15

or

t² +2t-15 = 0

t² +(5-3)t-15 = 0

t² +5t-3t-15 = 0

t(t+5)-3(t+5) = 0

(t-3)(t+5) = 0  

t = 3 , -5  

as t cannot be -ve as given ( t≥0)

so,

t = 3

Now,

the velocity when p(t)=11

As we know velocity = \frac{position}{time}

therefore to get the value of velocity from  function p(t)

we have to differentiate the function with respect to time

\frac{d(p(t))}{dt} =\frac{d}{dt} (t^{2} +2t-4)

v(t) = 2t + 2  

where v(t) = velocity at that time

as t = 3 for  p(t)=11  

so ,

v(t) = 2t + 2  

v(t) = 2*3 + 2  

v(t) = 8

Hence, the velocity when function p(t)=11 is 8 .

To know  more about function here:

brainly.com/question/12431044

#SPJ4

4 0
1 year ago
Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a c
Sidana [21]

Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

7 0
2 years ago
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