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nalin [4]
3 years ago
5

58.54

Physics
1 answer:
Lilit [14]3 years ago
7 0

Answer:

6m/s^2

Explanation:

at=Vf-Vi

a=? a(5)=65-35

t=5 a(5)=30

Vi=35 a(5)/5=30/5

Vf=65 a=6m/s^2

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When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these
Zigmanuir [339]

Answer:

a)  v = 0.7071 v₀, b) v= v₀, c)  v = 0.577 v₀, d)   v = 1.41 v₀, e)  v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

          v = \sqrt{ \frac{T}{ \mu } }

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

          m = 2m₀

let's find the new linear density

          μ = m / l

iinitial density

          μ₀ = m₀ / l

final density

          μ = 2m₀ / lo

          μ = 2 μ₀

we substitute in the equation for the velocity

initial            v₀ = \sqrt{ \frac{T_o}{ \mu_o} }

with the new dough

                    v = \sqrt{ \frac{T_o}{ 2 \mu_o} }

                    v = 1 /√2  \sqrt{ \frac{T_o}{ \mu_o} }

                    v = 1 /√2 v₀

                    v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

              μ = μ₀

   in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

           μ = 3m₀ / l₀

           μ = 3 m₀/l₀

           μ = 3 μ₀

            v = \sqrt{ \frac{T_o}{ 3 \mu_o} }

            v = 1 /√3  v₀

            v = 0.577 v₀

d) l = 2l₀

            μ = m₀ / 2l₀

            μ = μ₀/ 2

           v = \sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }

           v = √2 v₀

            v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

             μ = 10 m₀/2l₀

             μ = 5 μ₀

we look for speed

             v = \sqrt{ \frac{T_o}{5 \mu_o} }

             v = 1 /√5  v₀

             v = 0.447 v₀

5 0
3 years ago
A 49 kg person is being dragged in their sleeping bag to the lake by a 593 N
crimeas [40]

Answer:

f = 485.62 N

Explanation:

Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:

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Unbalanced Force = Fx - f

But, from Newtons Second Law of Motion:

Unbalanced Force = ma

comparing the equations:

ma = Fx - f

f = F Cos θ - ma

where,

f = frictional force  = ?

F = Applied force  = 593 N

m = mass of person = 49 kg

a = acceleration = 0.57 m/s²

θ = Angle with horizontal = 30°

Therefore,

f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)

f = 513.55 N - 27.93 N

<u>f = 485.62 N</u>

6 0
3 years ago
Samanthawalks along a horizontal path in the direction shown the curved path is a semi circle with a radius of 2 m while the hor
Anna11 [10]

Answer:

Explanation:

Samantha walks along a horizontal path in the direction shown the curved path is a semi circle with a radius of 2 m while the horizontal part is for me what is the magnitude of displacement

Displacement is given by the straight line distance between P and Q. Displacement will be length of straight line joining P and Q

a semi circle with a radius of 2 m

Length of this straight line=4+diameter

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=8 m

7 0
4 years ago
an electrically charged particle with mass m, velocity v, and charge q moves in a circle in magnetic field b. what is the formul
kolbaska11 [484]

Answer:

r=\frac {mv}{qb}

Explanation:

In this case,  since the charged particle moves in circular motion,  the centripetal force is equivalent to the magnetic force.

\frac {mv^{2}}{r}=qvb\\\frac {mv}{r}=qb\\r=\frac {mv}{qb}

5 0
3 years ago
A sportscar has a mass of 1500 kg and accelerates at 5 meters per second squared. What is the magnitude of the force acting on t
Hatshy [7]

Answer:

7500 Newtons

Explanation:

Mass of the sportscar= 1500 kg

Acceleration of the sportscar= 5m/s^2

Hence, let the Force acting on it be F

We\ know\ that,\\Force=Mass*Acceleration\\F=ma\\\\Here,\\F=1500*5\\=7500 kg m/s^2\ or\ 7500\ Newtons

4 0
3 years ago
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