Ask question

Ask question

All categories

4 years ago

12

some amusement parks have a ride where people are attached to a long cable, pulled back, and let go, like a pendulum. if the cab

le is 45.5 m long, what will the period of oscillation be?(unit=s)

1 answer:

stira [4]4 years ago

3 0

Answer:

The period is $T = 13.53 \ sec$

Explanation:

From the question we are told that

The length of the cable is $L = 45.5 \ m$

Let take acceleration due to gravity as $g = 9.8 \ m/s^2$

Now the period of the oscillation is mathematically represented as

$T = 2 \pi \sqrt{\frac{L}{g} }$

substituting values

$T = 2 * 3.142 \sqrt{\frac{45.5}{9.8} }$

$T = 13.53 \ sec$

You might be interested in

Give two examples of contact force and non contact force

Crank

Answer:

Explanation:

The two examples of contact forces are frictional force and applied force. The two examples of noncontact forces are gravitational force and magnetic force.

5 0

2 years ago

A beam of protons moves in a circle of radius 0.20 m. The protons move perpendicular to a 0.36-T magnetic field. (a) What is the

gregori [183]

Answer:

a) $v=6.898\times 10^{6}\ m.s^{-1}$ is the speed of each proton

b) $F_c=3.97\times 10^{-13}\ N$

Explanation:

Given:

radius of path of motion, $r=0.2\ m$

we know charge on protons, $q=1.6\times 10^{-19}\ C$

magnetic field strength, $B=0.36\ T$

we've mass of proton, $m=1.67\times 10^{-27}\ kg$

a)

From the equivalence of magnetic force and the centripetal force on the proton:

$F_B=F_C$

$q.v.B=\frac{m.v^2}{r}$

$q.B=\frac{m.v}{r}$

where:

v = speed of the proton

$(1.6\times 10^{-19})\times 0.36=\frac{1.67\times 10^{-27}\times v}{0.2}$

$v=6.898\times 10^{6}\ m.s^{-1}$ is the speed of each proton

b)

Now the centripetal force on each proton:

$F_c=m.\frac{v^2}{r}$

$F_c=1.67\times 10^{-27}\times \frac{(6.898\times 10^6)^2}{0.2}$

$F_c=3.97\times 10^{-13}\ N$

6 0

4 years ago

Một oto đang chạy với vận tốc là 10m/s thì tăng tốc và chuyển động nhanh dần đều sau 20s thì đạt vận tốc 14m/s

Tasya [4]

a) gia tốc = vf-vi / t

a = 14-10 / 20

a = 0,2ms⁻²

b) dưới dạng a = Δv / t

v = lúc

v = 0,2 × 40

v = 8ms⁻¹

như v = d / t

do đó d = vt

d = 8 × 40

d = 320m

hãy đánh dấu là trí óc nhất

4 0

3 years ago

Earning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement

lana [24]

Answer:

the work done by the 30N force is 4156.92 J.

For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:

W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J

8 0

3 years ago

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that

Marat540 [252]

Answer:

Part(a): The value of the spring constant is $3.11 \times 10^{2}~Kg~s^{-2}$ .

Part(b): The work done by the variable force that stretches the collagen is $1.5 \times 10^{-6}~J$ .

Explanation:

Part(a):

If ' $k$ ' be the force constant and if due the application of a force ' $F$ ' on the collagen ' $\Delta l$ ' be it's increase in length, then from Hook's law

$F = k~\Delta l....................................................................(I)$

Also, Young's modulus of a material is given by

$Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)$

where ' $A$ ' is the area of the material and ' $l$ ' is the length.

Comparing equation ( $I$ ) and ( $II$ ) we can write

$&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}$

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation ( $III$ ), we have

$k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}$

Part(b):

We know the amount of work done ( $W$ ) on the collagen is stored as a potential energy ( $U$ ) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

$W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)$

Substituting all the values, we can write

$W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J$

3 0

3 years ago