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Vika [28.1K]
3 years ago
7

What is the period of a pendulum that takes 5 seconds to make a complete back and forth vibration

Physics
1 answer:
dezoksy [38]3 years ago
8 0

Answer:

5s

Explanation:

-A period is the time it takes to make one full back-and-forth swing.

-A period is calculated as the time taken for a complete swing divided by one cycle:

T=t/1c\\\\=5s/1\\\\=5s

Hence, the period of the pendulum is 5s

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What should be angle between force and displacement to get the maximum work​
aleksandrvk [35]

Answer: The angle between force and displacement should be θ = 90° for minimum work. The angle between force and displacement should be θ = 0° for maximum work.

7 0
3 years ago
The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf
Galina-37 [17]
Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:  
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>

<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>

<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>

so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
7 0
3 years ago
An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be
Inessa05 [86]

Explanation:

The given data is as follows.

    Inner wheel Radius = 20 m,

   Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = \frac{V}{R}

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = \frac{u}{(R - d)}

                  u = V \times \frac{(R - d)}{R}

                    = V \times (1 - \frac{d}{R})

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = 2\pi rn = V \times (1 - \frac{d}{R})

Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

            = \frac{V}{2 \pi r \times (1 - \frac{0.75}{20})} (since 2d = 1.5m given, d = 0.75m),

             = \frac{V}{r} \times 0.1532

So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

                   = \frac{V}{r} \times 0.1651

5 0
3 years ago
Technician A says that a power-assisted brake system reduces stopping distances compared to a nonpower-assisted brake system. Te
yanalaym [24]

Answer:

Technician B

Explanation:

here on analyzing both the statements from technician A and technician B. The Statement from Technician B is more logical and correct. That the power-assisted brake system reduces the force that the driver must exert on the brake pedal.

The power-assisted brake system does not reduce the distance of stopping. What it does is it reduces the force to be applied by the driver. Thus, making the drive more comfortable.

4 0
3 years ago
A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
alex41 [277]

Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1

(b) The velocity of the car before the driver begins braking is

v_1 = 1.76*20 = 35.2m/s

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

We can use the following equation of motion to calculate how far the car has travel since braking to stop

s_2 = v_1t_2 + a_2t_2^2/2

s_2 = 35.2*5 - 7.04*5^2/2 = 88 m

Also the distance from start to where the driver starts braking is

s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0
3 years ago
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