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trapecia [35]
2 years ago
9

Geoff counts the number of oscillations of a simple pendulum at a location where the acceleration due to gravity is 9.80 m/s2, a

nd finds that it takes 25.0 s for 15 complete cycles. 1) calculate the length of the pendulum. (express your answer to three significant figures.
Physics
1 answer:
loris [4]2 years ago
7 0
Period of a simple pendulum = 2π √(L/G)

(25 sec/15) = 2π √(L / 9.8 m/s²)

5/3 sec  = 2π √(L/9.8 m/s²)

5 sec / 6π = √ (L/9.8 m/s²)

(5sec · √9.8m/s²) / 6π = √L

Square each side:

(25 s²) · (9.8 m/s²) / 36π²  =  L

L =  (25 · 9.8) / (36 π²) meters

L = 0.69 meter 
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An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele
leva [86]

To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.

Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.

The average angular acceleration

\alpha = \frac{\omega_f - \omega_0}{t}

Here

\alpha = Angular acceleration

\omega_{f,i} = Initial and final angular velocity

There is not initial angular velocity,then

\alpha = \frac{\omega_f}{t}

We know that the relation between the tangential velocity with the angular velocity is given by,

v = r\omega

Here,

r = Radius

\omega = Angular velocity,

Rearranging to find the angular velocity

\omega = \frac{v}{r}}

\omega = \frac{30}{0.20} \rightarrow Remember that the radius is half te diameter.

Now replacing this expression at the first equation we have,

\alpha = \frac{30}{0.20*6}

\alpha = 25 rad /s^2

Therefore teh average angular acceleration of each wheel is 25rad/s^2

3 0
2 years ago
A test car carrying a crash test dummy accelerates from 0 to 30 m/s and then crashes into a brick wall. Describe the direction o
earnstyle [38]

Answer:

the direction of acceleration of the vehicle is the same direction of its velocity of car

s acceleration has the opposite direction to the car speed.

Explanation:

The initial acceleration of the car can be calculated with

          v = v₀ + a t

          a = (v-v₀) t

       

indicate that the initial velocity is zero (v₀ = 0 m / s)

         a = v / t

         a = 300 / t

the direction of acceleration of the vehicle is the same direction of its acceleration movement.

When the car collides with the wall, it exerts a force in the opposite direction that stops the vehicle, therefore this acceleration has the opposite direction to the car speed. But your module must be much larger since the distance traveled to stop is small

5 0
2 years ago
A carousel that is 5.00 m in radius has a pair of 600-Hz sirens mounted on posts at opposite ends of a diameter. The carousel ro
Gelneren [198K]

Answer:

59cm

Explanation:

angular velocity = 0.8 rad/s

linear velocity = angular velocity * radius

                        =0.8rad/s * 5m

                        = 4 m/s

wavelength = (V + U)/F

where,

V is the velocity of the wave

U is the velocity of the source

F is the frequency of the source.

wavelength = (350 m/s + 4 m/s ) / 600 Hz

Wavelength = 0.59m or 59 cm

4 0
2 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
1 year ago
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.55 x 10-4 T) at a distance of 25
antiseptic1488 [7]

we are given in the problem the following dimensions or specifications 
B = 0.000055 T r = 0.25 m constant mu0 = 4*pi*10-7 

The formula that is applicable from physics is 
B = mu0*I/(2*pi*r) I = 2*B*pi*r/mu0 I = 68.75 Amperes 
7 0
1 year ago
Read 2 more answers
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