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3 years ago

Why is the mechanical advantage of a bottle opener always greater than 1?

1 answer:

6 0

For a lever, the mechanical advantage is the ratio of the lever arms. Since the lever arm of the opener is always greater than that of the output arm, the mechanical advantage is greater than 1.

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A plane is traveling with an air velocity of 720 kilometers/hour due east. It experiences a headwind of 16 kilometers/hour. Find

balandron [24]

If it is a headwind it means it's travelling against the motion of the plane. This means it's velocity is simply v=720-16=704 km/h due east.

7 0

3 years ago

Read 2 more answers

Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a

ANTONII [103]

Answer:

Explanation:

a )

Each blade is in the form of rod with axis near one end of the rod

Moment of inertia of one blade

= 1/3 x m l²

where m is mass of the blade

l is length of each blade.

Total moment of moment of 3 blades

= 3 x $\frac{1}{3}$ x m l²

ml²

2 )

Given

m = 5500 kg

l = 45 m

Putting these values we get

moment of inertia of one blade

= 1/3 x 5500 x 45 x 45

= 37.125 x 10⁵ kg.m²

Moment of inertia of 3 blades

= 3 x 37.125 x 10⁵ kg.m²

= 111 .375 x 10⁵ kg.m²

c )

Angular momentum

= I x ω

I is moment of inertia of turbine

ω is angular velocity

ω = 2π f

f is frequency of rotation of blade

d )

I = 111 .375 x 10⁵ kg.m² ( Calculated )

f = 11 rpm ( revolution per minute )

= 11 / 60 revolution per second

ω = 2π f

= 2π x 11 / 60 rad / s

Angular momentum

= I x ω

111 .375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s

= 128.23 x 10⁵ kgm² s⁻¹ .

4 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

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

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

3 years ago

An airplane flies at 40 m/s at an altitude of 50 meters. The pilot drops a heavy package which falls to the ground. Where, appro

igor_vitrenko [27]

Answer:

128 m

Explanation:

From the question given above, the following data were obtained:

Horizontal velocity (u) = 40 m/s

Height (h) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Horizontal distance (s) =?

Next, we shall determine the time taken for the package to get to the ground.

This can be obtained as follow:

Height (h) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

50 = ½ × 9.8 × t²

50 = 4.9 × t²

Divide both side by 4.9

t² = 50 / 4.9

t² = 10.2

Take the square root of both side

t = √10.2

t = 3.2 s

Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:

Horizontal velocity (u) = 40 m/s

Time (t) = 3.2 s

Horizontal distance (s) =?

s = ut

s = 40 × 3.2

s = 128 m

Therefore, the package will land at 128 m relative to the plane

6 0

3 years ago

The speed of a 500 g cricket ball changes from 10m/s to 30m/s in just 7 seconds. What is the force acting on the cart?

9966 [12]

The force acting on the cart is 1.43 N.

<h3>What is force?</h3>

Force can be defined as the product of mass and acceleration.

To calculate the force acting on the cart, we use the formula below.

Formula:

F = m(v-u)/t................. Equation 1

Where:

F = Force acting on the cart
m = mass of the cart
v = Final velocity
u = initial velocity
t = time

From the question,

Given:

m = 500 g = 0.5 kg
v = 30 m/s
u = 10 m/s
t = 7 seconds

Substitute these values into equation 1

F = 0.5(30-10)/7
F = 10/7
F = 1.43 N.

Hence, the force acting on the cart is 1.43 N.

Learn more about force here: brainly.com/question/13370981

7 0

2 years ago