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3 years ago

You drop a stone from rest into a well that is 5.2m deep. How long does it take before you hear the splash?

Physics

1 answer:

Leto [7]3 years ago

7 0

Answer:

So after 1.0301 sec we hear the splash

Explanation:

We have given that height of the well h = 5.2 m

As the stone is dropped from rest so initial velocity u = 0 m/sec

Acceleration due to gravity $g=9.8m/sec^2$

We have to find the time

From second equation of motion we know that

$h=ut+\frac{1}{2}gt^2$

$5.2=0\times t+\frac{1}{2}\times 9.8\times t^2$

$t^2=1.061$

t = 1.0301 sec

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What causes the mechanical advantage of the inclined plane to be less than that of the lever? 30 points !!!

gayaneshka [121]

Moving an object up an inclined plane requires lessforce than lifting it straight up, at a cost of an increase in the distance moved. The mechanical advantageof an inclined plane, the factor by which the force is reduced, is equal to the ratio of the length of the sloped surface to the height it spans.

8 0

4 years ago

A 1.0-in.-diameter hole is drilled on the centerline of a long, flat steel bar that is 1 2 thick and 4 in. wide. The bar is subj

Dominik [7]

Answer:

The answers are

The average stress = 20000 lb/in²

The maximum tensile stress immediately adjacent to the hole

= 31076.92 lb/in²

Explanation:

To solve the question we have

Weight of tensile load = 30,000 lb

Width of steel bar = 4 in

Thickness of steel bar = 1/2 in

Average Stress = Force/Area

Size of hole drilled = 1.0 in diameter

Available width at cross section where the 1.00 in diameter hole is drilled =

(4 - 1) in = 3 inches

Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)

= 3 in × 1/2 in = 1.5 in²

Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²

the maximum tensile stress immediately adjacent to the hole.

Bending stress = $\sigma_B= \frac{M_y}{I}$ where $I = \frac{(0.5^2 + 4^2)}{12}$

0.5*30000/I = 11076.92 lb/in²

Max stress = 31076.92 lb/in²

8 0

3 years ago

Read 2 more answers

A loudspeaker, mounted on a tall pole, is engineered to emit 75% of its sound energy into the forward hemisphere, 25% toward the

Naily [24]

Answer:

"0.049 W" is the correct answer.

Explanation:

According to the given question,

$r = \sqrt{(3.5)^2+(2.5)^2}$

$=\sqrt{8.5}$

$SL=85$

As we know,

⇒ $SL=10 \ log(\frac{I}{I_o} )$

$85=10 \ log(\frac{I}{10^{-12}} )$

$I=3.162\times 1^{-4} \ W/m^2$

Now,

⇒ $P_{front} = I(2\pi r^2)$

$=(3.162\times 10^{-4})(2\pi\times 18.5)$

$=0.0368 \ W$

$=0.75 \ P$

or,

$=0.049 \ W$

7 0

3 years ago

Find the acceleration for a force of 2 N acting on 2 kg.

igomit [66]

Answer:

The acceleration of a 2 kg box acted on by a net force of 2 N is solved by using Newton’s second law of motion in the form acceleration = F_net / mass. Since the mass of the box is 2 kg and the net force is 2 N, the acceleration of the 2 kg box is 1 m/s^2.

4 0

3 years ago

Consider a lawnmower of mass m which can slide across a horizontal surface with a coefficient of friction μ. In this problem the

inna [77]

Answer:

Fh = u*m*g / (cos(θ) - u*sin(θ))

Explanation:

Given:

- The mass of lawnmower = m

- The angle the handle makes with the horizontal = θ

- The force applied along the handle = Fh

- The coefficient of friction of the lawnmower with ground = u

Find:

Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Solution:

- Construct a Free Body Diagram (FBD) for the lawnmower.

- Realize that there is horizontal force applied parallel to ground due to Fh that drives the lawnmower and a friction force that opposes this motion. We will use to Newton's law of motion to express these two forces in x-direction as follows:

F_net,x = m*a

- Since, the lawnmower is to move with constant speed then we have a = 0.

F_net,x = 0

- The forces as follows:

Fh*cos(θ) - Ff = 0

Where, Ff is the frictional force:

Fh = Ff /cos(θ)

Similarly, for vertical direction y the forces are in equilibrium. Using equilibrium equation in y direction we have:

- W - Fh*sin(θ) + Fn = 0

Where, W is the weight of the lawnmower and Fn is the contact force exerted by the ground on the lawnmower. Then we have:

Fn = W + Fh*sin(θ)

Fn = m*g + Fh*sin(θ)

The Frictional force Ff is proportional to the contact force Fn by:

Ff = u*Fn

Ff = u*(m*g + Fh*sin(θ))

Substitute this expression in the form derived for Fh and Ff:

Fh*cos(θ) = u*(m*g + Fh*sin(θ))

Fh*(cos(θ) - u*sin(θ)) = u*m*g

Fh = u*m*g / (cos(θ) - u*sin(θ))

5 0

3 years ago