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2 years ago

Two girls stand 500m away from a tall wall

Physics

1 answer:

Mama L [17]2 years ago

4 0

From the calculations, the speed of sound in this case is 16.9 m/s.

The term echo has to do with the reflection of sound waves. Sound is a mechanical wave.

we know that the speed of sound is obtained from;

V = 2x/t

x = distance covered

t = time taken

V = 2(500)/59

v = 16.9 m/s

The error in the experiment could come from;

Lack of precise time measurement
Error can also arise from the environment of the experiment

Learn more about echo:brainly.com/question/9527413

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Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch i

Olegator [25]

Answer:

14.1 m/s

Explanation:

From the question,

μk = a/g...................... Equation 1

Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.

make a the subject of the equation

a = μk(g).................. Equation 2

Given: μk = 0.160, g = 9.8 m/s²

Substitute into equation 2

a = 0.16(9.8)

a = 1.568 m/s²

Using,

F = ma

Where F = force, m = mass.

Make m the subject of the equation

m = F/a................... Equation 3

m = 160/1.568

m = 102.04 kg.

Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.

Assuming the initial velocity of the skier to be zero

Fd+mgμ = 1/2mv²........................Equation 4

Where v = speed of the skier after crossing the patch, d = distance/width of the patch.

v = √2(Fd+mgμ)/m)................ Equation 5

Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16

Substitute these values into equation 5

v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]

v = √197.57

v = 14.1 m/s

v = 9.86 m/s

4 0

3 years ago

A thin rod (length = 2.97 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.

nlexa [21]

Answer:

a) w = 2.57 rad / s , b) α = 3.3 rad / s²

Explanation:

a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground

Initial. Higher

Em₀ = U = m g h

Final. Touching the ground

$Em_{f}$ = K = ½ I w²

How energy is conserved

Em₀ = $Em_{f}$

mg h = ½ I w2

The moment of specific object inertia

I = m L²

We replace

m g h = ½ (mL²) w²

w² = 2g h / L²

The height of the object is the length of the bar

h = L

w = √ 2g / L

w = √ (2 9.8 / 2.97)

w = 2.57 rad / s

b) the angular acceleration can be found from Newton's second rotational law

τ = I α

W L = I α

mg L = (m L²) α

α = g / L

α = 9.8 / 2.97

α = 3.3 rad / s²

3 0

3 years ago

Who developed radiometric dating? What age did he assign to the oldest rocks?

Mekhanik [1.2K]

Ernest Rutherford
don't know the age sorry

5 0

3 years ago

A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a d

pentagon [3]

Answer:

The block would have slid <u>12.6 m</u> if its initial velocity were increased by a factor of 2.8.

Explanation:

Given:

Mass of the block (m) = 3.6 kg

Initial velocity (u) = 1.7 m/s

Final velocity (v) = 0 m/s

Displacement (S) = 1.6 m

First we will find the acceleration of the block.

Using the equation of motion, we have:

$v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}$

Now, plug in the given values and solve for 'a'. This gives,

$a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2$

The acceleration is negative as it is resisting the motion.

Now, the initial velocity is increased by a factor of 2.8. So,

New initial velocity = 2.8 × 1.7 = 4.76 m/s

Again using the same equation of motion and expressing the result in terms of 'S'. This gives,

$v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}$

Now, plug in the given values and solve for 'S'. This gives,

$S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m$

Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8

6 0

3 years ago

A 500g cart moving at 0.25 m/s collides and sticks to a stationary 750g cart. How fast do the two carts

Tresset [83]

Answer:

0.1 m/s

Explanation:

Please see attached photo for explanation.

Mass of 1st cart (m₁) = 500 g

Initial velocity of 1st cart (u₁) = 0.25 m/s

Mass of 2nd cart (m₂) = 750 g

Initial velocity of 2nd cart (u₂) = 0 m/s

Velocity (v) after collision =.?

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(500 × 0.25) + (750 × 0) = v(500 + 750)

125 + 0 = v(1250)

125 = 1250v

Divide both side by 1250

v = 125 / 1250

v = 0.1 m/s

Thus, the two cart will move with a velocity of 0.1 m/s after collision.

3 0

3 years ago