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2 years ago

15

a powerboat accelerates along a straight path from 0 km/hr to 99.8 km/hr in 10.0 s.Find the average acceleration of the boat in

meters per second squared.

1 answer:

Alex777 [14]2 years ago

3 0

(27.72-0)/10
= 2.772 m/s2

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Waves hitting at an angle and then bending around features of the coast is known as

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When waves hitting a specific angle, some part of the waves will be closer to the shallow part of the water and some part will be closer to the deeper part of the water, which makes the wave became somehow bent around the shore.</span>

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You must determine the length of a long, thin wire that is suspended from the ceiling in the atrium of a tall building. A 2.00-c

AleksAgata [21]

Answer:

Explanation:

Let L be the length of the wire.

velocity of pulse wave v = L / 24.7 x 10⁻³ = 40.48 L m /s

mass per unit length of the wire m = 14.5 x 10⁻⁶ x 10⁻³ / 2 x 10⁻² kg / m

m = 7.25 x 10⁻⁷ kg / m

Tension in the wire = Mg , M is mass hanged from lower end.

= .4 x 9.8

= 3.92 N

expression for velocity of wave in the wire

$v = \sqrt{\frac{T}{m} }$ , T is tension in the wire , m is mass per unit length of wire .

40.48 L = $\sqrt{\frac{3.92}{7.25\times10^{-7}} }$

1638.63 L² = 3.92 / (7.25 x 10⁻⁷)

L² = 3.92 x 10⁷ / (7.25 x 1638.63 )

L² = 3299.64

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2 years ago

3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charg

Anni [7]

Answer:

$\mathbf{1.51\times10^{-15}N}$

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

$q_{paper}=-4.1\times10^{-15}C$

Now the distance from the charge is

$r=1cm=0.01m$

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

$mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}$

$\Rightarrow W=mg$

$=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}$

$=\mathbf{1.51\times10^{-15}N}$

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2 years ago

In the produce section of a supermarket, five pears are placed on a spring scale. The placement of the pears stretches the sprin

tankabanditka [31]

Answer:

The displacement of the spring due to weight is 0.043 m

Explanation:

Given :

Mass $m = 2$ Kg

Spring constant $k = 450 \frac{N}{m}$

According to the hooke's law,

$F = -kx$

Where $F =$ force, $x =$ displacement

Here,

$F = mg$ ( $g = 9.8 \frac{m}{s^{2} }$ )

$F = 2 \times 9.8 = 19.6$ N

Now for finding displacement,

$x = \frac{F}{k}$

Here minus sign only represent the direction so we take magnitude of it.

$x = \frac{19.6}{450}$

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Therefore, the displacement of the spring due to weight is 0.043 m

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inysia [295]

Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

τ = I α

T r = I α

T gR = I α

Alf = T R / I (1)

T = α I / R

Now let's use Newton's second law in the mass that descends

W- T = m a

a = (m g -T) / m

The two accelerations need related

a = R α

α = a / R

a = (m g - α I / R) / m

R α = g - α I /m R

α (R + I / mR) = g

α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

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2 years ago