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Oksi-84 [34.3K]

2 years ago

8

A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po

int she is 0.8 m from the ground. What is her maximum speed?The acceleration of gravity is 9.8 m/s². Answer in units of m/s.

1 answer:

Umnica [9.8K]2 years ago

8 0

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + $0.5mV1^{2}$ = mgh2 + $0.5mV2^{2}$

where

speed at highest point = V2

speed at lowest point = V1

mass of the girl and swing = m

at the highest point, the speed is minimum (V1 = 0)

at the lowest point the speed is maximum (V2 is the maximum speed)

therefore the equation becomes mgh1 + $0.5mV1^{2}$ = mgh2

m(gh1 + $0.5V1^{2}$ ) = m(gh2)

gh1 + $0.5V1^{2}$ = gh2

V1 = $\sqrt{\frac{gh2 - gh1}{0.5}}$

now we can substitute all required values into the equation above.

V1 = $\sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}$

V1 = $\sqrt{\frac{32.34}{0.5}}$

V1 =8.1 m/s

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Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

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2 years ago

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

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2 years ago

would an elephant standing on one leg exert a higher force on a scale than an elephant on four legs. why

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Answer:

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7 0

2 years ago

A particle of charge 2.3 ✕ 10−8 C experiences an upward force of magnitude 4.6 ✕ 10−6 N when it is placed in a particular point

Marysya12 [62]

<h2>Answer:</h2>

(a) +2 x 10² N/C (upwards)

(b) -2.2μN or -2.2 x 10⁻⁶N (downwards)

<h2>Explanation:</h2>

The force (F) acting on a particle of charge (Q) at a particular point is related to its electric field (E) by the following;

F = Q x E ----------------------(i)

This means that the force acting on the charged particle is the product of its charge and the electric field at that point.

<em>(a) Given</em>;

Q = charge of the particle = 2.3 x 10⁻⁸ C

F = force acting on the particle = 4.6 x 10⁻⁶N

<em>Substitute these values into equation (i) as follows;</em>

=> 4.6 x 10⁻⁶ = 2.3 x 10⁻⁸ x E

=> E = 4.6 x 10⁻⁶ ÷ 2.3 x 10⁻⁸

=> E = 2 x 10² N/C

Since the value is positive, the electric field is directed upwards.

Therefore, the electric field at that point is +2 x 10² N/C

<em>(b) If a charge of q = -1.1 x 10⁻⁸ is placed there, the force (F) acting is calculated as follows;</em>

<em>Substitute Q = q into equation (i) as follows;</em>

F = q x E

<em>Substitute the value of q and E = 2 x 10² N/C into the equation above as follows;</em>

F = -1.1 x 10⁻⁸ x 2 x 10²

F = -2.2 x 10⁻⁶ N

F = -2.2μN

Since the value of the force, F, is negative, it means it is directed downwards.

Therefore the force on the charge is -2.2μN

3 0

2 years ago

1. Each year at a college, there is a tradition of having a hoop rolling competition. Alex rolls his 0.350 kg hoop down the cour

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Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197 $kg \cdot cm^2$

Explanation:

<u>Given</u>:

Mass of the hoop = 0.350 g

Radius of the hoop = 75.0 cm

<u>To Find:</u>

The moment of inertia of Alex's rolling hoop = ?

<u>Solution</u><u>:</u>

The moment of inertia = $mr^2$

where

m is the mass

r is the radius

Converting cm to m, we get

75.0 cm = 0.75 m

Now substituting the values,

=> moment of inertia = $(0.350)(0.75)^2$

=> moment of inertia = $(0.350)(0.5625)$

=> moment of inertia = $(0.197)$

Question 2:

Answer:

The combined angular momentum of the masses is 1.76 $kg m^2 s^{-1}$

If she pulls her arms in to 0.12 m, her new linear speed is $18.33 m/s^2$

Explanation:

Given:

Mass = 2.0 kg

Radius = 0.8 m

Velocity = 1.2 m/s

a.The combined angular momentum of the masses:

$L = r \cdot m \cdot v_1$

Substituting the values,

$L = 0.8 \cdot 2.0 \cdot 1.1$

L= 1.76 $kg m^2 s^{-1}$

b. If she pulls her arms in to 0.12 m, what is her new linear speed

$0.12 \cdot 0.8 \cdot v_2 = 1.76$

$0.096 cdot v_2 = 1.76$

$v_2 = \frac{1.76}{0.096}$

$v_2 = 18.33 m/s^2$

6 0

1 year ago