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Alisiya [41]
3 years ago
13

A ball of mass 0.5 kg is released from rest at a height of 30 m. How fast is it going when it hits the ground? Acceleration due

to gravity is g = 9.8 m/s²

Physics
1 answer:
MArishka [77]3 years ago
4 0
The answer is attached.

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A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr
nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

 a= μ g                          ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

\omega=\sqrt{\dfrac{k}{M+m}}

By putting the values

\omega=\sqrt{\dfrac{150}{5.67+1}}

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2

kX^2=(m+M)v^2

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0
3 years ago
Read 2 more answers
A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

\mathrm{m}_{1}=600 \mathrm{kg}

\mathrm{m}_{2}=900 \mathrm{kg}

\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

600 \times 5=900 \times v_{2}

3000=900 \times v_{2}

\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

5 0
3 years ago
When a hot and cold object are placed in contact, the hot one loses energy. Does this violate energy conservation? Why or why no
blsea [12.9K]

Answer:

This does not violate the conservation of energy.

Explanation:

This does not violate the conservation of energy because the hot body gives energy in the form of heat to the colder body, this second absorbs energy. This will be the case until both bodies reach the same temperature, reaching thermal equilibrium and reducing the transfer of thermal energy. In this way the energy was only transferred from one body to another but the total energy of the system (body 1 plus body 2) will be the same as in the beginning, respecting the principle of conservation of energy or also called the first principle of thermodynamics .

The part of physics that studies these processes is in turn called heat transfer or heat transfer or thermal transfer. Heat transfer occurs whenever there is a thermal gradient or when two systems with different temperatures come into contact. The process persists until thermal equilibrium is reached, that is, until temperatures are equalized. When there is a temperature difference between two objects or regions close enough, the heat transfer cannot be stopped, it can only be slowed down.

8 0
3 years ago
While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 5.65 m/s. The stone subs
xxTIMURxx [149]

Answer:

1. 20.54m/s

2. 1.52s

Explanation:

QUESTION 1:

The speed the stone impact the ground is the final speed/velocity, which can be calculated using the formula:

v² = u² + 2as

Where;

v = final velocity (m/s)

u = initial velocity (m/s)

a = acceleration due to gravity (m/s²)

s = distance (m)

From the provided information, u = 5.65m/s, v = ?, s = 19.9m, a = 9.8m/s²

v² = 5.65² + 2 (9.8 × 19.9)

v² = 31.9225 + 2 (195.02)

v² = 31.9225 + 390.04

v² = 421.9625

v = √421.9625

v = 20.5417

v = 20.54m/s

QUESTION 2:

Using v = u + at

Where v = final velocity (m/s) = 20.54m/s

t = time (s)

u = initial velocity (m/s) = 5.65m/s

a = acceleration due to gravity (m/s²)

v = u + at

20.54 = 5.65 + 9.8t

20.54 - 5.65 = 9.8t

14.89 = 9.8t

t = 14.89/9.8

t = 1.519

t = 1.52s

3 0
3 years ago
A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of gla
Diano4ka-milaya [45]

Complete Question

A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted,  

A :

a force repels the glass out of the capacitor.  

B :

a force attracts the glass into the capacitor.    

C :

no force acts on the glass.      

D :

a net charge appears on the glass.      

E :

the glass makes the plates repel each other.

Answer:

The correct option is B

Explanation:

Generally when the glass dielectric is slowly inserted between the plated,

The positive plate of the capacitor will induce a negative charge on the glass while the negative  plate of the capacitor will induce a positive charge on glass which a electric field that posses an electric force that will attract the glass

3 0
3 years ago
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