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2 years ago

I need help please I’ll give BRAINLIST for correct answers!!

Physics

1 answer:

Juli2301 [7.4K]2 years ago

4 0

Answer:

1. the graph is showing the speed of the object at different times.

Explanation:

2. the object is slowing down

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g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide

vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring $\left ( m_1\right )=0.5 kg$

Compression in the spring $\left ( x\right )=20 cm$

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block $\left ( m_1\right )$

$\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2$

$v=x\sqrt{\frac{k}{m_1}}$

now conserving momentum

$m_1v=\left ( m_1+m_2\right )v_0$

$v_0=\frac{m_1}{m_1+m_2}v$

where $v_0$ is the final velocity

3 0

3 years ago

A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A (N/m²)

P1=P2

$\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }$

$F_{2} = \frac{F_{1}*A_{2} }{A_{1}}$ Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight (W₂) can the large piston support

We replace data in the equation (1)

$F_{2} = \frac{(500)*(40) }{2}$

F₂ = 10000 N

W₂= F₂= 10000 N

6 0

3 years ago

A 25.0-kg child is standing at the edge of a horizontal merry-go-round with a radius of 2.40 m and a moment of inertia of 356 kg

motikmotik

Answer:

$\omega_{f}=1.634\ rad/s$

Explanation:

given,

diameter of merry - go - round = 2.40 m

moment of inertia = I = 356 kg∙m²

speed of the merry- go-round = 1.80 rad/s

mass of child = 25 kg

initial angular momentum of the system

$L_i = I\omega_i$

$L_i =356\times 1.80$

$L_i =640.8\ kg.m^2/s$

final angular momentum of the system

$L_f = (I_{disk}+mR^2)\omega_{f}$

$L_f = (356 + 25\times 1.2^2)\omega_{f}$

$L_f= (392)\omega_{f}$

from conservation of angular momentum

$L_i = L_f$

$640.8= (392)\omega_{f}$

$\omega_{f}=1.634\ rad/s$

8 0

2 years ago

Read 2 more answers

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33Ã—Ã—10âˆ

AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, $F=3.33\times 10^{-21}\ N$

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

$F=k\dfrac{q^2}{r^2}$

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}$

$q=1.21\times 10^{-16}\ C$

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

$n=\dfrac{q}{e}$

$n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}$

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0

3 years ago

During the flight of the ball, what is the direction of its Acceleration due to gravity

vekshin1

Regardless of what direction an object is moving, the acceleration
due to gravity is always directed toward the center of the Earth.
That's the direction commonly known as "down".

4 0

3 years ago