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motikmotik
3 years ago
14

Imagine a rock is dropped from the top of a tall building. After 2 seconds of falling, the rock’s instantaneous speed is approxi

mately 20m/s. does anyone know is it true or false
Physics
2 answers:
pshichka [43]3 years ago
6 0

Answer:

TRUE

Explanation:

As we know that stone is falling freely from rest

so here we can use kinematics to find the final speed of the stone

v_f = v_i + at

here we know that

v_i = 0

a = 9.81 m/s^2

t = 2 s

now we have

v_f = 0 + (9.81)(2)

v_f = 19.62 m/s

so final speed of the rock is nearly 20 m/s

shepuryov [24]3 years ago
3 0
True
It's in free fall( 9.8 m/s), 2 seconds have passed.
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A 10.00 kg mass is moving to the right with a velocity of 14.0 m/s. A 12.0 kg mass is moving to the left with a velocity of 8.00
Basile [38]

Answer:

2 m/s

Explanation:

From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.

Let the 10.00 kg mass be m_1 and the 12.0 kg mass be m_2. When they collide and stick, they have a combined mass of m_1+m_2.

Momentum is given by p=mv. Set up the following equation:

\displaystyle m_1v_1+m_2v_2=(m_1+m_2)v_f, where v_f is the desired final velocity of the masses.

Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.

Solving yields:

10\cdot 14 + 12\cdot (-8)=(10+12)v_f\\\implies v_f=\boxed{2 \text{ m/s}}

4 0
2 years ago
A square plate of copper with 47.0 cm sides has no net charge and is placed ina region of uniform electric field of 75.0 kN/C di
timurjin [86]

Answer:

(a) Charge density σ=6.6375×10²nC/m²

(b) Total charge Q=1.47×10²nC

Explanation:

Given Data

A=47.0 cm =0.47 m

Electric field E=75.0 kN/C

To find

(a) Charge density σ

(b)Total Charge Q

Solution

For (a) charge density σ

From Gauss Law we know that

Φ=Q/ε₀.......eq(i)

Where

Φ is electric flux

Q is charge

ε₀ is permittivity of space

And from the definition of flux

Φ = EA

The flux is  electric field passing  perpendicularly through the surface

Put the this Φ in equation(i)

EA =Q/ε₀

where Q(charge)=σA

EA=(σA)/ε₀

E=σ/ε₀

σ=ε₀E

=(8.85*10^{-12} )*(75.0*10^{3} )\\=6.6375*10^{-7} C/m^{2}\\=6.63*10^{2}nC/m^{2}

σ=6.6375×10²nC/m²

For (b) total charge Q

Q=σA

Q=(6.6375*10^{2} nC/m^{2} )(0.47m)^{2}\\ Q=1.47**10^{2}nC

6 0
3 years ago
Glycerin at 30°C has a density of 1,260 kg/m3 and a viscosity of 0.630 Pa s. In a laboratory experiment, some glycerin is forced
Alekssandra [29.7K]

Answer:

Explanation:

Rate of flow of liquid through a tube can be expressed by the following expression

V = π P r⁴ / 8ηl

P is pressure difference between end of tube = 618 Pa

r , radius of tube = .5 x 10⁻²

η is viscosity of liquid flowing = .63  

l is length of tube = .10 m

V = 3.14 x 618 x (  .5 x 10⁻² )⁴ / (8 x .63 x .10 )

= 240.64 x 10⁻⁸ m³ /s

mass = 240.64 x 1260 x 10⁻⁸ kg / s

= 3.03 x 10⁻³ kg /s

= 3.03 gram /s .

5 0
3 years ago
Physics. 40 points.<br><br><br>Help please :) (View attached image)
laiz [17]

Answer:

1) 10 m/s  2) 14.1 m/s  3) 6.7 m  4) 11.25 m  5) 4.5 m/s  6) 12 joules

Explanation:

7 0
2 years ago
What would the density of an object be with a mass of 0.8g
True [87]

The density of the object is 0.032 g/cm^3.

<h3>What is density?</h3>

The term density means the ratio of the mass to the volume of an object. Now we know that when an object is dense that it would sink to the bottom of a fluid.

Now given that;

Mass = 0.8 g

volume= 25 cm^3

Density = mass/volume

Density =  0.8 g / 25 cm^3

= 0.032 g/cm^3

Learn more about density:brainly.com/question/15164682

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MISSING PARTS

What is the density of an object having a mass of

0.8g and a volume of 25cm^3

4 0
2 years ago
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