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kipiarov [429]
3 years ago
9

How many cubic objects of volume 2cm cube can be started in a room of dimension 2m by 3m by 4m​

Physics
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer:

12,000,000 boxes

Explanation:

the volume of the room can be found by using the equation for volume of a rectangular box:V=LxWxH

where:

L=2m

W=3m

H=4m

(it doesn't really matter which is which since it is multiplication)

when we multiply our values (2m*3m*4m) we get 24cubic meters

now we need to convert cubic meters to cubic centimeters

each cubic meter is 1,000,000 cubic centimeter we multiply 24 by 1,000,000 and we get: 24,000,000 cubic centimeters (cc)

dividing 24,000,000 by 2 (since each box is 2cc) we get 12,000,000

so, we know we can fit 12,000,000, 2 cubic centimeter boxes in this room

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One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t
Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2\frac{m}{s^{2} }

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=FΔx=(0.400m)\\

W_{ext}=20.4 Nm

U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2}  \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}

To determine, the acceleration we solve the following equation for a:

F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }

8 0
4 years ago
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsice force
Dmitriy789 [7]

Answer:

Q = 1.095 x 10^-9 C

Let the force experienced by the top piece of tape be F

F = kQ²/r²

r = distance between the two pieces tape = 1.00cm = 1.00 x 10^ -2 m

1/4(pi)*Eo = k = 8.99 x 10^9 Nm²/C²

The electric force of repulsion between the two charges and the weight of the top piece of tape are equal so

F = kQ²/r² = mg

Where m is the mass of the top piece of tape and g is the acceleration due to gravity

On re-arranging the equation above,

Q² = mgr²/k

Q² = ((11.0 x 10^-6) x 9.8 x (1.00x10^-2)²)/(8.99 x 10^9)

Q = 1.095x10^-9 C

Explanation:

The charge Q on both pieces of tape are equal and both act with a force of repulsion on each other.

The force of repulsion between both tapes pushes the top piece of tape upwards. The weight of the top piece of tape acts vertically downward. Since the top tape is in a position of equilibrium, the two forces acting on the top piece of tape must be equal to each other. This assumption is backed up by newton's first law of motion which states that the summation of all forces acting on a body at rest must be equal to zero. That is

Fe (electric force) - Fg (gravitational force) = 0

Fe = Fg

kQ²/r² = mg

On substituting the respective values for all variables except Q and rearranging the equation Q = 1.09 x 10^-9

6 0
4 years ago
Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th
aev [14]

Answer:

A) G = m³/kg.s²

B) E = kg.m²/s²

Explanation:

A)

The given relation is:

F = Gm₁m₂/r²

where, the units of all variables are:

F = Force = kg.m/s²

m₁ = m₂ = mass = kg

r = distance = m

G = Gravitational Constant = ?

Therefore,

kg.m/s² = G(kg)(kg)/m²

(kg.m/s²)(m²/kg²) = G

<u>G = m³/kg.s²</u>

<u></u>

B)

The given equation is:

E = mc²

where, the units of all variables are:

m = mass = kg

c = speed = m/s

E = Energy = ?

Therefore,

E = (kg)(m/s)²

<u>E = kg.m²/s²</u>

This is the correct answer, which is not present in any option.

4 0
4 years ago
What are the importance of physics
Alexus [3.1K]

Answer:

Physics contributes to the technological infrastructure and provides trained personnel needed to take advantage of scientific advances and discoveries. Physics is an important element in the education of chemists, engineers and computer scientists, as well as practitioners of the other physical and biomedical sciences.

Explanation:

4 0
3 years ago
Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.
Aleonysh [2.5K]

De broglie wavelength, \lambda = \frac{h}{mv}, where h is the Planck's constant,  m is the mass and v is the velocity.

h = 6.63*10^{-34}

Mass of hydrogen atom,  m = 1.67*10^{-27}kg

v = 440 m/s

Substituting

   Wavelength \lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m

1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm

So  the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0
3 years ago
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