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ra1l [238]
3 years ago
9

In order to find density of an object, maria is trying is trying to measure its volume however the object doesn't not fit the to

ol she is using to solve this problem maria decides filling to break apart the object. will maria be able to find density following this method
Physics
2 answers:
MA_775_DIABLO [31]3 years ago
8 0

Answer:

Yes

Explanation:

The density of the object does not change when it is broken in more pieces, therefore Maria will be able to find the density of the material.

In fact, by breaking the object, then she can take only one piece, and by measuring the mass of this piece and its volume she will be able to calculate the density of the material by using the formula:

d=\frac{m}{V}

where

m is the mass of the piece

V is its volume

d is the density

And the density will be the same for every piece, since they are made by the same material, and the density depends only on the properties of the material itself.

aivan3 [116]3 years ago
6 0
Provided she breaks it apart cleanly (that is, no crumbles or pieces of the object lost), and adds up the volume of the objects that she broke apart, then the number should be no different than if she were to weigh the object whole.
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A solenoid with 435 turns has a length of 7.50 cm and a cross-sectional area of 3.50 ✕ 10−9 m2. Find the solenoid's inductance a
OLga [1]

Answer:

Solenoid's inductance is 1.11 × 10^-8H

The average emf around the solenoid is 1.3 × 10^-5V

Explanation: Please see the attachments below

4 0
3 years ago
A furlong is an old british unit of length equal to 0.125 mi, derived from the length of a furrow in an acre of ploughed land. a
castortr0y [4]

The speed of light is: c = 3x10^8 m/s <span>

or 

c = 186,000,000 miles/sec = 1.86x10^8 mi/s 

1 furlong = 0.125 mile 

1 fortnight = 2 weeks(7d/wk)(24h/d)(3600s/h) 
= 1209600s = 1.2096x10^6 s 

Therefore, 

c =1.86x10^8 mi/s(1furl/0.125mi)(1.2096x10^6s/fort) 

<span>c = 18x10^14 furlong/fortnight = 18x10^8 Mfurlong/fortnight</span></span>

4 0
3 years ago
An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an
kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

v_{f1} = \sqrt{2*a_{1}*d_{1}  }  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

v_{f1} = \sqrt{2*3.2*3}  }

v_{f1} =\sqrt{2*3.2*3}

v_{f1} = 4.38 m/s

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0
3 years ago
a 360 mile trip began on a greeway in a car traveling at 62 mph. Once the road became a 2 lane highway, the car slowed to 54 mph
Daniel [21]

Answer:

Time spent on the greenway road  = 4.5 hours

Time spent on the 2 lane road = 1.5 hours

Explanation:

The distance of the trip  is 360 miles and the initial speed of the car is 62 miles/hr and after the road became 2 lane highway the car slowed to 54 miles/hr.

Let us divide the trip into two

Greenway

speed = distance/time

speed = 62 mph

time = a

distance = speed × time

distance = 62a

2 lane highway

speed = distance/time

speed = 54 mph

time = b

distance = speed × time

distance = 54b

Total distance

62a + 54b = 360......................(i)

Total time

a + b = 6..............................(ii)

a = 6-b

insert a in equation (i)

62(6-b) + 54b = 360

372  - 62b + 54b = 360

-8b = 360-372

-8b = - 12

b = 12/8

b = 1.5

from equation (ii)

a + 1.5 = 6

a = 6 - 1.5

a = 4.5

8 0
2 years ago
A student compared some soccer players to the atoms in the gaseous state. Which of the following activities were the soccer play
tensa zangetsu [6.8K]

Answer:

The answer will be <em>D</em>

Explanation:

I choose answer <em>C</em> and got it wrong on the test. Also Liquid atoms vibrate fast and slide past each other. what answer <em>C </em>shows is a gas.

8 0
3 years ago
Read 2 more answers
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