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2 years ago

Independent practice

Physics

1 answer:

coldgirl [10]2 years ago

4 0

Explanation:

formula: Mass

Density x volume

2a) m=10kg v=0.3m³

10÷0.3=33.3 kg/m

2b) m = 160 kg V=0.1m³

160÷0.1=1600 kg/m

2c) m = 220 kg V = 0.02m³

220÷0.02=11000 kg/m

A wooden post has a volume of 0.025m³ and a mass of 20kg. Calculate its density in kg/m.

density = volume ÷ mass

20÷ 0.025=800 kg/m

Challenge: A rectangular concrete slab is 0.80m long, 0.60 m wide and 0.04m thick. Calculate its volume in m³.

Formula : Length x width x height = Volume

0.80 x 0.60 x 0.04 = 0.0192m³

B) The mass of the concrete slab is 180 kg. Calculate its density in kg/m.

density = volume ÷ mass

180 ÷ 0.0192 = 9375 kg/m

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Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

The lightbulb is an example of

AlexFokin [52]

A object that has been reinvented so it is more energy efficient

4 0

3 years ago

A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

amm1812

Answer:

The velocity of the tennis racket after the collision 14.966 m/s.

Step-by-step explanation:

let the following:

m₁ = mass of tennis racket = 0.311 kg

m₂ = mass of the ball = 0.057 kg

u₁ = velocity of tennis racket before collision = 30.3 m/s

u₂ = velocity of the ball before collision = -19.2 m/s

v₁ = velocity of tennis racket after collision

v₂ = velocity of the ball after collision

Right (+) , Left (-)

An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same.

So, the total kinetic energy before collision = the total kinetic energy after collision.

So, 0.5 m₁ u₁² + 0.5 m₂ u₂² = 0.5 m₁ v₁² + 0.5 m₂ v₂² ⇒ (1)

Also, the total momentum before collision = the total momentum after collision.

So, m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂ ⇒ (2)

Solving (1) and (2):

∴ v₁ = [ u₁ * (m₁ - m₂) + u₂ * 2m₂ ]/ (m₁ + m₂)

= ( 30.3 * (0.311 - 0.057) - 19.2 * 2 * 0.057 ) / ( 0.311 + 0.057)

= 14.966 m/s.

So, the velocity of the tennis racket after the collision 14.966 m/s.

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3 years ago

Use the slit example to explain why you can hear a noise in another room through an open door.

defon

Because the wall reflects sound waves to your ears bouncing off of the walls, even if it's in another room.

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3 years ago

What is the theory behind a ticker tape lab?

Lana71 [14]

Purpose: experiments will use it to measure the straight-line accelerated motion of a human hand. The displacement data will be measured and velocity and acceleration will be calculated, run the ticker tape under the guides on the timer and under the carbon circle.Hope this helps! ; )

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3 years ago