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KiRa [710]
3 years ago
6

A poster is 0.95m high and 1.0m wide how many digits follow the decimal point when the perimeter of the poster is expressed with

the correct number of significant figures
A) zero digits
B) one digit
C)two digits
D)three digits
Physics
1 answer:
STALIN [3.7K]3 years ago
6 0

Explanation:

It is given that,

Length of the poster, l = 0.95 m

Breadth of the poster, b = 1 m

The poster is in the shape of rectangle. For a rectangular shape, perimeter is equal to :

P = 2(l+b)

P=2(0.95+1)

P=3.9 m

(b) One digits, P = 4 m

(c) Two digits, P = 3.9 m

(d) Three digits, P = 3.90 m

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The displacement volume of an automobile engine is 167 in3. what is the displacement volume in liters?
forsale [732]

The displacement volume in liters is 2.74 liters.

<h3>What is displacement volume?</h3>

Displacement volume is the quantity of solvent that will be displaced by a specified quantity of a solid during dissolution.

It can also be defined as the volume displaced by the piston as it moves between top dead center and bottom dead center in a car engine.

<h3>Displacement volume in liters</h3>

1 liter = 61.02 in³

? = 167 in³

= 167/61.02

= 2.74 liters

Thus, the displacement volume in liters is 2.74 liters.

Learn more about displacement volume here: brainly.com/question/1945909

#SPJ1

4 0
1 year ago
A mass of 250 N is on a piston of 2.0 m^2. What force is needed to lift this piston if the area of the second piston is 0.5 m^2?
prohojiy [21]

Answer:

<h3>62.5N</h3>

Explanation:

The pressure at one end of the piston is equal to the pressure on the second piston.

Pressure = Force/Area

F1/A1 = F2/A2

Given

F1 = 250N

A1 = 2.0m²

A2 = 0.5m²

F2 = ?

Substituting the given values in the formula;

250/2 = F2/0.5

cross multiply

250*0.5 = 2F2

125 = 2F2

F2 = 125/2

F2 = 62.5N

Hence the  force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N

8 0
3 years ago
car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw
sergejj [24]

Answer:

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = m \times  v_f - m \times  v_0

Where;

v_f = The final velocity

v_0 = The initial velocity

For the the 265 kg mass, we have;

v_f = 0.1627 m/s

v_0 = 1.5 m/s

Which gives the impulse a s F×Δt =  265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

8 0
3 years ago
Calculate the density of a tin of mass 100g whose dimensions are 2cmx5cmx​
Luda [366]

Answer:

your question in not complete.

you need to the high too.

7 0
3 years ago
Homogous chromosomes separate during.....?
stepladder [879]
Not entirely sure if you're saying Homologous , but assuming you do , the homologous chromosomes seperate in the anaphase stage of Mitsosis of the Cell cycle
3 0
3 years ago
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