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2 years ago

Please guys help me plz for god sake plz plz guys plz plz plz i will mark u as a brainlist plz

Physics

1 answer:

victus00 [196]2 years ago

8 0

Answer:

Explanation:

The complete sequence of motions and activities required to complete one work cycle.

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Choose the statement(s) that is/are true about the ratio \frac{C_p}{C_v} C p C v for a gas? (Ii) This ratio is the same for all

Blababa [14]

Answer:

(i) false

(ii) true

(iii) true

(iv) false

Explanation:

(i) The ratio of Cp and Cv is not constant for all the gases. It is because the value of cp and Cv is different for monoatomic, diatomic and polyatomic gases.

So, this is false.

(ii) For monoatomic gas

Cp = 5R/2, Cv = 3R/2

So, thier ratio

Cp / Cv = 5 / 3 = 1.67

This statement is true.

(iii) for diatomic gases

Cp = 7R/2, Cv = 5R/2

Cp / Cv = 7 / 5 = 1.4

This statement is true.

(iv) It is false.

6 0

3 years ago

What are 3-D glasses made of?

Firlakuza [10]

Plastic is what they are made of

5 0

3 years ago

Find the current passing through a circuit consisting of a battery and one resistor. The resistance has a retance of 2ohms and t

Bingel [31]

The current passing through a circuit consisting of a battery of 12 V and resistor of 2 ohms is 6 Ampere .

Explanation:

Assume the wires are ideal with zero resistance.

The current passing through the circuit will be

I = V/R = 12/2 = 6.000 A.

5 0

2 years ago

You are riding a bicycle. If you apply a forward force of 150 N, and you and the bicycle have a combined mass of 90 kg, what wil

bekas [8.4K]

<h3>Hello There!!</h3>

<h3><u>Given</u>,</h3>

Force(F) = 150N

Mass(m) = 90kg

Acceleration(a) = ?

F= m×a

$150 = 90 \times \text{a} \\ \\ \text{a} = \frac{150}{90} \\ \fbox{cancelling by 3} \\ \\ \text{ a} = \cancel \frac{150}{90} \\ \\ \text{ a} = \frac{5}{3} = 1.67 \text{m/s} {}^{2}$

$\therefore \text{Option A= 1.67 m/s² is the correct answer}$

<h3>Hope this helps</h3>

6 0

2 years ago

Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is 1s

jok3333 [9.3K]

The initial height of the first body is given by:
$h_1 = \frac{1}{2}gt^2$
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
$h_1 = \frac{1}{2}(9.81 m/s^2)(1 s)^2=4.9 m$

The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
$h_2 = \frac{1}{2}(9.81 m/s^2)(2 s)^2=19.6 m$

The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.

6 0

3 years ago

Please guys help me plz for god sake plz plz guys plz plz plz i will mark u as a brainlist plz​

Please guys help me plz for god sake plz plz guys plz plz plz i will mark u as a brainlist plz