# A10-kg isto a of 10 m at aa0.9 kg/m.30 kg of ,at aas10-m .. ( 9.8 m/s2 g.)

1 to 5 is letter A

6 to 8 is letter D

9.) B

10.) C

11.) D

12.) A

13.) D

14.) D

15.) B

16.) D

17.) B

18.) B

19.) 8.750N

20.) B

these are my answers for your ff. questions

Acceleration = 1.428m/s2

Tension = 102.85N

Explanation:

The detailed solution is attached

Mass of an object, m = 7.3 kg

Explanation:

Given that,

Net force acting on an object, F = 8 N

Acceleration of the object, a = 1.1 m/s²

We have to find the mass of an object. It can be calculated suing second law of motion as :

F = m a

m = 7.272 kg

or

m = 7.3 kg

Hence, the mass of object is 7.3 kg

N=0

Explanation:

Given

both blocks experiencing free fall so net weight of block during free fall is zero thus there is no normal reaction between them.

N=0

The acceleration of the system is 1.401 m/s² and

The tension in the cord is 100.902 N

Explanation:

Let the 9 kg mass be m

Let the 12 kg mass be M

By Newton's second law of motion we have

For the 9 kg mass, T - mg = ma and for the 12 kg mass we have T - Mg = -Ma

Here we took the upward acceleration as positive a of the 9 kg mass and the downward acceleration of the 12 kg mass as -a

Solving for T for the 9 kg mass we have

T = mg + ma

Substituting the value of T in to the 12 kg mass equation, we have

mg + ma - Mg = -Ma or a = therefore the acceleration is

1.401 m/s²

and the tension is T = mg + ma = 9×(9.81+1.401) = 100.902 N

The anser is B) 7.3 kg, you get this by multiplying times pi (3.14) and then dividing the sum.

Hope this helped!

v = 1176.23 m/s

y = 741192.997 m = 741.19 km

Explanation:

Given

M₀ = 9 Kg (Initial mass)

me = 0.225 Kg/s (Rate of fuel consumption)

ve = 1980 m/s (Exhaust velocity relative to rocket, leaving at atmospheric pressure)

v = ? if t = 20 s

y = ?

We use the equation

v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt where t ∈ (0, t)

⇒ v = - ve*Ln ((M₀ - me*t)/M₀) - g*t

then we have

v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)

v = 1176.23 m/s

then we apply the formula

y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt

⇒ y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt

⇒ y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀ - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)

For t = 20 s we have

y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)

⇒ y = 741192.997 m = 741.19 km

The graphs are shown in the pics.