X<br> x<br> 2.5 N<br> 1.7 N<br> 2.5 N

What is the formula that describes the magnitude of impulse on an object?

vladimir1956 [14]

Answer:

Option C.

Impulse = mass × change in velocity

Explanation:

Impulse is defined by the following the following formula:

Impulse = force (F) × time (t)

Impulse = Ft

From Newton's second law of motion,

Force = change in momentum /time

Cross multiply

Force × time = change in momentum

Recall:

Impulse = Force × time

Thus,

Impulse = change in momentum

Recall:

Momentum = mass x velocity

Momentum = mv

Chang in momentum = mass × change in velocity

Change in momentum = mΔv

Thus,

Impulse = change in momentum

Impulse = mass × change in velocity

8 0

3 years ago

The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa.

scoundrel [369]

Answer:

108.37°C

Explanation:

P₁ = Initial pressure = 101 kPa

V₁ = Initial volume = 530 m³

T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K

P₂ = Final pressure = 101 kPa (because it is open to atmosphere)

V₂ = Final volume = 530 m³

P₁V₁ = n₁RT₁

⇒101×530 = n₁RT₁

⇒53530 J = n₁RT₁

P₂V₂ = n₂RT₂

⇒53530 J = n₂RT₂

$\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347$

Dividing the first two equations we get

$1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K$

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C

8 0

3 years ago

1.3 kg of water, at an initial temperature of 25oC, is heated at a rate of 100 W in a well-insulated container. How much time (i

Leviafan [203]

Answer:

68.25 minutes.

Explanation:

Power = Energy/time or

Power = Heat/time

P = Q/t....................... Equation 1

Q = Pt ........................ Equation 2

Where Q = quantity of heat, P = power, t = time.

Also,

Q = cm(t₂-t₁) ................. Equation 3

Where c = specific heat capacity of water, m = mass of water, t₁ = initial temperature of water, t₂ = final temperature of water.

substitute equation 2 into equation 3

Pt = cm(t₂-t₁) ............... Equation 4

make t the subject of the equation

t = cm(t₂-t₁)/P................ Equation 5

Given: P = 100 W, m = 1.3 kg, t₁ = 25 °C, t₂ = 100 °C.

Constant: 4200 J/kg.K

Substitute into equation 5

t = 1.3(4200)(100-25)/100

t = 409500/100

t = 4095 seconds

t = (4095/60) minutes = 68.25 minutes.

Hence the time it will take the water to reach boiling point = 68.25 minutes

6 0

3 years ago

Please answer this question , I’m not solve

umka2103 [35]

Answer:

makinig ka kasi sa titser mo tas intindihin mo bobo

7 0

3 years ago

PLEASE HELP 15 POINTS!!!!

goldfiish [28.3K]

Answer:

15 degrees Celsius

Explanation:

8 0

2 years ago

X x 2.5 N 1.7 N 2.5 N