What is formula for finding period of a planet if its mass is found by sending a spacecraft

Two forces with magnitudes of 6 pounds and 18 pounds are applied to an object. The magnitude of the resultant is 13 pounds. Find

Nikolay [14]

Answer: the angle between the resultant vector and the vector of the 18 pound is 28°

Explanation:

given that data in the question; as its interpreted in the diagram below;

from the cosine rule, we know that;

a² = b² + c² - 2bc

so

(13)² = (6)² + (18)² - (2 × 6 × 18 ) cos∅

169 = 36 + 324 - 216cos∅

169 = 360 - 216cos∅

216cos∅ = 360 - 169

216cos∅ = 191

cos∅ = 0.8842

∅ = cos⁻¹ ( 0.8842 )

∅ = 27.8° ≈ 28° {nearest whole number}

Therefore the angle between the resultant vector and the vector of the 18 pound is 28°

6 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

2 years ago

Mrs. Rohaley gives you a piece of iron and asks you to determine the physical and chemical properties of the metal. She wants yo

Fofino [41]

Answer:The answer is B

Explanation:

5 0

3 years ago

the idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse

Finger [1]

Forward thrust has positive values and reverse thrust has negative values.

Thrust is a sudden push or pull in a certain direction.

a)

Flight speed u = 150 km/h

1 km/h = $\frac{1}{3.6}$ km/s

therefore, 150 km/h = 41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

$T = m_{exhaust} * U_{exhaust} - U_{flight}$

T = 50 * (150 - 41.67)

T = 5416.67 N

Therefore, the value of forward thrust is 5416.67 N.

b)

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component,

thus thrust equation is:

$T = m_{exhaust} * U_{exhaust} - U_{flight}$

T = 50 * (0 - 41.67)

T = -2083.5 N

Therefore, the thrust force T is -2083.5 N in the reverse direction.

c)

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero because $U_{exhaust} = U_{flight} = 0\\$

T = 0

Therefore, there is no difference in two velocities in x direction.

The given question is incomplete, the complete question is,

"The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.

a. What is the forward thrust of this engine?

b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?

c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?"

To know more about thrust,

brainly.com/question/14552836

#SPJ1

3 0

1 year ago

An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull

puteri [66]

Answer:

$\mu=0.0049Kg/m$

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

$f_n=\frac{nv}{2L}$

Where $L$ is the length of the string and $v$ the velocity of propagation. Use this expression to find the value of $v$ .

$f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s$

The velocity of propagation is given by the expression:

$v=\sqrt{\frac{T}{\mu }$

Where $\mu$ is the desirable variable of the problem, the linear mass density, and $T$ is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

$T=W=mg=(5)(9.81)=49.05N$

With the value of the tension and the velocity you can find the mass density:

$v=\sqrt{\frac{T}{\mu}$

$v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m$

6 0

2 years ago