Each mass in the figure is 3 kg. Find the magnitude and direction of the net gravitational force on mass A due to the other masses.A. 2.45 × 10–7 N toward B
B. 3.75 × 10–7 N toward C
C. 2.00 × 10–7 N toward D
D. 1.15 × 10–7 N toward D

Answers

Answer 1

Answer:

The magnitude and direction of the net gravitationalforce on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this case, all of the masses are equal to 3 kg, and the distance between mass A and mass D is 3 m.

The gravitational force between mass A and mass D is therefore:

F = G * m_A * m_D / r²

= 6.674 × 10⁻¹¹ N m² / kg² * 3 kg * 3 kg / 3 m²

= 1.15 × 10⁻⁷ N

The direction of the gravitational force is towards mass D.

Therefore, the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

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Answer 2

Answer:

Answer:

THE ANSER IS B

Explanation:

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How much longer will it taketo travel a distance of 6ookm at
a speed of 50kmh than it
would at a
speed of 6okmh?

Answers

Answer:

2hr much longer

Explanation:

Given parameters

Distance = 600km

Speed 1 = 50km/h

Speed 2 = 60km/h

Unknown:

How much longer will it take to travel a distance = ?

Solution:

Speed is the distance divided by time;

Speed = $(distance)/(time)$

Now;

Time taken = $(Distance)/(Speed)$

Time 1;

= $(600)/(50)$

= 12hr

Time 2;

= $(600)/(60)$

= 10hr

To find how much more time;

Time 1 will take 12hr - 10hr, 2hr much longer to travel the distance at that rate.

wo balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L = 0.550 m. One is attached at one end of the rod and the other at the middle of the rod. If the rod is held by the open end and rotates in a circular motion with angular speed of 45.6 revolutions per second,

Answers

Answer:

$T_1 =677224.40\ N$

Explanation:

given,

mass of the both ball = 5 Kg

length of rod = 2 L

where L = 0.55 m

angular speed = 45.6 rev/s

ω = 45.6 x 2 π

ω = 286.51 rad/s

v₁ = r₁ ω₁

v₁ =0.55 x 286.51 = 157.58 m/s

v₂ = r₂ ω₂

v₂ = 1.10 x 286.51 = 315.161 m/s

finding tension on the first half of the rod

r₁ = 0.55 r₂ = 2 x r₁ = 1.10

$T_1 = m ((v_1^2)/(r_1)+(v_2^2)/(r_2))$

$T_1 = 5 ((157.58^2)/(0.55_1)+(315.161^2)/(1.1))$

$T_1 =677224.40\ N$

A straight trail with a uniform inclination of 16° leads from a lodge at an elevation of 600 feet to a mountain lake at an elevation of 7,000 feet. What is the length of thetrail (to the nearest foot)?
O A. 6,658 ft
OB. 25,396 ft
OC. 7,282 ft
OD. 23,219 ft

Answers

I believed the answer is d

A ball is thrown into the air with 100 J of kinetic energy, which is transformed to gravitational potential energy at the top of its trajectory.When it returns to its original level after encountering air resistance, its kinetic energy is __________.

A) more than 100 J.

B) Not enough information given.

C) less than 100 J.

D) 100 J.

Answers

To solve this problem we could apply the concepts given by the conservation of Energy.

During the launch given in terms of kinetic energy and reaching the maximum point of the object, the potential energy of the body is conserved. However, part of all this energy is lost due to the work done by the friction force due to friction with the air, therefore

$E_T = PE + KE -W_f$

The potential and kinetic energy are conserved and are the same PE = KE and this value is equivalent to 100J, therefore

$E_T = 100-W_f$

The kinetic energy will ultimately be less than 100J, so the correct answer is C.

What is the potential energy of a spring that is compressed 0.65 m by a 25 kg block if the spring constant is 95 N/m?A. 1.6J
B. 7.9J
C. 15J
D. 20J

Answers

Answer:

D. 20J

Explanation:

Answer:

20 J

Explanation:

yes

An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.

Answers

The new length of aluminum rod is 17.435 cm.

The linear expansion coefficient is given as,

$\alpha=(L_(1)-L_(0))/(L_(0)(T_(1)-T_(0)))$

Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.

and linear expansion coefficient is $25*10^(-6)C^(-1)$

Substitute, $L_(0)=17.400cm,T_(1)=100,T_(0)=20,\alpha=25*10^(-6)C^(-1)$

$25*10^(-6)C^(-1) =(L_(1)-17.400)/(17.400(100-20))\n\n25*10^(-6)C^(-1) = (L_(1)-17.400)/(1392) \n\nL_(1)=[25*10^(-6)C^(-1) *1392}]+17.400\n\nL_(1)=17.435cm$