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belka [17]
3 years ago
5

A 3.0 kg box is suspended by a series of ropes as shown below. The tension force in the horizontal rope is -40 N. What is the te

nsion force in the rope attached to the ceiling?
40 N
30 N
60 N
50 N

Physics
1 answer:
Nitella [24]3 years ago
6 0

Answer:

50 N

Explanation:

Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:

The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.

The horizontal component of the forces:

F₁ + F₂ = -40N + F₂ = 0

F₂ = 40N

The vertical component of the forces:

F₁ + F₂ - mg = 0 + F₂ - mg = 0

F₂ = mg

If I assume the gravitational constant g = 10 m/s²:

F₂ = (3 kg) * (10 m/s²) = 30N

Adding the horizontal and vertical components of the force F₂:

F₂ = √((40N)² + (30N)²) = 50N

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Paracetamol is grouped within the 'painkillers' category. 
6 0
3 years ago
Three bullets are fired simultaneously by three guns aimed toward the center of a circle where they mash into a stationary lump.
earnstyle [38]

Answer:

2,87 * 10^{-3}

Explanation:

When the bullets meet at the center and collide, since momentum is a vectoral quantity, their momentum vectors even up and are sumof zero. Formula of momentum is P = m.v , where m is mass and v is velocity. Let’s name the first two bullets as x,y and the one which mass is unknown as z. Then calculate momentum of x and y:

Px= 5,30 * 10^{-3} * 301 = 1,5953 kg*m/s

Py= 5,30 * 10^{-3} * 301 = 1,5953 kg*m/s

The angle between x and y bullets is 120°, and we know that if the angle between two equal magnitude vectors is 120°, the magnitude of the resultant vector will be equal to first two and placed in exact middle of two vectors. So we can say total momentum of x and y (Px+Py) equals to 1,5953 kg*m/s as well (Shown in the figure).  

For z bullet to equalize the total momentum of x and y bullets, it needs to have the same amount of momentum in the opposite way.

Pz = 1,5953 = m * 554

m = 2,87 * 10^{-3} kg

8 0
3 years ago
How long will it take an object to move 100 meters if the object is traveling with and average speed of 0.5 meters per second
gogolik [260]
100\0.5 = 200 seconds
4 0
3 years ago
A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler
Pepsi [2]

Answer:

331665750000\ m/s^2

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = 1.989\times 10^{30}\ kg

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2

The gravitational acceleration at the surface of such a star is 331665750000\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s

The velocity of the object would be 3257806.62409 m/s

8 0
3 years ago
A watermelon is thrown down from a skyscraper with a speed of 7.0\,\dfrac{\text m}{\text s}7.0 s m ​ 7, point, 0, space, start f
Degger [83]

Answer:

y = 17,89 m

Explanation:

Let us fixate the reference point in top of the building, from where the watermelon is thrown down. We will assume also that the positive axis of our system points up. We describe the watermelon’s motion with the equation:

 v_y^2 =v_0^2 + 2ay

Clearing the equation so we isolate y we have that:

 y = (v_y^2 - v_0^2 )/2a

Making a substitution with the values from the statement we have:

y = ((20 m/s)^2 - (7 m/s)^2)/(2*9,81 m/s^2) = 17,89 m]

So, this skyscraper is about 17,89 m tall; which is not very tall for a skyscraper but who am I to judge.  17,89 m is also the displacement of the watermelon from the point it was thrown down.

I hope everything was clear with my explanation. If I can help with anything else, just let me know. Have an awesome day :D

7 0
3 years ago
Read 2 more answers
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