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4 years ago

Science cannot be used to answer questions about _____.

Physics

2 answers:

aksik [14]4 years ago

8 0

Science cannot be used to answer questions about Moral judgements

nexus9112 [7]4 years ago

8 0

Moral judgements. !!!!!!!!!

You might be interested in

A 5.0 kg block is pushed 2.0 m at a con-

Alona [7]

Answer:

$W_F=127.64283 J$

Explanation:

Information Given:

$m = 5kg$ $v=constant$

Key: μ = Kinetic Friction (Kf) θ = Theta α = 180° N = Normal Force

$W_F=F_ydcos$ θ

$W_F=Fdsin$ θ

$_{net}F_y = sin$ θ-μ $N-mg=0$

$_{net}F_x = 0$

$N=Fcos$ θ

$Fsin$ θ-μ $N=mg$

$Fsin$ θ-μ $Fcos$ θ $=mg$

$F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}$ → $W_F=\frac{mgdsin(theta)}{sin(theta)-(Kf)cos(theta)}$

$W_F=\frac{(2)(9.81)(2)sin(30)}{sin(30)-(0.40)cos(30)}$

$W_F=127.64283 J$

3 0

3 years ago

A 1.0-kg standard cart collides on a low-friction track with cart A. The standard cart has an initial x component of velocity of

Andru [333]

Answer:

Explanation:

Mathematically, linear momentum is expressed as the product of mass and velocity. Linear momentum conservation law states that a body or system of bodies retains its total momentum unless an external force is applied to the system.

In this case, the system consists of two carts.

At the start, the linear momentum (P) of the system is equal to:

$P=1.0kg*0.4m/s=0.4kg*m/s$

It's only composed of linear momentum of the standard cart because cart A doesn't have any linear momentum at that moment.

After the collision, linear momentum has to be the same

$P=0.4kg*m/s=1.0kg*0.20m/s+m_{A} *0.70m/s$

where m_A is the mass of the cart A.

Solving for m_A

$m_{A} =0.28kg$

After the cart A rebounds, the linea momentum of the system has changed (because of the force present in the rebound). The new linear momentum is:

$P=1kg*0.2m/s+0.29kg*(-0.7m/s)=-0.003kg*m/s$

Then, the lump of putty is added to the system, but the linear momentum has to be the same, because we added a mass, not a force. The mass of that putty (m_p) has to be added to the equation of the system

$-0.003kg*m/s=1.0kg*(-0.2m/s)+(0.29kg+m_{p} )(0.4m/s)$

Solving for m_p

$m_{p}=0.20kg$

6 0

3 years ago

What creates more pressure?Check all that apply A.Gas molecules moving more quickly B.More gas molecules C. Going to a higher al

Shkiper50 [21]

It's A.Gas molecules moving more quickly..... hope it helps you dear

7 0

3 years ago

I NEED HELP PLEASE THANKS! :)

Nady [450]

Answer:

0.962 s

Explanation:

Speed = frequency × wavelength

v = fλ

2.60 m/s = f (2.50 m)

f = 1.04 Hz

Period = 1 / frequency

T = 1/f

T = 1 / (1.04 Hz)

T = 0.962 s

7 0

3 years ago

A 5 kg

kotykmax [81]

Answer:

<em>The object-Earth system is open</em>

$F_n=ma=5\ (7.2)=36\ N$

Explanation:

<u>Accelerated Motion </u>

When an object is released in free air (with no other forces than the gravity), it describes a free-fall motion and the formulas include the acceleration of gravity as part of the calculations. But when there is another external force, then the acceleration is not the gravity, but the result of the net force exerted on the mass of the object.

By definition, an open system includes the exchange of energy from and to the surroundings, that is why all systems surrounding our planet are considered as open systems. In our case, the object is interacting with the planet's gravity and there is some other external force, which will be computed later. The object-Earth system is open.

If the object starts from rest, its initial speed is zero, and

$v_f=a\ t$