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Alekssandra [29.7K]
3 years ago
8

Similarity of gravity and electromagnetic force

Physics
1 answer:
Mashutka [201]3 years ago
5 0

Answer:

Gravity is an attractive force as well as electromagnetic, but electromagnetic attracts and repels.

Explanation:

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What well u can use to make a shelter but that's all I can think of ??
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19. A 0.0340 kg bullet traveling at 120 m/s embeds itself in a 1.24 kg wooden block which
Blababa [14]

Answer: x ≈ 36.3 cm

Explanation:

Conservation of momentum during the collision

0.0340(120) + 1.24(0) = (0.0340 + 1.24) v

v = 3.2025 m/s

The kinetic energy of the block/bullet mass will convert to spring potential

½kx² = ½mv²

x = √(mv²/k)

x = √(1.274(3.2025²) / 99.0)

x = 0.363293... ≈ 36.3 cm

3 0
2 years ago
At a particular instant, a moving body has a kinetic energy of 295 J and a momentum of magnitude 25.1 kg · m/s.(a)What is the sp
motikmotik

Answer:

a) 23.51 m/s

b) 1.07 kg

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

K = \frac{1}{2} mv^2

where m = mass of the body and v = speed of the body

We know that momentum is given as:

p = mv

Therefore:

K = 1/2 * pv

=> v = 2K / p

v = (2 * 295) / 25.1 = 23.51 m/s

The velocity of the body at that instant is 23.51 m/s.

b) Momentum is given as:

p = mv

=> m = p / v

m = 25.1 / 23.51  = 1.07 kg

The mass of the body at that instant is 1.07 kg

5 0
3 years ago
Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support
alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

r_{M\perp}(F_M)-r_{W\perp}(W)=0

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

r_{M\perp}(F_M)=r_{W\perp}(W)

Now, we divide by the distance of the muscle:

(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}

Now, we substitute the values:

F_M=\frac{(2.4cm)(50N)}{5.1cm}

Now, we solve the operations:

F_M=23.53N

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

\Sigma F_v=F_j-F_M-W

Since there is no vertical movement the sum of vertical forces is zero:

F_j-F_M-W=0

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

F_j=F_M+W

Now, we plug in the values:

F_j=23.53N+50N

Solving the operations:

F_j=73.53N

Therefore, the force is 73.53 Newtons.

8 0
7 months ago
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