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alexandr1967 [171]

3 years ago

14

The four best metallic conductors are: A. gold B. cadmium C. alloys D. copper E. silver F. sodium G. aluminum H. iron I have to

pick 4

1 answer:

RoseWind [281]3 years ago

7 0

Answer:

Explanation:

I can give you four for certain.

Copper

Gold

Silver

Aluminum

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You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand at 30 m/s. What is the kinetic energy (KE) of the volley

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Explanation:

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3 years ago

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prohojiy [21]

A. The magnitude of the spring force (in N) acting upon the object is 15.9 N

B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²

C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).

<h3>A. How to determine the force </h3>

Extension (e) = 0.150 m
Spring constant (K) = 106 N/m
Force (F) = ?

F = Ke

F = 106 × 0.15

F = 15.9 N

<h3>B. How to determine the acceleration</h3>

Mass (m) = 0.52 Kg
Force (F) = 15. 9 N
Acceleration (a) =?

F = ma

Divide both sides by m

a = F / m

a = 15.9 / 0.52

a = 30.58 m/s²

<h3>C. How to determine the direction of the acceleration vector</h3>

Considering the diagram, we can see that the spring was pulled away from the equilibrium point.

Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.

Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.

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2 years ago

What unit of time on Earth is based on the revolution of the Earth around the Sun?

ollegr [7]

C.) Year

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3 years ago

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Which elements are not likely to be formed in the sun sometime during its life cycle?

Sladkaya [172]

Answer:

correct answer is iron

Explanation:

solution

as Iron is only formed in the cores of the star that is more massive than the sun because it takes very much energy as gravitational force push inward on the star

and the sun can form the helium and carbon and the oxygen and also other elements

but not the iron or any iron group element

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8 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.

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1 year ago