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Ivanshal [37]
3 years ago
8

Explain the 3 factors that affect the Speed of Sound

Physics
1 answer:
Alexxandr [17]3 years ago
3 0
Nature of media,Temperature frequency
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Marianne and her family are driving from West Virginia to California. While driving through the Rocky Mountains, they encounter
Elena-2011 [213]

Answer:

Point A

Explanation:

Point C is way to close to where the force would need to be.

Point B is close but not as close, the reasonable answer would be point A.

7 0
2 years ago
A girl is whirling a ball on a string around her head in
Thepotemich [5.8K]

Answer:

Answered

Explanation:

The girl whirling the ball should let go off ball when the ball is at a position such that tangent to the circle is in the direction of the target.

the tangent at any point in a circular path indicates the direction of velocity at that point. And the moment when the centripetal force is removed the ball will follow the tangential path at that moment.

8 0
3 years ago
Q 1: Calculate the pressure.
Oksana_A [137]

Answer:

1: 300pa

2: 10cm2

3: 430pa

4: a: 1.6667pa

b:2.5m2

c:20pa

Explanation:

7 0
3 years ago
What is the net force on a car with a mass of 1000 kg if its<br> acceleration is 35 m/s^2?
VashaNatasha [74]

Answer:

3000N

Explanation:

divided to get answer

the force needed to accelerate the 1000kg car by 3m/s2 is 3000N

7 0
2 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
11 months ago
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