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Alika [10]
4 years ago
9

What is the average acceleration of the object represented in the graph above over the eight seconds? 0.5 cm/s2 1.0 cm/s2 1.5 cm

/s2 2.0 cm/s2
Physics
1 answer:
Olenka [21]4 years ago
3 0

Answer:

Draw tangents to the graph at 1.0 s and 5.0 s. 0. 1. 2. 3. 4. 5. 6. 0 ... When the velocity component is negative, the extra displacement during ... 2 - 3 min. constant acceleration component until the car stops. ..... ii). The gravitational force and the drag force are doing work on the object. .... N) × (0.5 m) = 5 × 10. 2.

Explanation:

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An object or living organism that is stationary(not moving) for example the car is stopped at the top of the hill
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A swimmer travels from the south end to the north end of a 30.0 m pool in 15.0 s and makes the return trip to the starting posit
JulsSmile [24]

Answer:

Zero

Explanation:

Average velocity is given by:

v=\frac{d}{t}

where

d is the displacement of the trip

t is the time it takes for the trip to complete

In this problem, the net displacement of the swimmer is zero. In fact:

- First, he swims 30.0 m in the north direction

- Then, he travels back (-30.0 m) in the south direction, to the starting position

Since the final position is equal to the starting position, the displacement is zero:

d = 0

And therefore, the average velocity is also zero.

8 0
4 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
1 year ago
Calculate the individual positive plate capacity in motive power cell that has 15 plates and a copa of 595 Ah A. 110 Ah B. 75 Ah
Sphinxa [80]

Answer:

The individual positive plate capacity is 85 Ah.

(D) is correct option.

Explanation:

Given that,

Number of plates = 15

Capacity = 595 Ah

We need to calculate the individual positive plate capacity in motive power cell

We have,

15 plates means 7 will make pair of positive and negative.

So, there are 7 positive cells individually.

The capacity will be

capacity =\dfrac{power}{number\ of\ cells}

Put the value into the formula

capacity =\dfrac{595}{7}

capacity =85\ Ah

Hence, The individual positive plate capacity is 85 Ah.

6 0
4 years ago
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An alpha particle consists of two protons and two neutrons with a mass of 6.64 x 1027 kg and charge 3.2 x 10-19 C. If two of the
Dafna11 [192]

Answer:  ratio = 3.14 × 10³⁵

Explanation:

From Newtons law of Universal Gravitation;

Every object in the universe is attracted to every other object

F = (Gm₁m₂)/r²

G = 6.67 X 10⁻¹¹ N-m²/kg²

M₁= mass of one object

M₂= mass of second object

r = distance from center of objects

q = charge = 3.2 x 10⁻¹⁹C.

  • Calculating the gravitational force;

Fg = (Gm₁m₂)/r² = (Gm²)/r²

which is = (6.67×10⁻¹¹ × (6.64×10⁻²⁷)²) / r²

= (2.94 × 10⁻⁶³) / r²

  • Calculating the electric force;

Fe = (Kq₁q₂) / r² = K q² / r² = (9×10⁹ × (3.2 × 10⁻¹⁹)² / r²

Fe = 9.22 × 10⁻²⁸ / r²

comparing the ratio of both forces we have;

Fe/Fg = (9.22×10⁻²⁸/r²) ÷ (2.94×10⁻⁶³/r²)

r² cancels from the above expression, which gives;

ratio as 3.14 × 10³⁵

8 0
3 years ago
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