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laiz [17]
2 years ago
6

Because the time was the same for each segment you know the speed was the same for each segment

Physics
1 answer:
netineya [11]2 years ago
3 0

Answer:

Market segmentation helps you send the right message, every time, by efficiently targeting specific groups of consumers.

Explanation:

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A force of 0.2 Newtons is required to slide a book across the table. The book accelerates at 0.11 m/s squared. What is the mass
adelina 88 [10]
My calculations state, not rounding, the mass is 1.8
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3 years ago
What is the displacement for the couple's hiking trip? Give your answer as an ordered pair, with magnitude first, followed by a
nekit [7.7K]

Answer:

Explanation:

See the attached

7 0
3 years ago
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Dragsters can actually reach a top speed of 145.0 m/s in only 4.45 s. (a) Calculate the average acceleration for such a dragster
astraxan [27]

Answer:

a) 32.58 m/s²

b) 161.84 m/s

Explanation:

Initial velocity = u = 0

Final velocity = v = 145 m/s

Time taken = t = 4.45 s

s = Displacement of dragster = 402 m

a = Acceleration

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{145-0}{4.45}\\\Rightarrow a=32.58\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 32.58\times 402-0^2}\\\Rightarrow v=161.84\ m/s

The final velocity is greater than the velocity used to find the average acceleration due to the gear changes. The first gear in a dragster has the most amount of toque which means the acceleration will be maximum. The final gears have less torque which means the acceleration is lower here. The final gears have less acceleration but can spin faster which makes the dragster able to reach higher speeds but slowly.

7 0
3 years ago
An object is attached to a hanging unstretched, ideal and massless spring and slowly lowered to its equilibrium position, a dist
Ad libitum [116K]

Answer:

10.6cm

Explanation:

We are given 5.3cm below the starting point (spring extension).

Therefore, to find static vertical equilibrium, we use the equation:

kx = mg

Where:

k = spring constant =

=mg/5.3 kg/s²

We are told the object was dropped from rest.

Therefore:

loss in potential energy = gain in spring p.e

Let's use the expression:

mgx = ½kx²

We are asked to find the stretch at maximum elongation x.

To find x, we make x subject of the formula.

Therefore, we have:

x = 2mg/k (after rearranging the equation above)

x = (2mg) / (mg/5.3)

x = 10.6cm

3 0
3 years ago
If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m
AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}

We can cancel the common factor of m which leaves us with

gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}

Lets solve for v_f

gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}

Subtract gh_f from both sides of the equation.

gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}

Multiply both sides of the equation by 2.

2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}

Simplify the left side.

Apply the distributive property.

2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}

Cancel the common factor of 2.

2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}

Take the square root of both sides of the equation to eliminate the exponent on the right side.

{v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}

We are given g,v_{i},h_{i},h_{f}.

We can now solve for the final velocity.

{v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)

Anything multiplied by 0 is 0.

{v_{f}=\sqrt{2*9.81*2.41

{v_{f}=\sqrt{47.2842

v_f=6.88

7 0
1 year ago
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