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Pavlova-9 [17]
2 years ago
13

A body with a uniform acceleration travels distances of 24m and 64m during the first two equal consecutive intervals of time, ea

ch of duration 4s. Determine the initial velocity and acceleration of the moving body.​
Physics
1 answer:
Dvinal [7]2 years ago
3 0

Answer:

Explanation:

Average velocity in the 24 m interval is 24 / 4 = 6 m/s

Average velocity in the 64 m interval is 64 / 4 = 16 m/s

There is a 4 second interval between the two points where average velocity equals actual velocity

a = Δv/t = (vf - vi) / t = (16 - 6) / 4 = 2.5 m/s²

s = v₀t + ½at²

24 = v₀(4) + ½(2.5)4²

4v₀ = 24 - 20

v₀ = 1 m/s

Not asked for but the velocity at the end of the first segment and beginning of the second segment is 11 m/s and final velocity is 21 m/s

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A. If the rock weighs 1050 N, how far does the girl move her end of the lever if the lever is
3241004551 [841]

student is removing rocks from a garden. She exerts a force of 218 N on a lever to raise one rock a distance of 11.0 cm.

Answer:

52.98 cm

Explanation:

The ratio of forces and distance moved will be the same hence we apply cross multiplication rule

If 218 N can only raise the rock to a distance of 11 cm then we check the distance it can move when force is increased to 1050 N

218 N=11 cm

1050 N=?

By cross multiplication, the distance will be equivalent to \frac {1050\times 11 cm}{218}=52.981651376146\ cm\approx 52.98\ cm

8 0
3 years ago
One cubic meter (1.0 m3
hammer [34]
Because the masses that you give are for blocks that are 1 cubic meter in volume, they also serve as the densities for the two metals that you are comparing. 

<span>mass = density*volume </span>

<span>volume = (4/3)*pi*r^3 </span>

<span>volume of iron sphere = (4/3)*3.14*0.0201^3 = 3.40*10^-5 m^3 </span>

<span>mass of iron sphere = 7860* 3.40*10^-5 m^3 = 0.27 kg = mass of Aluminum Sphere </span>

<span>Volume of Al Sphere = 0.27/2700 = 9.90*10^-5 m^3 </span>

<span>Radius = cube root (volume / (4/3) / pi) = 2.87 cm. </span>

<span>I did this using the MS calculator, and I'm not 100% sure on the numerical answer, but the process is what you need to do to solve the problem. You should double check my answer.

hope this helped :)


</span>
8 0
2 years ago
A box sits on the back of a flatbed truck. If the coefficient of static friction between the box and the truck bed is μs = 0.400
slava [35]

Answer:

28,699m

Explanation:

The force to make the box move should be <u>μs.N=μs.m.g=m.</u><u>|</u><u>a</u><u>|</u>

then,

|a|=μs.g

Being

μs coefficient of static friction,

N the force made by the truck on the box caused by the gravity force,

m the mass,

g the acceleration of gravity

and a the acceleration of the truck.

x = v \times t  +  \frac{1}{2} \times a \times   {t}^{2}

as the truck is stopping, the acceleration is negative. then,

x = v \times t  -  \frac{1}{2} \times  |a| \times   {t}^{2}

|a|  = v  \div  t \\ t = v \div   |a|

x = v \times (v \div   |a| ) -  \frac{1}{2} \times  |a| \times   {(v \div   |a|)}^{2}

x = v \times (v \div  μs.g ) -  \frac{1}{2} \times  |a| \times   {(v \div   μs.g)}^{2}

x =   \frac{{v}^{2}}{μs \times g} -  \frac{1}{2} \times  \frac{{v}^{2}}{μs \times g} \\ x = \frac{1}{2} \times  \frac{{v}^{2}}{μs \times g} \\ x = 0.5 \times  \frac{{(15m/s) }^{2} }{0.4 \times 9.8m/ {s}^{2} }  = 28.699m

28,699m

7 0
3 years ago
You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 2.1 m/s. The
garri49 [273]

Answer:

4.6 Joules

Explanation:

K=1/2*MV^2

1/2 * 2.1kg * 2.1^2m/s

==4.6305  Joules

simplified to 4.6 Joules

7 0
2 years ago
Determine weather or not the equation below is balanced.
Liula [17]

Answer:

Yep.. It's balanced and its a combination reaction

Explanation:

Reactants : S₈ + 24F₂

Product 8SF₆

8 0
3 years ago
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