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katen-ka-za [31]
3 years ago
7

An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result

to the nearest tenth.
Physics
1 answer:
mylen [45]3 years ago
8 0

The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

s=ut+\frac{1}{2}gt^{2}

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

   s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}

So the distance covered by the acorn is 3.136 m.

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Worth 15 points<br><br> Please answer which one matches for each conversion
FrozenT [24]

Answer:

distance - meters

speed - meters/seconds

time - seconds

velocity - meters/seconds

acceleration - meters/seconds²

5 0
2 years ago
A car tire rotates with an average angular speed of 29 rad/s. In what time interval will the tire rotate 3.5 times? ​
meriva

Answer:

"0.758315 sec" is the appropriate choice.

Explanation:

The given values are:

Angular speed,

= 29 rad/s

or,

= \frac{29}{2 \pi} \ rps

Tire rotates,

= 3.5 times

Now,

The time interval will be:

⇒  \Delta t=\frac{3.5}{\frac{29}{2 \pi} }

⇒       =\frac{2 \pi\times 3.5}{29}

⇒       =\frac{21.99}{29}

⇒       =0.758315 \ sec

8 0
3 years ago
A particle moves along the x axis according to the equation x = 6t², where x is in meters and t is in seconds. Therefore:
vladimir2022 [97]
Answer is E

time can be negative.
A is not true because <span>a=<span><span><span>d2</span>x</span><span>d<span>t2</span></span></span>=12 m/<span>s2</span></span>
C: question already said that particle move along x-axis, which is not parabola path.
D: velocity is <span><span><span>dx</span><span>dt</span></span>=12t</span>, therefore velocity changes by 12 m/s and not 9.8 m/s
So we are left with E. <span>
</span>
7 0
3 years ago
7. A spring balance reads force in Newton. The scale is 20 cm long and reads from 0 to
GarryVolchara [31]

Answer:

The potential  energy when it reads 40 N is PE  = 5.33 \ J

Explanation:

From the question we are told that

   The lowest reading of the spring balance is  0 N and this is at  0 cm = 0 m

   The height reading of the spring balance is 60 N  and this is at 20 cm =  0.20 m

   Generally the length corresponding to the reading of 40 N is mathematically represented as

       d = \frac{40 * 0.20 }{60 }

=>    d = 0.133 \  m

Generally the potential  energy is mathematically represented as

      PE = m * g * d

Here

     F = mg  =  40 N

So  

     PE = 40 * 0.133

=>  PE  = 5.33 \ J

7 0
3 years ago
Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the
lesya [120]

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration  of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

F_{net} = - W_x - f_s

The W_x is a x-component of force due to gravity (W) and, in this case, is given by: W_x = W.sin(14)

W is described as: W = m.g

Force due to friction (f_s) is given by: f_s = μs.N

N is the normal force and, in the system, is equivalent of W_y, so:

W_y = m.g.cos(14)

Therefore, the formula will be:

F_{net} = - W_x - f_s

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

3 0
3 years ago
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