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3 years ago

13

It requires 0.30 kJ of work to fully drive a stake into the ground. If the average resistive force on the stake by the ground is

828 N, how long is the stake?

1 answer:

LenaWriter [7]3 years ago

6 0

Answer:

Length of the stake will be 0.3623 m

Explanation:

We have given energy required to fully drive a stake into ground = 0.30 KJ = 300 J

Average resistive force acting on the floor is equal to F = 828 N

We have to find the length of the stake

We know that work done is given by

W = Fd, here W is work done , F is average force and d is the length of the stake

So 300 = 828×d

d = 0.3623 m

So length of the stake will be 0.3623 m

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An organ pipe open at both ends has two successive harmonics with frequencies of 220 Hz and 240 Hz. What is the length of the pi

Harman [31]

Answer:

The length is $l = 8.6 \ m$

Explanation:

From the question we are told that

The frequencies of the two successive harmonics are $f_1 = 220 \ Hz$ , $f_2 = 240 \ Hz$

The speed of sound in the air is $v_s = 343 \ m/s$

Generally the frequency of a given harmonic is mathematically represented as

$f_n = \frac{n v }{2l}$

Here n defines the position of the harmonics

Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

$220 = \frac{n v}{2l}$

and

$240 = \frac{(n+1) v}{2l}$

So

$\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220$

=> $\frac{v}{2l} = 20$

=> $l = 8.6 \ m$

7 0

3 years ago

CaHeK987 [17]

Potential energy is the answer

6 0

3 years ago

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

3 years ago

When the pendulum bob reaches the mean position, the net force acting on it is zero. Why then does it swing past the mean positi

ryzh [129]

Answer:

<u>The pendulum bob swing past the mean position because:</u>

When a pendulum's bob is accelerating at its extreme position its velocity is zero. Due to the restoring toque the bob starts to accelerates towards its mean postion. The maximum acceleration of the pendulum's bob is $-w^{2} A$ and the the acceleration decreases as $-w^{2} x$ towards the mean position.

The acceleration at the mean position becomes zero but the velocity remains maximum. Hence the bob continues to move and does not stops.Thus it can summarised as the force decreases ,acceleration decreases and velocity increases at slow rate.

6 0

3 years ago

What are successfulness of the Competition policy in South Africa?

Mars2501 [29]

Answer:

The product choices along with its competitive prices were provided to the consumers. Practices such as horizontal collusion and resale price maintenance was declared unlawful in 1984. Prevention of monopoly growth was the aim of the competition policy act.

Explanation:

hope this helps you

6 0

3 years ago