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2 years ago

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A. 24.89 B. 25.89 C. 17.74 D. 19.73

1 answer:

Veronika [31]2 years ago

5 0

Answer: D

Explanation:

Just did it got an 100

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The electric potential at a point equidistant from two particles that have charges +Q and –Q is larger than zero. a. smaller tha

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Answer:

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3 years ago

How does inertia affect a person who is not wearing a seatbelt during a collision?

Vadim26 [7]

When someone fails to wear a seat belt the passenger becoming a projectile the force a person will be subjected to for a passenger weighting 100 pounds, and the car is traveling at 60 mph would be the same as 6000 pounds. thats like hitting a brick wall. so is other words put ur seat belt on.

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3 years ago

What is the frequency of a wave whose wavelength is 0.7m and whose velocity is 120m/s?

ANEK [815]

Answer:

f=171.43Hz

Explanation:

Wave frequency is the number of waves that pass a fixed point in a given amount of time.

The frequency formula is: f=v÷λ, where <em>v</em> is the velocity and <em>λ</em> is the wavelength.

Then replacing with the data of the problem,

f= $\frac{120\frac{m}{s} }{0.7m}$

f=171.43 $\frac{1}{s}$

f=171.43 Hz (because $\frac{1}{s} = Hz$ , 1 hertz equals 1 wave passing a fixed point in 1 second).

5 0

3 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

2 years ago

Paul lifts a sack weighing 245 newtons vertically from the ground and places it on a platform at a height of 0.7 meters. If he t

d1i1m1o1n [39]

Answer:

B. 17.15 watts

Explanation:

Given that

Time = 10 seconds

height = distance = 0.7 meters

weight of sack = mg = F = 245 newtons

Power = work done/ time taken

Where work done = force × distance

Substituting the given parameters into the formula

Work done = 245 newton × 0.7 meters

Work done = 171.5 J

Recall,

Power = work done/time

Power = 171.5 J ÷ 10

Power = 17.15 watts

Hence the power expended is B. 17.15 watts

6 0

3 years ago