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3 years ago

If a force does a positive amount of work on an object, does the object's speed

Physics

1 answer:

Alexxandr [17]3 years ago

8 0

The speed of the object increases

Explanation:

We can answer this question by applying the work-energy theorem, which states that the work done on an object is equal to the change in kinetic energy of the object. Mathematically:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

W is the work done on the object

$K_f, K_i$ are the final and initial kinetic energy of the object, respectively

m is the mass of the object

v is its final speed

u is its initial speed

In this case, the force does a positive amount of work on the object, so

$W>0$

This also implies that

$K_f > K_i$

And so

$\frac{1}{2}mv^2 > \frac{1}{2}mu^2$

And therefore

$v>u$

which means that the speed of the object increases.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? explain.

choli [55]

Answer

Hi,

The forces are; weight (gravity), Normal/centripetal force and friction. Force due to gravity is constant where as friction and centripetal are not.

Explanation

Weight is constant, given by the force of gravity on the object. The centripetal force is a function of the angles occurring between the velocity vector and the weight vector that is at right angle with the perpendicular line drawn from the surface. Friction is a function of the centripetal force thus it also varies.

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3 years ago

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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

3 years ago

A time-varying horizontal force F(t) = At4 + Bt2 acts for 0.500 s on a 12.25-kg object, starting attime t = 1.00 s. In the SI sy

PSYCHO15rus [73]

Answer:

3.82 Ns

Explanation:

Time varying horizontal Force is given as

F(t) = A t⁴ + B t²

F(t) = 4.50 t⁴ + 8.75 t²

Impulse imparted is given as

$I = \int_{0}^{t}Fdt$

$I = \int_{0}^{1}Fdt$

$I = \int_{0}^{1}(4.50 t^{4} + 8.75 t^{2})dt$

$I = ((0.9) (1)^{5} + (2.92) (1)^{3})$

$I = 3.82 N-s$

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3 years ago

Read 2 more answers

A fire helicopter carries a 580-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee

inn [45]

Answer:

F = 41,954 N

Explanation:

given,

mass of bucket = 580 Kg

length of the cable = 20 m

velocity = 40 m/s

angle made = 38.0°

T cos 38° = m g..............(1)

T sin 38^0 = \dfrac{mv^2}{l} + F......(2)

dividing equation (2) by (1)

$tan 38^0 = \dfrac{\dfrac{mv^2}{l} + F}{mg}$

$tan 38^0 = \dfrac{\dfrac{580\times 40^2}{20} + F}{580 \times 9.81}$

$4445.36 = \dfrac{580\times 40^2}{20} + F$

F = -46400 + 4445.36

F = -41,954 N

hence, the force is acting in the opposite direction as assumed.

F = 41,954 N

4 0

3 years ago