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3 years ago

What is the instantaneous speed of the object after the five seconds?

Physics

1 answer:

OverLord2011 [107]3 years ago

7 0

Answer:

12.5 m/s

Explanation:

In a acceleration time graph the area under the curve gives the change in velocity of the object. Here object starts at rest and therefore initial velocity is 0. After 5 seconds acceleration is 5m/s2.

change in velocity=area under the curve

change in velocity= 0.5*acceleration* change in time

v-0=0.5*5*5

v=12.5 m/s

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Many physical quantities are connected by inverse square laws, that is, by power functions of the form f(x)=kx^(-2). In particul

goldenfox [79]

Answer:

4 times

Explanation:

Since the equation for the illumination of an object, i.e. the brightness of the light, is <em>inversely proportional to the square of the distance from the light source</em>, the form of the function is:

f(x) = k.x⁻²

Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.

Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :

$F=k(x/2)^{-2}=\frac{k}{(x/2)^2}= \frac{k}{x^2}. 4=f(x).4$

Then, you have found that the light is 4 times as bright as it originally was.

3 0

3 years ago

A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0Â° with the horizontal. The coeff

natulia [17]

Answer:

$H =212.6 \ J$

$v = 7.647 \ m/s$

Explanation:

From the question we are told that

The child's weight is $W_c = 287 \ N$

The length of the sliding surface of the playground is $L = 7.20 \ m$

The coefficient of friction is $\mu = 0.120$

The angle is $\theta = 31.0 ^o$

The initial speed is $u = 0.559 \ m/s$

Generally the normal force acting on the child is mathematically represented as

=> $N = mg * cos \theta$

Note $m * g = W_c$

Generally the frictional force between the slide and the child is

$F_f = \mu * mg * cos \theta$

Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as

$F =m* g sin(\theta) - F_f$

Here F is the resultant force and it is represented as $F = ma$

=> $ma = m* g sin(31.0) - \mu * mg * cos (31.0)$

=> $a = g sin(31.0)- \mu * g * cos (31.0)$

=> $a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)$

=> $a = 4.039 \ m/s^2$

$F_f = 0.120 * 287 * cos (31.0)$

=> $F_f = 29.52 \ N$

Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as

$H = F_f * L$