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3 years ago

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A dog starts from rest to chase after a ball. If the dog reaches a maximum speed of 5.4 m/s in 8.2 seconds, what is the accelera

tion of the dog?

1 answer:

Lelechka [254]3 years ago

4 0

Answer:

0.66 m/s²

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 0

Final velocity (v) = 5.4 m/s

Time (t) = 8.2 secs

Acceleration (a) =....?

The acceleration of the dog can be obtained as follow:

v = u + at

5.4 = 0 + a × 8.2

5.4 = 8.2a

Divide both side 8.2

a = 5.4/8.2

a = 0.66 m/s²

Therefore, the acceleration of the dog is 0.66 m/s².

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How can hypotheses best be tested

Lerok [7]

In order for a hypothesis to be tested, and experiment needs to be designed.

Many trials need to be runned in order to get accurate results and to form a valid conclusion that can prove or disprove the hypothesis

3 0

4 years ago

Suppose that a block of mass 2 kg is pulled to the right with a force of 10 N, and the friction force on the block is directed t

Law Incorporation [45]

Answer:

The block has an acceleration of $3 m/s^{2}$

Explanation:

By means of Newton's second law it can be determine the acceleration of the block.

$\sum F_{r} = ma$ (1)

Where $\sum F_{r}$ represents the net force, m is the mass and a is the acceleration.

$F_{x} + F{y} = ma$ (2)

The forces present in x are $F = 10 N$ and $f = 4 N$ (the friction force):

$F_{x} = 10 N - 4 N$

Notice that $f$ subtracts to $F$ since it is at the opposite direction.

$F_{x} = 6 N$

The forces present in y balance each other:

$F_{y} = 0$

Therefore:

$6 + 0 = ma$

$6 N = (2kg)a$ (3)

But $1 N = 1 Kg.m/s^{2}$ and writing (3) in terms of a it is get:

$a = \frac{6 Kg.m/s^{2}}{2 Kg}$

$a = 3 m/s^{2}$

So the block has an acceleration of $a = 3 m/s^{2}$ .

4 0

3 years ago

You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

stellarik [79]

A) The change in internal chemical energy is $1.15\cdot 10^7 J$

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

$P=Fv$

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

$P=(80)(8.0)=640 W$

The energy output is related to the power by the equation

$P=\frac{E}{t}$

where:

P = 640 W is the power output

E is the energy output

$t = 30 min \cdot 60 = 1800 s$ is the time elapsed

Solving for E,

$E=Pt=(640)(1800)=1.15\cdot 10^6 J$

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

$\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J$

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

$E=1.15\cdot 10^7 J$

Here we want to find the time t' needed to convert an amount of chemical energy of

$E'=3.8\cdot 10^5 J$

So we can setup the following proportion:

$\frac{t}{E}=\frac{t'}{E'}$

And solving for t',

$t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min$

Learn more about power and energy:

brainly.com/question/7956557

#LearnwithBrainly

3 0

3 years ago

De donde eres responde y te doy corona

irakobra [83]

soy de texas, united states

8 0

3 years ago

In his motorboat, Bill Ruhberg travels upstream at top speed to his favorite fishing spot, a distance of 120120 mi, in 33 hr.

photoshop1234 [79]

Answer:

The rate of the boat in still water is 44 mph and the rate of the current is 4 mph

Explanation:

x = the rate of the boat in still water

y = the rate of the current.

Distance travelled = 120 mi

Time taken upstream = 3 hr

Time taken downstream = 2.5 hr

Speed = Distance / Time

Speed upstream

$\frac{120}{3}=x-y\\\Rightarrow 40=x-y$

Speed downstream

$\frac{120}{2.5}=x+y\\\Rightarrow 48=x+y$

Adding both the equations

$48+40=x-y+x+y\\\Rightarrow 88=2x\\\Rightarrow 44=x$

$40=44-y\\\Rightarrow 40-44=-y\\\Rightarrow y=4$

The rate of the boat in still water is <u>44 mph</u> and the rate of the current is <u>4 mph</u>

8 0

3 years ago