4 years ago

When children use the word “doggie” to refer to every four-legged animal they see, __________ occurs.

Physics

1 answer:

Oduvanchick [21]4 years ago

7 0

This is known as overextension! (the correct answer is B.)

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Which example possesses mechanical potential energy?. A. a taut guitar string. B. an oscillating pendulum. C. a roller coaster r

NeTakaya

The taut guitar string haspotencial energy which we can see in action. · so option a is correct.

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4 years ago

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A 1-kg rock is suspended by a massless string from one end of a

maxonik [38]

Answer:

The weight of measuring stick is 9.8 N

Explanation:

given information:

the mass of the rock, $m_{r}$ = 1 kg

measuring stick, x =1 m

d = 0.25 m

to find the weight of measuring stick, we can use the following equation:

τ = Fd

τ = 0

$F_{r} d$ - $F_{s}d$ = 0

F_{r} = the force of the rock

F_{s} = the force of measuring stick

$F_{s} =F_{r}$

= m g

= 1 kg x 9.8 m/s

= 9.8 N

thus, the weight of measuring stick is 9.8 N

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3 years ago

during a test crash an airbag inflates to stop a dummies forward motion. the dummies mass is 75 kg. if the net force on the dumm

den301095 [7]

Use Force=Mass x Acceleration (newtons second law states force is directly proportional to the acceleration) so you can say that the force is negative and solve for Acceleration.

5 0

3 years ago

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A car is traveling 35 mph on a smooth surface. If a balanced force is applied to the car, what happens?

zepelin [54]

Answer:

Here is the answer.

Explanation:

Balanced forces- they are those forces that produce 0 resultant forces.

therefore, on applying a balanced force on the object, it wouldn't result in any change, as resultant force would be 0.

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3 years ago

The sun rotates once in 25.05 days. That is 601.2 hours. so the sun's rotational speed is 0.0017 rotations per hour. what is the

Margaret [11]

Answer:

$v=2019.09\ m.s^{-1}$

Explanation:

Given:

time taken by the sun to complete one revolution, $t=601.2\ hr$

radial distance of the sunspot, $r=6.955\times 10^8\ m.s^{-1}$

Therefore, angular speed of rotation of sun:

$\omega=\frac{2\pi}{601.2\times 3600} \ rad.s^{-1}$

Now the tangential velocity of the sunspot can be given by:

$v=r.\omega$

$v=6.955\times 10^8\times \frac{2\pi}{601.2\times 3600}$

$v=2019.09\ m.s^{-1}$

4 0

3 years ago