All categories

3 years ago

According to Newton’s first law of motion, what will an object in motion do when no external force acts on it?

2 answers:

7 0

According to Newton’s first law, the object will either continue at a constant velocity/speed with an acceleration of 0 or remain at rest.

Zigmanuir [339]3 years ago

5 0

Answer:

It remains in motion.

Explanation:

According to the Newton's first law of motion, if a body is in motion it remains in motion or if a body is at rest it remains at rest until and unless an external force is not applied on the body.

So, here the object is in motion and no external force is applied, so remains in motion as the same.

You might be interested in

Density of water is 1000 kg/m^3. What will be the volume of 35000 kg water?

d1i1m1o1n [39]

Volume = mass/density
Volume = 35000/1000
Volume = 35m^3

8 0

2 years ago

A bat emitts a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away

Varvara68 [4.7K]

The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.

From the question given above, the following data were obtained:

Velocity of sound (v) = 343 m/s

Distance (x) = 8.42 m

Time (t) =?

We can obtain obtained the time as illustrated below:

v = 2x / t

343 = 2 × 8.42 / t

343 = 16.84 / t

Cross multiply

343 × t = 16.84

Divide both side by 343

t = 16.84/343

t = 0.05 s

Thus, the time between when the bat emits the sound and when it hears the echo is 0.05 s.

<h3>How does a bat know how far away something is?</h3>

A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.

Learn more about time elapses between when the bat emits the sound :

<u>brainly.com/question/16931690</u>

#SPJ4

Correction question:

A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)

8 0

2 years ago

The distance between adjacent nodes in a standing wave pattern in a length of string is 25.0 cm:A. What is the wavelength of wav

mina [271]

A) 50 cm

B) 10000 cm/s

Explanation

Step 1

If you know the distance between nodes and antinodes then use this equation:

$\begin{gathered} \frac{\lambda}{2}=D \\ \text{where}\lambda\text{ is the wavelength} \\ D\text{ is the distance betw}een\text{ nodes} \end{gathered}$

then, let

$D=\text{ 25 cm }$

now, replace to find the wavelength

$\begin{gathered} \frac{\lambda}{2}=25 \\ \text{Multiply both sides by 2} \\ \frac{\lambda}{2}\cdot2=25\cdot2 \\ \lambda=50\text{ Cm} \end{gathered}$

so, the wavelength is

A) 50 cm

Step 2

The speed of a wave can be found using the equation

$v=\lambda f$

or velocity = wavelength x frequency,

then,let

$\begin{gathered} \lambda=50\text{ cm} \\ f=200\text{ Hz} \end{gathered}$

replace and evaluate

$\begin{gathered} v=\lambda f \\ v=50\text{ cm }\cdot200\text{ HZ} \\ v=10000\text{ }\frac{\text{cm}}{s} \end{gathered}$

B) 10000 cm/s

I hope this helps you

6 0

1 year ago

A. B. C. D.<br> -___________

mixas84 [53]

Where is the data for this question? what is the purpose ?

6 0

3 years ago

What's the weight of a 30x30x50 cm body with the density of 1.8/cm cube?

grin007 [14]

Answer:

The weight of the body, W = 793.8 m/s²

Explanation:

Given,

The volume of the body, v = 45,000 cm³

The density of the body, ρ = 1.8 g/cm³

The mass of the body is given by the formula,

m = ρ x v

= 1.8 g/cm³ x 45,000 cm³

= 81,000 g

Hence, the mass of the body is m = 81 kg

The weight of the body,

W = m x g

= 81 kg x 9.8 m/s²

= 793.8 m/s²

Hence, the weight of the body, W = 793.8 m/s²

3 0

3 years ago