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2 years ago

Newton’s Third Law:

1 answer:

5 0

If one force acting in one direction is greater than the force acting in the opposite direction, the object will move.

<h3>What is Newton's third law of motion?</h3>

Newton's third law of motion state that every action has an equal and opposite reaction. It is an action-reaction principle. It stated that the force always exists in a pair.

Hence,if one force acting in one direction is greater than the force acting in the opposite direction, the object will move.

To learn more about Newton's third law of motion, refer to the link;

brainly.com/question/1077877

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To convert centimeters to kilometers, which conversion factors would you need?

Lelechka [254]

Answer:

The last option is the only correct one if you like to multiply

The second last option is good if you like to divide.

Explanation:

Each fraction in the last two options has a value of 1

example

dividing by 1

15 cm /(100 cm/ 1 m) = 0.15 m 0.15 m / (1000 m/ 1km) = 0.00015 km

and

multiplying by 1

15 cm(1 m / 100cm) = 0.15 m 0.15m(1 km/1000m) = 0.00015 km

only one of the two fractions in each of the top two options has a value of 1.

3 0

2 years ago

Hello, I want to ask. . anyone knows the answer.

stealth61 [152]

I would say D. because you round to the nearest whole number and 0.04 is way less than 0.5 which is a good rounding up number.

5 0

2 years ago

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ ...Newtons second law Fnet = ma

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ ...f is the friction which is zero here.

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0

3 years ago

Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer dista

nadezda [96]

Distance = (speed) x (time)

Car A: Distance = (8 m/s) x (43 s) = 344 meters

Car B: Distance = (7 m/s) x (50 s) = 350 meters

350 meters is a longer distance than 344 meters.

Car-B traveled a longer distance than Car-A did.

5 0

2 years ago

Read 2 more answers

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

5 0

2 years ago