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Jobisdone [24]
3 years ago
12

What is air A. A Buchner substance B. A compound C. An element D. A mixture

Physics
1 answer:
bezimeni [28]3 years ago
7 0
D. A mixture (mark brainliest)
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What is the formular for lmpulse​
Brums [2.3K]

Answer:

the change in momentum = Force x change in time

3 0
2 years ago
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Select the correct answer.
natali 33 [55]

The wave speed completely depends on the characteristics and properties of the medium  . . . physical properties for mechanical waves, electrical properties for electromagnedtic waves.

So if you want to change the speed of a wave, you have to change the medium . . . shoot it through some different kind of stuff. <em>(B) </em>

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3 years ago
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A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° 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       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

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

⇒ 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
Q. A train accelerates from 36 km/h to 54 km/h in 10 sec. (i) Acceleration (ii) The distance travelled by car.
Brrunno [24]

u=10m/s

v=15m/s

acceleration=

v_u/ t

5/10

0.5

5 0
3 years ago
A 2.10 X 10^3 kg car starts from rest at the top of a 5.0
xeze [42]

Answer:

3.80 m/s

Explanation:

Solution: From the question we have:

A 2.10 x 10^3 kg car starts from rest at the top of a 5.0 m long driveway that is sloped at 20 degrees with the horizontal.

average friction force of 4.0 x 10^3 N impedes the motion

To find out:  find the speed of the car at the bottom of the driveway.

=>h = 5×sin20 = 1.71 m.

PE = mgh = 2100×9.8×1.71 = 35,192 J.

KE = PE-work done by frictional force

KE=PE-4000×5

0.5mV^2 = 35192 - 4000×5

1050V^2 = 15,192

V^2 = 14.47

V = 3.80 m/s.

4 0
3 years ago
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