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3 years ago

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The area of the velocity time graph gives displacement of the body true or false?

2 answers:

mote1985 [20]3 years ago

8 0

The area of the velocity time graph gives displacement of the body true or false?

The answer this is true.

zaharov [31]3 years ago

6 0

the correct one is the first option

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50 POINTS‼️‼️‼️‼️‼️

Inessa05 [86]

Answer:

Explanation:

the yellow box is -5

3 0

3 years ago

Read 2 more answers

A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and t

Readme [11.4K]

Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s

Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis ( tan^-1 (13.5 / 16.4)

5 0

3 years ago

NEED HELP ASAP<br><br>ONLY ANSWER IF YK THE ANSWERS

nalin [4]

Answer:

there yah go that's the answer

6 0

3 years ago

A solid, uniform disk of radius 0.250 m and mass 45.2 kg rolls down a ramp of length 5.40 m that makes an angle of 17.0° with th

serious [3.7K]

Explanation:

Given that,

Radius of the disk, r = 0.25 m

Mass, m = 45.2 kg

Length of the ramp, l = 5.4 m

Angle made by the ramp with horizontal, $\theta=17^{\circ}$

Solution,

As the disk starts from rest from the top of the ramp, the potential energy is equal to the sum of translational kinetic energy and the rotational kinetic energy or by using the law of conservation of energy as :

(a) $mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

h is the height of the ramp

$sin\theta=\dfrac{h}{5.4}$

$h=sin(17)\times 5.4=1.57\ m$

v is the speed of the disk's center

I is the moment of inertia of the disk,

$I=\dfrac{1}{2}mr^2$

$\omega=\dfrac{v}{r}$

$mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}\dfrac{1}{2}mr^2\times (\dfrac{v}{r})^2$

$gh=\dfrac{1}{2}v^2+\dfrac{1}{4}v^2$

$gh=\dfrac{3}{4}v^2$

$9.8\times 1.57=\dfrac{3}{4}v^2$

v = 4.52 m/s

(b) At the bottom of the ramp, the angular speed of the disk is given by :

$\omega=\dfrac{v}{r}$

$\omega=\dfrac{4.52}{0.25}$

$\omega=18.08\ rad/s$

Hence, this is the required solution.

4 0

3 years ago

¿Cuál de las siguientes no es un tipo de fuerza de roce

Degger [83]

Enertia es el answer de tu question

8 0

3 years ago