How many times is larger than a centigram is a dekagram

A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular t

Lisa [10]

Explanation:

Let us assume that piece 1 is $m_{1}$ is facing west side. And, piece 2 is $m_{2}$ facing south side.

Let $m_{1} \text{and} m_{2}$ = m and $m_{3}$ = 2m. Hence,

$2mv_{3y} = m(21)$

$v_{3y} = \frac{21}{2}$

= 10.5 m/s

$2mv_{3x} = m(21)$

$v_{3x} = \frac{21}{2}$

= 10.5 m/s

$v_{3} = \sqrt{v^{2}_{3x} + v^{2}_{3y}}$

= $\sqrt{(10.5)^{2} + (10.5)^{2}}$

= 14.84 m/s

or, = 15 m/s

Hence, the speed of the third piece is 15 m/s.

Now, we will find the angle as follows.

$\theta = tan^{-1} (\frac{v_{3y}}{v_{3x}})$

= $tan^{-1}(\frac{10.5}{10.5})$

= $tan^{-1} (45^{o})$

= $45^{o}$

Therefore, the direction of the third piece (degrees north of east) is north of east.

5 0

3 years ago

A student says that a speed of 50 m/s is faster than a speed of 140 km/h because the number is bigger. What would you say to the

Yuri [45]

if you convert these into miles per hour 50 m/s would be higher, since

50 m/s = 111.85 mph and

140 km/h = 86.99 mph

3 0

3 years ago

A bicycle is ridden along a horizontal road with a driving force of 400, n,400n. its speed is constant at 12, m, slash, s,12m/s.

mr_godi [17]

The magnitude of the sum of the frictional forces acting on the bike and its rider is 400N.

<h3>What is friction force?</h3>

The friction force is the opposing force which acts on the object which is in relative motion.

The driving force is equal and opposite to the friction force acting between road and bicycle.

Friction force = 400N

The friction force between rider and bike is zero.

So the magnitude of sum of friction force = 400N +0 = 400N

Thus, the magnitude of the sum of the frictional forces acting on the bike and its rider.

Learn more about friction force.

brainly.com/question/1714663

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4 0

1 year ago

A mason is shot with a constant speed of 7.5 x 10²m/sinto a region, when an electric field produces acceleration on the mason of

nadya68 [22]

Answer:

Distance, d = 0.1 m

It is given that,

Initial velocity of meson,

Finally, the meson is coming to rest v = 0

Acceleration of the meson, (opposite to initial velocity)

Using third equation of motion as :

s is the distance the meson travelled before coming to rest.

So,

s = 0.1 m

The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.

5 0

2 years ago

State Newton's three laws of motion.

tresset_1 [31]

Answer:

Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Newton's third law states that if an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A. This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself.

Explanation:

6 0

3 years ago