What is the speed of a 0.145kg baseball if its kinetic energy is 109J?

A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-

Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

$v_{s_{x}}=0$

$v_{s_{y}}=2.0\ m/s$

The velocity of the boat in term x and y coordinate

For x component,

$v_{b_{x}}=v_{b}\cos\theta$

Put the value into the formula

$v_{b_{x}}=5.60\cos19$

$v_{b_{x}}=5.29\ m/s$

For y component,

$v_{b_{y}}=v_{b}\sin\theta$

Put the value into the formula

$v_{b_{y}}=5.60\sin19$

$v_{b_{y}}=1.82\ m/s$

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

$v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}$

Put the value into the formula

$v_{sb_{x}=0-5.29$

$v_{sb}_{x}=-5.29\ m/s$

For y component,

$v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}$

Put the value into the formula

$v_{sb_{x}=2.-1.82$

$v_{sb}_{x}=0.18\ m/s$

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

7 0

3 years ago

A body weighing 108N moves with speed of 5m/s in a horizontal

poizon [28]

Answer:

i) 5 $m$ / $s^{2}$ ii) 54 $N$ iii) 54 $N$

Explanation:

i) $a = \frac{v^{2}}{r}$ ⇒ $a$ = 5² ÷ 5 = 5 $m/s^{2}$

ii) $m = \frac{W}{g}$ ⇒ $m$ = 108 ÷ 10 = 10.8 $kg$ , $F = ma$ ⇒ $F =$ 10.8 × 5 = 54 $N$

iii) F1 = F2 = 54 $N$

4 0

3 years ago

What requirement must a force acting on a object satisfy in order for the object to undergo simple harmonic motion?

viktelen [127]

Answer:

Simple harmonic motion is the movement of a body or an object to and from an equilibrium position. In a simple harmonic motion, the maximum displacement (also called the amplitude) on one side of the equilibrium position is equal to the maximum displacement.

The force acting on an object must satisfy Hooke's law for the object to undergo simple harmonic motion. The law states that the force must be directed always towards the equilibrium position and also directly proportional to the distance from this position.

6 0

3 years ago

If you are an astronaut in the middle of the near side of the moon during a full moon,how would the ground around you look?How w

Radda [10]

The moon would be bright and the earth would be darker because the sun is on the opposite side of the earth at that time and the light from it is reflecting off the moon to produce light upon the nigh also.......

You wouldn’t see the sun a night...

Unless you lived in the north/south pole

3 0

3 years ago

Read 2 more answers

.

beks73 [17]

Answer:

2 m/s²

Explanation:

the equations of motion are

S= ut +½at²

v² = u²+ 2as

v = u + at

s = (u+v)/2 × t

From the parameters given

u = 0m/s this is because it starts from rest

Distance (s) = 9m

Time (t) = 3s

Based on this the first equation would be used

s = ut + ½at²

Input values

9 = 0×3 + ½ × a x 3²

9 = 0 + 9a/2

9 = 4.5a

Divide both sides by 4.5

a = 9 / 4.5 m/s²

a = 2 m/s²

I hope this was helpful, please mark as brainliest

3 0

4 years ago