Fitness can be achieved only through

Help with i) and ii) pls >_

alina1380 [7]

Answer:

4 it the corrcert anser

Explanation:

4 0

2 years ago

Read 2 more answers

An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 k

vichka [17]

Answer:

1.) 0 kgm/s

2) 6.3 kg

3) -0.0978 m/s

4)

5)

6)

An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of θ1 = 23° above the –x axis with a speed of v1 = 25.4 m/s. A second piece with mass m2 = 5.2 kg moves down and to the right an angle of θ2 = 28° to the right of the -y axis at a speed of v2 = 23.8 m/s.

2) What is the mass of the third piece?

3) What is the x-component of the velocity of the third piece?

4) What is the y-component of the velocity of the third piece?

5) What is the magnitude of the velocity of the center of mass of the pieces after the collision?

6) Calculate the increase in kinetic energy of the pieces during the explosion

Explanation:

Since explosions and collisions follow the law of conservation of Momentum.

1) Magnitude of the final momentum of the system = Magnitude of the initial momentum of the system

Since the body was initially at rest,

Magnitude of the initial momentum of the system = 0 kgm/s

Hence, Magnitude of the final momentum of the system is also equal to 0 kgm/s.

2) mass of the third piece

Sum of all the masses = 16.2 kg

4.7 + 5.2 + c = 16.2

c = 6.3 kg

3) doing an x-component balance on momentum

(4.7)×(-25.4 cos 23) + (5.2)×(23.8 cos 28) + (6.3)(v) = 0

-0.6163 + 6.3v = 0

v = -0.0978 m/s

Hope this Helps!!!

4 0

2 years ago

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

Talja [164]

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

$F_{g}$ = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

$T$ = Tension force in upward direction

$F_{d}$ = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

$F_{g}$ - $F_{d}$ - $T$ = 0

7000 - 1800 - $T$ = 0

$T$ = 5200 N

$T$ = 5.2 x 10³ N

Part B)

$F_{g}$ = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

$T$ = Tension force in upward direction

$F_{d}$ = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

$T$ - $F_{g}$ - $F_{d}$ = 0

$T$ - 7000 - 1800 = 0

$T$ = 8800 N

$T$ = 8.8 x 10³ N

4 0

2 years ago

66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown st

kozerog [31]

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t Equation 2

y: The vertical distance the ball moves at time t

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t

Known information

We know the following data:

Vi=15 m / s

$g =9.8 \frac{m}{s^{2} }$

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)

a) t=1s

$y = 15*1 - ½ 9.8*1^{2}$ =15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

$y = 15*1.5 - ½ 9.8*1.5^{2}$ =22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

$y = 15*2 - ½ 9.8*2^{2}$ = 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

3 0

2 years ago

A cat jumps from 2.5-meter-tall bookshelf to a 1.3-meter-tall countertop. If the cat

UkoKoshka [18]

The change in Potential energy of the cat is 176.4 J.

<h3 /><h3>Potential Energy:</h3>

This is the energy due to the position of a body. The S.I unit is Joules (J)

The formula for change in potential energy.

<h3 /><h3>Formula:</h3>

ΔP.E = mg(H-h).............. Equation 1

<h3>Where:</h3>

ΔP.E = Change in potential energy
m = mass of the cat
g = acceleration due to gravity
H = First height
h = second height.

From the question,

<h3>Given:</h3>

m = 15 kg
H = 2.5 m
h = 1.3 m
g = 9.8 m/s²

Substitute these values into equation 1

ΔP.E = 15×9.8(2.5-1.3)
ΔP.E = 15×9.8×1.2
ΔP.E = 176.4 J.

Hence, The change in Potential energy of the cat is 176.4 J

Learn more about Potential energy here: brainly.com/question/1242059

5 0

1 year ago