A high powered projectile is fired horizontally from the top of a cliff at a speed of 638.6 m/s. Determine the magnitude of the velocity (in m/s) after 5 seconds.

Take gravitational acceleration to be 9.81 m/s2.

2-

A man throws a ball with a velocity of 20.9 m/s upwards at 33.2° to the horizontal. At what vertical distance above the release height (in metres) will the ball strike a wall 13.0 m away ?

Take gravitational acceleration to be 9.81 m/s2.

3-

A particle is moving along a straight path and its position is defined by the equation s = (1t3 + -5t2 + 3) m, where t is measured in seconds. Determine the average velocity (in m/s) of the particle when t = 5 seconds.

4-

A particle has an initial speed of 26 m/s. The particle undergoes a deceleration of a = (-9t) m/s2, where t is measured in seconds. Determine the distance (in metres) the particle travels before it stops. When t = 0, s = 0.

Answer:

**Answer:**

1**.**V= 640.48 m/s :total velocity in t= 5s

2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. v =25m/s

4. s= (-1.5t³+26t ) m

**Explanation:**

1. Parabolic movement in the x-y plane , t=5s

V₀=638.6 m/s=Vx :Constant velocity in x

Vy=V₀y +gt= 0+9.8*5 = 49 m/s : variable velocity in y

**V= 640.48 m/s **: total velocity in t= 5s

2.

x=v₀x*t

13=v₀x*t

13=17.49*t

**t=13/17.49=0.743s : time for 13.0 m away**

th=v₀y/g=11.44/9.8= 1,17s :time for maximum height

at t=0.743 sthe ball is going up ,then g is negative

y=v₀y*t - 1/2 *g¨*t²

y=11.44*0.743 -1/2*9.8*0.743²

**y= 5.79m **: vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. s = (1t3 + -5t2 + 3) m

v=3t²-10t=3*25-50=75-50=25m/s

at t=0, s=3 m

at t=5s s=5³-5*5²+3

4. a = (-9t) m/s2

a=dv/dt=**-**9t

dv=-9tdt

v=∫ **-**9tdt

v=**-**9t²/2 + C1 equation (1)

in t=0 , v₀=26m/s ,in the equation (1) C1= 26

v=**-**9t²/2 + 26=ds/dt

ds=( **-**9t²/2 + 26)dt

s= ∫( **-**9t²/2 + 26)dt

s= **-**9t³/6+26t+C2 Equation 2

t = 0, s = 0 , C2=0

s= (**-**9t³/6+26t ) m

**s= (-1.5t³+26t ) m**

The position of a particular particle as a function of is given by r=(8.5t i+5.6j-2tk )m (a) Determine the particle’s velocity and acceleration as a function of time.(b) Find the instantaneous velocity and acceleration of the particle at t=3.0 s.

When a box is placed on an inclined surface with no friction, it will:

A celestial body moving in an ellipical orbit around a star

An object is constrained by a cord to move in a circular path of radius 1-m on a frictionless, horizontal surface. The cord will break if the tension exceeds 26.9-N. The maximum kinetic energy that this object can have is _____ J. Round your answer to the nearest whole number.

Instantaneous speed is...a) A speed of 1000 km/hb) The speed attained at a particular instant in time.c) The speed that can be reached in a particular amount of time.PLEASE HURRY

When a box is placed on an inclined surface with no friction, it will:

A celestial body moving in an ellipical orbit around a star

An object is constrained by a cord to move in a circular path of radius 1-m on a frictionless, horizontal surface. The cord will break if the tension exceeds 26.9-N. The maximum kinetic energy that this object can have is _____ J. Round your answer to the nearest whole number.

Instantaneous speed is...a) A speed of 1000 km/hb) The speed attained at a particular instant in time.c) The speed that can be reached in a particular amount of time.PLEASE HURRY

**Answer:**

projectiles

electromagnetic

**Answer:**

**Explanation:**

física cuántica y Quantum Moves

B. It won't move.

C. It will move at the same speed that the large object was moving initially.

D. It will move slower than the large object was moving initially.

**Answer:**

a ut will move faster than the large object was moving initially

**Answer: It will move faster than the large object was moving initially.**

**Explanation:**

B) must be accelerating

C) may be slowing down

D) may be moving at constant speed

**Answer:**

hmmm thats too hard for me.

**Explanation:**

The area of this triangle can be calculated using herons formula since th three sides are given. It is expressed as:

A=sqrt( s(s-a)(s-b)(s-c))

where s is equal to a+b+c / 2

s=15 +20 +25 /2=30

A=sqrt( 30(30-15)(30-20)(30-25))

A= 150 cm^3

A=sqrt( s(s-a)(s-b)(s-c))

where s is equal to a+b+c / 2

s=15 +20 +25 /2=30

A=sqrt( 30(30-15)(30-20)(30-25))

A= 150 cm^3

**Answer:**

11.625 m.

**Explanation:**

Difference of pressure will be due to hydro-static pressure due to column of water of height L.

Pressure of water column = L d g , where L is depth ,

d is density of water = 1000kg /m³

g = 9.8 ms²

Pressure difference = 9.3 kPa = 9300 Pa

So Ldg = 9300

L X 1000 X 0.8 =9300

800 L = 9300

L = 11.625 m.

1. Which terms (if any) are positive?

2. Which terms (if any) are negative?

3. Which terms (if any) are zero?

b) Determine the energy output by the athlete in SI unit.

c) Determine his metabolic power in SI unit.

d) Another day he performs the same task in 1.2 s.

1. Is the metabolic energy that he expends more, less, or the same?

2. Is his metabolic power more, less, or the same?

**Answer:**

**Explanation:**

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

Positive, negative, and zero terms in the **energy equation**. Calculation of energy output and metabolic power. Comparison of metabolic energy and power for different time durations.

To apply the energy equation to the system, we need to determine whether each term is positive, negative, or zero:

- Positive terms:

- ΔUg - the change in gravitational potential energy is positive as the barbell is lifted vertically from the ground.
- ΔUs - the change in elastic
**potential**energy is positive if there is any stretch or compression in the system.

- ΔK - the change in kinetic energy is negative as the barbell is lifted at a constant speed, so there is no change in velocity.
- ΔEch - the change in chemical potential energy is negative if the athlete is not ingesting any food or drinks during the exercise.

- ΔEth - the change in thermal energy is zero if there is no heat transfer in the system.

To determine the energy output by the athlete, we can calculate the work done on the barbell using the formula W = ΔUg. In this case, the work done is equal to the change in **gravitational **potential energy, which is equal to mgh. Thus, W = 400 N × 2.0 m = 800 J. So the energy output by the athlete is 800 J.

The metabolic power can be calculated using the equation P = W / t, where P is the power, W is the work done, and t is the time taken. Substituting the given values, P = 800 J / 1.6 s = 500 W. Therefore, the **metabolic** power of the athlete is 500 W. If the task is performed in a faster time, the metabolic energy expended will be the same. However, the metabolic power will be greater as the work is done in less time.

#SPJ11