A wagon is pulled at a speed of 0.40metets/seconds by a horse exerting an 1,800 Newton's horizontal force. what is the power of this horse?

1. How fast is the blue car going 1.8 seconds after it starts?

Recall this kinematic equation:

Vf = Vi + aΔt

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.

Given values:

Vi = 0 m/s (the car starts at rest)

a = 3.7 m/s² (this is the acceleration between t = 0s and t = 4.4s)

Δt = 1.8 s

Substitute the terms in the equation with the given values and solve for Vf:

Vf = 0 + 3.7×1.8

Vf = 6.66 m/s

2. How fast is the blue car going 10.0 seconds after it starts?

The car stops accelerating after t = 4.4s and continues at a constant velocity for the next 8.3 seconds. This means the car is traveling at a constant velocity between t = 4.4s and t = 12.7s. At t = 10s the car is still traveling at this constant velocity.

We must use the kinematic equation from the previous question to solve for this velocity. Use the same values except Δt = 4.4s which is the entire time interval during which the car is accelerating:

Vf = 0 + 3.7×4.4

Vf = 16.28 m/s

The constant velocity at which the car is traveling at t = 10s is 16.28 m/s

3. How far does the blue car travel before its brakes are applied to slow down?

We must break down the car's path into two parts: When it is traveling under constant acceleration and when it is traveling at constant velocity.

Traveling under constant acceleration:

Recall this kinematic equation:

d = ×Δt

d is the distance traveled.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.

Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 16.28 m/s (determined from question 2).

Δt = 4.4 s

Substitute the terms in the equation with the given values and solve for d:

d = ×4.4

d = 35.8 m

Traveling at constant velocity:

Recall the relationship between velocity and distance:

d = vΔt

d is the distance traveled.

v is the velocity.

Δt is the amount of elapsed time.

Given values:

v = 16.28 m/s (the constant velocity from question 2).

Δt = 8.3 s (the time interval during which the car travels at constant velocity)

Substitute the terms in the equation with the given values:

d = 16.28×8.3

d = 135.1 m

Add up the distances traveled.

d = 35.8 + 135.1

d = 170.9 m

4. What is the acceleration of the blue car once the brakes are applied?

Recall this kinematic equation:

Vf²=Vi²+2ad

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration

d is the distance traveled.

Given values:

Vi = 16.28 m/s

Vf = 0 m/s

d = 216 m - 170.9 m = 45.1 m (subtracting the distance already traveled from the total path length)

Substitute the terms in the equation with the given values and solve for a:

0² = 16.28²+2a×45.1

a = -2.94 m/s²

5. What is the total time the blue car is moving?

We already know the time during which the car is traveling under constant acceleration and traveling at constant velocity. We now need to solve for the amount of time during which the car is decelerating.

Recall again:

d = ×Δt

Given values:

d = 45.1 m

Vi = 16.28 m/s (the velocity the car was traveling at before hitting the brakes).

Vf = 0 m/s (the car slows to a stop).

Substitute the terms in the equation with the given values and solve for Δt:

45.1 = ×Δt

Δt = 5.54s

Add up the times to get the total travel time:

t = 4.4 + 8.3 + 5.54 =

t = 18.24s

6. What is the acceleration of the yellow car?

Recall this kinematic equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.

Given values:

d = 216 m (both cars meet at 216m)

Vi = 0 m/s (the car starts at rest)

Δt = 18.24 s (take the same amount of time to reach 216m)

Substitute the terms in the equation with the given values and solve for a:

216 = 0×18.24 + 0.5a×18.24²

a = 1.3 m/s²

Answer:

Explanation:

When the craft was stationary , weight will be balanced by tension

T = mg

T = 7000 N

A)

when the craft was being lowered to the seafloor

drag force will act in upper direction , so

T₁ + 1800 = mg

T₁  = mg - 1800

= 7000 - 1800

= 5200 N

52 X 10² N

B)

when the craft was being raised from the seafloor , Tension will act in downward direction

T₂ = mg+ 1800

T₂  = 7000 - 1800

= 8800N

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

t=1s ,1.5s,2s

Development of problem

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

a) t=1s

=15-4.9=10.1m

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

b) t=1.5s

=22.5-11.025=11.475 m

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

c) t=2s

= 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)

Answer;

Proof Surrogate.

Explanation;

A claim masquerading as proof or evidence, when no such proof or evidence is actually being offered.

Proof Surrogate is an expression used to suggest that there is evidence for a claim without actually citing any evidence.  For example;

-“Scientists have known that aliens exist for years now.”

-“Everyone knows that bottled water is better for you than tap water.”