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Bess
9 days ago
5

Which graph represents the motion of an object traveling with a positive velocity and a negative acceleration?

Physics
2 answers:
Keith_Richards [2.2K]9 days ago
6 0
I think the right choice according to the illustration above is the second option. The graph demonstrating the motion of an object moving with a positive velocity and a negative acceleration is graph 2, which shows a curve that increases in y as x rises. I hope this resolves the question.
Yuliya22 [2.4K]9 days ago
5 0
Answer
graph 2

Explanation
Velocity refers to the speed at which displacement changes.
Velocity = (change in displacement)/(change in time)
In graph 2, the slope of the curve is positive, indicating a positive velocity.
Acceleration indicates how velocity varies.
The incline of the same curve (graph 2) is diminishing. This suggests that the velocity is falling, signifying negative acceleration.


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A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h
Yuliya22 [2420]

Answer:

Speeds of 1.83 m/s and 6.83 m/s

Explanation:

Based on the law of conservation of momentum,

mv_o=m(v_1 + v_2)where m represents mass, v_o is the initial speed before impact, v_1 and v_2 are the velocities of the impacted object after the collision and of the originally stationary object after the impact.

5m=m(v_1 +v_2)

Thus, v_1+v_2=5

After the collision, the kinetic energy doubles, therefore:

2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})

2v_o^{2}=v_1^{2} + v_2^{2}

Substituting the initial velocity of 5 m/s provides the equation needed to proceed.v_o

2*(5^{2})= v_1^{2} + v_2^{2}We know that v_1+v_2=5 leads to v_1=5-v_2

50=(5-v_2)^{2}+ v_2^{2}

50=25+v_2^{2}-10v_2+v_2^{2}

2v_2^{2}-10v_2-25=0

Using the quadratic formula leads us to solve for the speeds after the explosion, specifically where a=2, b=-10, and c=-25. v_2=6.83 m/s

By substituting the values, the solution yields results for the speeds of the blocks, which are ultimately 1.83 m/s and 6.83 m/s.

6 0
11 days ago
Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0
inna [2205]

Answer:

Distance: 4.6 km Displacement= -0.2 km

Explanation:

The overall distance covered: 1.5 + 2.4 + 0.7 = 4.6 km

Displacement calculation: 1.5 - 2.4 + 0.7 = -0.2 km

The displacement could also simply be stated as 0.2 km depending on whether negative value is preferred.

7 0
14 days ago
A camera operator is filming a nature explorer in the Rocky Mountains. The explorer needs to swim across a river to his campsite
inna [2205]

Answer:

a. Angle= 28.82°

b. Approved. Although he might feel cold, he should be able to cross.

Explanation:

Velocity Vector

Velocity is a measure of how quickly something is moving in a specific direction. It is represented as a vector that has both magnitude and direction. If an object can only move in one direction, then speed can serve as the scalar equivalent of that velocity (only focusing on magnitude).

a.

The explorer aims to swim across a river to reach his campsite, as depicted in the image below. The river's velocity is vr and the explorer's swimming speed in still water is ve. If he were to swim straight towards the campsite, he would end up downstream due to the river's current. Therefore, he must swim at an angle that allows him to overcome the current while still moving towards his goal. This angle relative to the shore is what we need to determine. The explorer's speed can be broken down into its horizontal (vx) and vertical (vy) components. In order to counteract the river's flow:

v_{ey}=v_r

We can calculate the vertical component of the explorer's swimming speed as

v_{ey}=|v_e|cos\alpha

Thus

v_r=|v_e|cos\alpha

Finding the value of \alpha

\displaystyle cos\alpha=\frac{v_r}{|v_e|}

\displaystyle cos\alpha=\frac{0.665}{0.759}=0.876

Then the angle is given by

\alpha=28.82^o

b.

The component of the explorer's velocity that goes horizontally is

v_{ex}=0.759sin28.82^o

v_{ex}=0.366\ m/s

This represents the actual velocity directed towards the campsite

Considering that

\displaystyle v=\frac{x}{t}

To find t

\displaystyle t=\frac{x}{v}

Calculating the duration for the explorer to cross the river

\displaystyle t=\frac{29.3}{0.366}

t=80\ sec

As this time is under the hypothermia threshold (300 seconds), the conclusion is

Approved. Although he will feel cold, he should manage to cross successfully.

3 0
27 days ago
The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect
serg [2593]
Rx= 3.5 km

Ry= 2.9 km
4 0
10 days ago
Read 2 more answers
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
Maru [2337]

1) 9.18 s

During the initial phase, the rocket accelerates at

a_1=13.5 m/s^2

over a time interval of

t_1=3.50 s

The final velocity after this period is found using the SUVAT formula:

v_1=u+a_1t_1

with initial velocity u = 0. Substituting a1 and t1 gives:

v_1=(13.5)(3.50)=47.3 m/s

Afterward, the rocket decelerates uniformly at

a_2 = -5.15 m/s^2

until it stops, meaning the final velocity is

v_2 = 0

Again using the SUVAT formula,

v_2 = v_1 + a_2 t_2

and knowing v_1 = 47.3 m/s, solve for t2, the time until the rocket halts:

t_2 = -\frac{v_1}{a_2}=-\frac{47.3}{5.15}=9.18 s

2) 299.9 m

Calculate distances covered in both phases.

Distance in the first phase:

d_1 = ut_1 + \frac{1}{2}a_1 t_1^2

Substituting values from part 1,

d_1 = 0 + \frac{1}{2}(13.5) (3.50)^2=82.7 m

Distance in the deceleration phase:

d_2= v_1 t_2 + \frac{1}{2}a_2 t_2^2

Substituting known values,

d_2 = (47.3)(9.18) + \frac{1}{2}(-5.15) (9.18)^2=217.2 m

Total distance traveled is

d = 82.7 m + 217.2 m = 299.9 m

6 0
1 month ago
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