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scoray [572]
3 years ago
10

For this car, predict the shape of a graph that shows distance (x) versus time (t). Note that time is the independent variable a

nd will be on the bottom axis.
NEED HELP ASAP
Physics
2 answers:
Mariana [72]3 years ago
6 0
Idk I haven’t learned this yet
nata0808 [166]3 years ago
5 0

Answer:

I expect this to be a linear graph with a constant, positive slope, where x increases linearly as t increases.

Explanation:

I out a different answer but this is the correct one. Hope this helps!

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A swimmer is capable of swimming at 1.4m/s in still water. a. How far downstream will he land if he swims directly across a 180m
ozzi

Answer:

t = 180 / 1.4 = 129 sec   (time to swim horizontally across river)

S = 129 sec * V     where V is speed of current and S is the distance he will be carried downstream

The problem does not specify V the speed of the river

8 0
3 years ago
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3. The coefficient of kinetic friction between an object and the surface upon which it is sliding is
Neporo4naja [7]

Answer:

2

Explanation:

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A bock of lead has deminsions of 4.50cm by 5.20cm by 6.00cm, what is the volume
lawyer [7]

Answer:

Look at this picture this is the answer

Explanation:

6 0
3 years ago
A girl starts from rest and reaches a walking speed of 1.4 m/s in 3.0 s. She walks at this speed for 6.0 s. The girl then slows
Alona [7]

Answer:

1) The girl's acceleration at time 't' is, a=0.47 m/s²

2) The girl's acceleration at time 't₁' is, a_{1}=0 m/s²

3) The girl's acceleration at time 't₂' is,  a_{2}=-0.14 m/s²

Explanation:

Given data,

The initial walking speed of the girl, u = 0

The speed of the girl at the time 't' 3 s is, v₁ = 1.4 m/s

The time period the girl walked at the speed 1.4 m/s is, t₁ = 6 s

The girl slows down and comes to a stop during a period, t₂ = 10 s

1) The girl's acceleration at time 't'

                                a=\frac{v-u}{t}

                                a=\frac{1.4-0}{3}

                               a=0.47 m/s²

2) The girl's acceleration at time 't₁'

                                a_{1} =\frac{v_{1} -v}{t_{1}}

                                a_{1}=\frac{1.4-1.4}{6}

                               a_{1}=0 m/s²

3) The girl's acceleration at time 't₂'

                                a_{2} =\frac{v_{2} -v_{1}}{t_{2}}

                                a_{2}=\frac{0-1.4}{10}

                               a_{2}=-0.14 m/s²

4 0
3 years ago
(Thanks in advance)
Dmitrij [34]

I would say it B) in the same direction feel free to ask me more question

5 0
3 years ago
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