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liubo4ka [24]
3 years ago
9

The graph below represents the relationship between speed and time for an object moving along a straight line. What is the total

distance traveled by the object during the first 4 seconds
Physics
1 answer:
KiRa [710]3 years ago
4 0

The distance travelled by the object during the first 4 seconds is 80 m

<h3>Definition of speed </h3>

Speed is defined as the distance travelled per unit time. Mathematically, it can be expressed as:

Speed = distance / time

With the above formula, we can obtain the distance travelled by the object in the first 4 seconds.

<h3>How to determine the distance travelled </h3>
  • Speed = 20 m/s
  • Time = 4 s
  • Distance =?

Speed = distance / time

20 = distance / 4

Cross multiply

Distance = 20 × 4

Distance = 80 m

Complete question:

See attached photo

Learn more about speed:

brainly.com/question/680492

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The velocity v of a particle moving in the xy plane is given by v = (6.0t -4.0t2 )i + 8.5j, in m/s. Here v is in m/s and t (for
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Answer:

Explanation:

Given

Velocity of the particle in vector form is given by

v=\left ( 6t-4t^2\right )\hat{i}+8.5\hat{j}

acceleration is rate of change of velocity thus acceleration is

a=\frac{\mathrm{d} v}{\mathrm{d} t}

a=(6-8t)\hat{i}+0\hat{j}

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Answer:

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3 0
3 years ago
Under what environmental conditions are you most likely to generate static electricity?
pogonyaev
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5 0
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Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
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