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eduard
3 years ago
5

HELPPPPPPPP PLSSSSSSSS A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2

= 66 m at t2 = 6.0 s . Answer the following by considering ratios, without computing the car's velocity.\ 1.What will be its position at t = 24 s ? 2.What is the car's position at t = 3.0 s ?
Physics
2 answers:
yawa3891 [41]3 years ago
6 0

Answer:

Explanation:

6.2 miles

vodka [1.7K]3 years ago
6 0

Answer: 1. 264, 2.33

Explanation: Explanation: <u>66m= 6s</u> so, to find the position at <u>3s</u> you just need to take 66<u>/2</u> = 33m cause <u>3 is half of 6</u>. & for 2 you will take 66<u>x4</u>= 264m cause it took <u>4s multiply by the original 6s to get 24s</u>. Answer: 1 is 264m & 2 is 33m.

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A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp
raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

5 0
3 years ago
The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figu
abruzzese [7]

Answer:

\boxed{\sf Kinetic \ energy \ (KE) = 85 \ J}

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 6.8 \times  {5}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 3.4 \times 25

\sf \implies KE =3.4 \times 25

\sf \implies KE = 85 \: J

8 0
3 years ago
Read 2 more answers
An orange light (f = 5.2 * 10'4Hz) is
KonstantinChe [14]

Answer:

2.145×10^-10 V or 0.2145nV

Explanation:

From hf=eV

h= Plank's constant = 6.6×10^-34JS

f= frequency of the electromagnetic wave = 5.2×10^4 Hz

e= electronic charge= 1.6×10^-19 C

V= voltage

V= hf/e

V= 6.6×10^-34JS × 5.2×10^4 Hz/ 1.6×10^-19 C

V= 2.145×10^-10 V or 0.2145nV

Therefore the voltage created is 2.145×10^-10 V or 0.2145nV

7 0
3 years ago
A high school bus travels 240 km in 6.0 h. What is its average speed for the trip? (in km/h).
KengaRu [80]

Answer:

40 km/h

Explanation:

First...

Look at the formula speed is equal to the distance over time or s = d/t.

Next...

Use the formula: 240/6.0

Finally...

Solve: 40

So the answer: 40 km/h

5 0
3 years ago
What force must the deltoid muscle provide to keep the arm in this position?
ruslelena [56]

Answer:

Deltoid Force, F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, T_{Deltoid}

2. The torque of the arm, T_{arm}  

Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

                       T_{arm} = 0

       ⇒           T_{Deltoid} = T_{arm}

                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

Where,

r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

                ∴            F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

7 0
3 years ago
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