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boyakko [2]
4 years ago
12

How is speed represented on a speedometer ? how is acceleration represented?

Physics
1 answer:
Marianna [84]4 years ago
8 0
Speed is scalar quantity that is defined only by its magnitude and is represented with one value. 
Velocity on the other side is a vector quantity that is defined with its magnitude and direction. The speedometer measures and displays the speed of the vehicle.
Where the needle points on the speedometer that is the speed of the vehicle. 
The velocity is not explicitly shown, but if f the needle is moving<span>, you can get a rough idea of acceleration. If the needle is going clockwise the vehicle is accelerating.</span>
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Determine the magnitude and direction of the resultant velocity of 75.0 m/s. 25.0 east of north, and 100.0 m/s, 25.0 east of sou
aleksandr82 [10.1K]

Answer:

77.35 m / s

Ф = -17° from + X axis or 343° from + X axis

Explanation:

v1 = 75 m/s 25° east of north

v2 = 100 m/s  25° east of south

Write the velocities in vector form ,we get

\overrightarrow{v_{1}}=75\left ( Sin25\widehat{i} +Cos25\widehat{j}\right )=31.7\widehat{i}+67.97\widehat{j}

\overrightarrow{v_{1}}=100\left ( Sin25\widehat{i} -Cos25\widehat{j}\right )=42.26\widehat{i}-90.63\widehat{j}

Now add the velocity vectors to get the resultant of the velocities.

\overrightarrow{v}=\overrightarrow{v_{1}}+\overrightarrow{v_{2}}

\overrightarrow{v}=\left (31.7+42.26  \right )\widehat{i}+\left ( 67.97- 90.63 \right )\widehat{j}

\overrightarrow{v}=73.96\widehat{i}-22.66\widehat{j}

magnitude of resultant velocity is \sqrt{\left ( 73.96 \right )^{2}+\left ( -22.66 \right )^{2}}

  = 77.35 m / s

The direction is Ф from X axis

tan\phi =\frac{-22.66}{73.96}=-0.306

Ф = -17° from + X axis or 343° from + X axis

7 0
4 years ago
A large box sits on a rough floor. A person pushes on the box with a horizontal force of magnitude 50 N. The box remains at rest
artcher [175]

Answer:

The magnitude of the frictional force is   F_f \ge 50 \ N

Explanation:

From the question we are told that

   The force exerted on the box is  F =  50 \  N

Generally for the box to remain at rest then it means that the frictional  force is greater than or equal to the force applied to move it i.e

         F_f \ge F

=>      F_f \ge 50 \ N

5 0
3 years ago
A scientific law seeks to explain why an event occurred. True or false
Nostrana [21]
The answer would be A) True.
8 0
4 years ago
Read 2 more answers
11) Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a speed of 3.0 m/s and Betty has a sp
sleet_krkn [62]

Answer:

100 m

Explanation:

Arthur and Betty should be walking the same amount of time if they start walking at the same time and stop when they meet = t

Speed of Arthur = 3 m/s

Speed of Betty = 2 m/s

Distance = Speed × time

Distance covered by Arthur = 3t

Distance covered by Betty = 2t

The distance covered by both of them will be 100 m

3t+2t=100\\\Rightarrow 5t=100\\\Rightarrow t=\frac{100}{5}\\\Rightarrow t=20\ s

The speed of dog is 5 m/s

Spot is running back and forth at 5 m/s for 20 seconds

In 20 seconds the distance covered by the dog is

\mathbf{5\times 20}=\mathbf{100\ m}

5 0
4 years ago
Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.99 J of work to set the ball spinnin
scoundrel [369]

To solve this problem it is necessary to apply the concepts related to rotational kinetic energy, the definition of the moment of inertia for a sphere and the obtaining of the radius through the circumference. Mathematically kinetic energy can be given as:

KE= I\omega^2

Where,

I = Moment of inertia

\omega = Angular velocity

According to the information given we have that the radius is

\Phi= 2\pi r

0.749m = 2\pi r

r = 0.1192m

With the radius obtained we can calculate the moment of inertia which is

I = \frac{2}{3}mr^2

I = \frac{2}{3}(0.624)(0.1192)^2

I = 5.91*10^{-3} kg \cdot m^2

Finally, from the energy equation and rearranging the expression to obtain the angular velocity we have to

\omega = \sqrt{\frac{2KE}{I}}

\omega = \sqrt{\frac{2(1.99)}{5.91*10^{-3}}}

\omega = 25.95rad/s

Therefore the angular speed will the ball rotate is 25.95rad/s

8 0
3 years ago
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