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Alina [70]
3 years ago
5

Where would a car traveling on a roller coaster have the most potential energy

Physics
1 answer:
sveta [45]3 years ago
6 0

At the top because it has the most potential to  go down. meaning it is a greater distance

c. At the top

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What is the resultant of a and b if a = 3i 3j and b = 3i − 3j?
ss7ja [257]
 <span>a+b= ? 
3i +3j + (3i -3j) = ? 
3i + 3j + 3i -3j =? 
= 6i + 0j</span>
3 0
3 years ago
Read 2 more answers
The strength of the force of friction depends on which two factors?
Stolb23 [73]

Options A and D are correct. The strength of the force of friction depends on the objects' sizes and weights and the heat generated by the friction and the types of surfaces involved.

<h3 /><h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration in the different components and balancing the equation gets. Components in the x-direction.

The strength of the force of friction depends on the two factors, as;

A. The objects' sizes and weights.

D. The heat generated by the friction and the types of surfaces involved.

Hence, options A and D are correct.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

#SPJ1

3 0
2 years ago
The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.
sattari [20]

Answer:

d=691.71km

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

t=\frac{d}{v_t}-\frac{d}{v_l}

Here d is the distance at which the earthquake take place and v_t, v_l is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km

8 0
3 years ago
You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just
Zepler [3.9K]

Answer:

The angular magnification is M = 2.808

Explanation:

From the question we are told

           The focal length is  f = 8.9cm

          The near point is n_p = 25.0cm

The angular magnification is mathematically represented as

                          M = \frac{n_p}{f}

Substituting values

                        M = \frac{25}{8.9}

                           = 2.808

4 0
3 years ago
The route followed by a hiker consists of three displacement vectors, X, Y and Z. Vector X is along a measured trail and is 1430
poizon [28]

Answer:

  • magnitude : 1635.43 m
  • Angle: 130°28'20'' north of east

Explanation:

First, we will find the Cartesian Representation of the \vec{X} and \vec{Y} vectors. We can do this, using the formula

\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | its the magnitude of the vector and θ the angle. For  \vec{X} we have:

\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )

\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )

where the unit vector \hat{i} points east, and \hat{j} points north. Now, the \vec{Y} will be:

\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )

Now, taking the sum:

\vec{X} + \vec{Y} + \vec{Z} = 0

This is

\vec{Z} = - \vec{X} - \vec{Y}

(Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )

(Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 2200 m \ - \ 956.86 m \ )

(Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 1243.14 m\ )

Now, for the magnitude, we just have to take its length:

|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}

|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}

|\vec{Z}| = 1635.43 m

For its angle, as the vector lays in the second quadrant, we can use:

\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m})

\theta = 180\° - arctan( -1.1720)

\theta = 180\° - 45\°31'40''

\theta = 130\°28'20''

5 0
3 years ago
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