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2 years ago

Is the frictional force the same as the applied force when the net force equals zero?

Physics

1 answer:

DerKrebs [107]2 years ago

6 0

Answer:

Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces.

Explanation:

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Of all the radio waves, these have the shortest wavelengths and the the highest frequency?

Julli [10]

If by chance you mean any wave, the wave with the shortest wavelength/highest frequency is gamma rays.

4 0

3 years ago

A child in a tree house uses a rope attached to a basket to lift a 24 N dog upward through a distance of 4.9 m into the house.

jenyasd209 [6]

<h3>Answer:</h3>

117.6 Joules

<h3>Explanation:</h3>

<u>We are given;</u>

Force of the dog is 24 N
Distance upward is 4.9 m

We are required to calculate the work done

Work done is the product of force and distance
That is; Work done = Force × distance
It is measured in Joules.

In this case;

Force applied is equivalent to the weight of the dog.

Work done = 24 N × 4.9 m

= 117.6 Joules

Hence, the work done in lifting the dog is 117.6 Joules

3 0

3 years ago

A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco

Sunny_sXe [5.5K]

Answer:

(a)

$x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km$

(b) $r=41.32km$

$\alpha =24.23^o$

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies $45^o$ south of east. The y-component will be negative and the x-component will be positive.

$x_1=25cos45^o=17.68km$

$y_1=-25sin45^o=-17.68km$

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

$x_2=40cos60^o=20km$

$y_2=40sin60^o=34.64km$

Therefore, total displacements:

$x=x_1+x_2=(17.68+20)km=37.68km$

$y=y_1+y_2=(-17.68+34.64)km=16.96km$

Magnitude of displacements,

$r=\sqrt{x^2+y^2}=41.32km$

Direction,

$\alpha =tan^{-1}(\frac{y}{x})=24.23^o$

6 0

3 years ago

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)