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tensa zangetsu [6.8K]
2 years ago
10

According to the graph, during which time interval are the particles in the air slowing down?

Physics
2 answers:
kenny6666 [7]2 years ago
4 0
Answer b like the last time should be it
eimsori [14]2 years ago
3 0
Answer should be b 6:00pm to 12:00 am
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Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.
Ad libitum [116K]

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

\theta= arctan(\frac{v_y}{v_x})=arctan(\frac{4.0 m/s}{6.0 m/s})=arctan(0.67)=33.7^{\circ}

relative to the direction of the river.


b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

t=\frac{S_y}{v_y}=\frac{360 m}{4.0 m/s}=90 s


c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

S_x = v_x t =(6.0 m/s)(90 s)=540 m

so, Dave's landing point is 540 m downstream.


d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

7 0
4 years ago
You weigh 716 newtons on Earth. You
Otrada [13]

Answer:

537 N

Explanation:

The force due to gravity of a planet is:

F = GMm / r²

where G is the universal gravitational constant

M is the mass of the planet

m is the mass of the object

and r is the distance between the object and the center of the planet

On Earth, you weigh 716 N, so:

716 N = GMm / r²

On planet X:

F = G (3M) m / (2r)²

F = 3/4 GMm / r²

F = 3/4 (716 N)

F = 537 N

8 0
4 years ago
2. Un ciclista que está en reposo comienza a pedalear hasta alcanzar los 16.6km/h en 6 minutos. Calcular la distancia total que
jek_recluse [69]

Answer:

s = 5822.28\,m\,(5.822\,km)

Explanation:

La distancia recorrida es la suma de las etapas de aceleración y velocidad constante (Travelled distance is the sum of the acceleration and constant speed stages). La aceleración es (The acceleration is):

a = \frac{4.611\,\frac{m}{s}-0\,\frac{m}{s}}{360\,s}

a = 0.013\,\frac{m}{s^{2}}

La distancia recorrida es (The travelled distance is):

s = \frac{1}{2}\cdot \left(0.013\,\frac{m}{s^{2}} \right)\cdot (360\,s)^{2} + \left(4.611\,\frac{m}{s} \right)\cdot (1080\,s)

s = 5822.28\,m\,(5.822\,km)

3 0
3 years ago
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0
3 years ago
Difference between interpolation and extrapolation.
AlladinOne [14]

Answer:

Interpolation gives more accurate data or points compared to extrapolation.

Interpolation can be a little easier than extrapolation.

Interpolation does not require one to extend the already existing data points, where extrapolation requires elaboration of the pattern, curve, or line.

Extrapolation:

This is a statistical method used to predict unknown values for points outside the range of the recorded data. It is used to extend a known sequence of values beyond the sampled area

Extrapolation: with Extrapolation, we can be less confident with the unknown values derived using this method. The method includes values from outside the sampled area, which makes the predictions diverge away from the true values. In curve fitting, this method is not preferred.

Extrapolation:

the ability to retrieve negative answers is one characteristic that makes it unique for interpolation. This statistical method allows one to extend the values in the range either forward or backward. This technique of extending values is responsible for attaining the negative values.

Explanation:

8 0
3 years ago
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