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soldi70 [24.7K]
3 years ago
10

A 700 kg car makes a turn going at 30 m/s with radius of

Physics
1 answer:
Gnoma [55]3 years ago
5 0

Answer:5250 N

Explanation: ig:iihoop.vince

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Which statement does NOT explain the concept of the Doppler Effect? A. As an observer moves away from a stationary sound, the so
Nookie1986 [14]

Answer:

  D

Explanation:

When the sound is moving away from the stationary object the pitch decreases according to the Doppler Effect.

4 0
3 years ago
A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig
atroni [7]

Answer:

The plane is 2353.7 mi from the starting position.

Explanation:

Please, see the attached figure for a graphic representation of the problem.

We have 2 displacement vectors "a" and "b" and a vector "c" that is the sum of vectors "a" plus "b" (c = a + b). The module of "c" will be the distance of the plane from the starting point.

vector a = (xa, ya)

vector b = (xb, yb)

where “xa” and “xb” are the horizontal components of the vectors “a” and “b” respectively and “ya” and “yb” are the vertical components of each vector.

Then, the vector c = a + b will be:

c = (xa + xb, ya + yb)

The module of a vector is calculated using the following expression for a vector “v”:

module of v = \sqrt{x^{2} + y^{2} }

Then, the module of c will be:

module of c = \sqrt{(xa + xb)^{2} + (ya + yb)^{2}} = distance from starting point

Then, we have to find the components of vectors “a” and “b”

The distance traveled during the first 1.5 hours of the trip is the module of the vector “a”. Then:

module of a = \sqrt{xa^{2} + ya^{2} } = distance traveled during the first 1.5 hours.

The distance can be calculated using the equation of the position of an object moving in a straight line at constant speed:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = speed

t = time

Considering x0 as the starting point (x0 = 0)

x = 675 mi/h * 1.5 h = 1012.5 mi

Then:

module of a = \sqrt{xa^{2} + ya^{2} } = 1012. 5 mi

Since the plane moves only on the horizontal (see figure), the "y" component of the vector, "ya", will be 0.

Then:

(1012.5 mi)² = xa²

xa = 1012. 5mi

a = (1012.5 mi, 0)

In the same way, we have fo find the components of the vector “b”. The module of “b” will be the distance traveled during this part of the flight:

module of b = \sqrt{xb^{2} + yb^{2} } = x = x0 + v * t

Considering x0 as the point at which the plane turns (x0 = 0)

x = 675 mi / h * 2 h = 1350 mi

Using trigonometry, we can calculate xb and yb (see figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In this case:

opposite = yb

adjacent = xb

hypotenuse = module of “b”

Then:

sin 10° = yb / module of “b”

sin 10° * module of “b” = yb

In the same way:

cos 10° * module of “b” = xb

Since module of “b” = 1350 mi

xb = 1329.5 mi

yb = 234.4 mi

b = (1329.5 mi, 234.4 mi)

The vector c = a+b can now be calculated:

c = (xa + xb, ya + yb)

c =(1012.5 mi + 1329.5 mi, 0 mi + 234.4 mi) = (2342 mi, 234.4 mi)

The module of c will be:

module of c = \sqrt{(2342 mi)^{2} + (234.4 mi)^{2} } = 2353.7 mi

The plane is 2353.7 mi from the starting position.

4 0
3 years ago
A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

4 0
4 years ago
Two forces and are applied to an object whose mass is 13.3 kg. The larger force is . When both forces point due east, the object
ANEK [815]

Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem.  Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them.  So let's call them F1 and F2, with F1 arger than F2.  Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.  

Can you solve this system of equations seeing them like this, or do you need more help?

6 0
3 years ago
Please, pretty please help me!
Andrews [41]

Answer:

umm

Explanation:

1223565 578633 =334675

8 0
3 years ago
Read 2 more answers
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