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1 year ago

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A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2 , which force is closest to the net force

acting on the vehicle?
Answer choices:
A. -9600N
B. -1300N
C. -900N
D. -94N

2 answers:

larisa [96]1 year ago

8 0

Answer:

D

Explanation:

boyakko [2]1 year ago

7 0

Answer:

Explanation:

D

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A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n

Y_Kistochka [10]

A. 90.1 m

The wavelength of a wave is given by:

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is its frequency

For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is

$\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m$

B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

$f=\frac{v}{\lambda}$

where for the dolphin:

v = 1531 m/s is the wave speed

$\lambda=1.50 cm=0.015 m$ is the wavelength

Substituting into the equation,

$f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz$

C. 13.6 m

Again, the wavelength is given by:

$\lambda=\frac{v}{f}$

where

v = 340 m/s is the speed of sound in air

f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

$\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m$

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

- Wavelength corresponding to the minimum frequency (f=39.0 Hz):

$\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m$

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

$\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m$

So the range of wavelength is 4.4-8.7 m.

E. 6.2 MHz

In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so

$\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m$

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

$f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz$

3 0

2 years ago

A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later

UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]

= (64.2 m/s) - [ 14.7 m/s ]

= 49.5 m/s . (upward)

7 0

2 years ago

A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average

frez [133]

(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
</span>
<span>0 = 20 + a(4)

</span><span>4 x a = -20
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>

8 0

2 years ago

Read 2 more answers

A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo

padilas [110]

Answer:

a) λ = 189.43 10⁻⁹ m b) λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

d sin θ= m λ

Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

tan θ = sin θ / cos θ = sin θ

We replace

d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

d = λ L / y

d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

λ = d y / L

λ = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

λ = 189.43 10⁻⁹ m

Part B

The compound wavelength B

λ = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

λ = 269.19 10⁻⁹ m

4 0

2 years ago

Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc

xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0

1 year ago