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jarptica [38.1K]

3 years ago

7

a car moving at 30.0 m/s slows uniformly to a speed of 10.0 m/s in a time of 5.00 s determine the. acceleration of the car

1 answer:

netineya [11]3 years ago

7 0

Answer: $-4m/s^{2}$

Explanation:

Acceleration $a$ is defined as the variation of Velocity $\Delta V$ in time $\Delta t$ :

$a=\frac{\Delta V}{\Delta t}$

In this case $\Delta V=10m/s - 30m/s$ and $\Delta t=5 s$ .

Therefore:

$a=\frac{10m/s - 30m/s}{5 s}$

$a=\frac{-20m/s}{5 s}$

$a=-4 m/s^{2}$ Note the negative sign indicates the car is slowing

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You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength

Afina-wow [57]

Answer:

λ = 451.7 nm

Explanation:

The expression for the constructive interference of the double diffraction experiment is

d sin θ = m λ

let's use trigonometry

tan θ = y / L

how the experiment occurs at very small angles

tan θ = sin θ / cos θ = sin θ

sin θ = y / L

we substitute

d y / L = m λ

λ = $\frac{d \ y}{m \ L}$

let's calculate

λ = $\frac{0.342 \ 10^{-3} \ 2.80 \ 10^{-3} }{1 \ 2.12 }$

λ = 4.51699 10⁻⁷ m

λ = 4.517 10⁻⁷ m (109 nm / 1m)

λ = 451.7 nm

5 0

3 years ago

What is the frequency of light with a wavelength of 7.9 x 10^-9 m? ( the speed of light is 3.00 x 10^8)

nlexa [21]

Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.

so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f

3 0

3 years ago

Jonas sees this symbol in a circuit diagram.

Alenkinab [10]

The circuit component the symbol represents is: C) Battery

3 0

4 years ago

Read 2 more answers

Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8

sergejj [24]

Answer:

The hottest temperature is $T_2 = 39^o C$

Explanation:

From the question we are given

$T_1 = 98 F$

$T_2 = 39^oC$

$T_3 = 303 \ K$

Generally converting $T_3$ to Fahrenheit

$T_3' = (T_3 -273 ) * \frac{9}{5} + 32$

=> $T_3' = (303 -273 ) * \frac{9}{5} + 32$

=> $T_3' = 86 F$

Converting $T_2$ to Fahrenheit

$T_2' = T_2 * \frac{9}{5} + 32$

=> $T_2' = 39 * \frac{9}{5} + 32$

=> $T_2' =102.2 F$

Now comparing the temperature in Fahrenheit we see that $T_2$ is the hottest

3 0

3 years ago

Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, in

olya-2409 [2.1K]

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N$

7 0

4 years ago