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2 years ago

12

One car travels 40. meters due east in 5.0 seconds, and a second car travels 64 meters due west in 8.0 seconds. During their per

iods of travel, the cars definitely had the same

1 answer:

KengaRu [80]2 years ago

5 0

Answer:

They had the same speed.

Explanation:

It won't be velocity, because velocity is a vector quantity. Speed is scalar.

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The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

2 years ago

When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters fro

NISA [10]

Remember your kinematic equations for constant acceleration. One of the equations is $x_{f} = x_{i} + v_{i}(t) + \frac{1}{2} at^{2}$ , where $x_{f}$ = final position, $x_{i}$ = initial position, $v_{i}$ = initial velocity, t = time, and a = acceleration.

Your initial position is where you initially were before you braked. That means $x_{i}$ = 100m. You final position is where you ended up after t seconds passed, so $x_{f}$ = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was $v_{i}$ = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
$350\:m = 100\:m + (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\
250\:m = 498\:m +34.445\:s^{2}(a)\\
-248\:m = 34.445\:s^{2}(a)\\
a \approx -7.2 \: m/s^{2}$

Your acceleration is approximately $-7.2 \: m/s^{2}$ .

4 0

2 years ago

A diver explores a shallow reef off the coast of Belize. She initially swims d1 = 74.8 m north, makes a turn to the east and con

Nataly [62]

Answer:R=1607556m

θ=180degrees

Explanation:

d1=74.8m

d2=160.7km=160.7km*1000

d2=160700m

d3=80m

d4=198.1m

Using analytical method :

Rx=-(160700+75*cos(41.8))= -160755.9m

Ry= -(74.8+75sin(41.8))-198.1=73m

Magnitude, R:

R=√Rx+Ry

R=√160755.9^2+20^2=160755.916

R=160756m

Direction,θ:

θ=arctan(Rx/Ry)

θ=arctan(-73/160755.9)

θ=-7.9256*10^-6

Note that θ is in the second quadrant, so add 180

θ=180-7.9256*10^6=180degrees

8 0

2 years ago

An ambulance driving 35.0 m/s emits a sound wave with a wavelength of 80.0 centimeters. As it drives away from a hospital, which

katen-ka-za [31]

Apparent frequency heard by the staff: 389 Hz

Explanation:

The phenomenon described in this situation is called Doppler effect.

Doppler effect occurs when there is a source emitting a wave in relative motion with respect an observer. In such situation, the frequency of the wave as perceived by the observer ("apparent frequency") is shifted from the real frequency of the sound ("proper frequency"). In particular:

- The observer perceives a higher frequency if the source is moving towards them

- The observer perceives a lower frequency if the source is moving away from them

The formula to calculate the apparent frequency in the Doppler effect is

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the proper frequency

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if they are moving towards the source, negative if moving away)

$v_s$ is the velocity of the source (positive if it is moving away, negative if moving towards the observer)

First of all, in this problem we have to calculate the proper frequency of the sound wave emitted from the ambulance; we have:

v = 343 m/s (speed of sound wave)

$\lambda=80 cm = 0.80 m$ (wavelength)

So the proper frequency is

$f=\frac{v}{\lambda}=\frac{343}{0.80}=429 Hz$

Now we can calculate the apparent frequency heard by the staff at the hospital when the ambulance moves away; we have:

$v_s = +35.0 m/s$ (velocity of the ambulance)

$v_o = 0$ (velocity of the staff)

Substituting,

$f'=\frac{343+0}{343+35}(429)=389 Hz$

Learn more about frequency and wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 0

2 years ago

would an elephant standing on one leg exert a higher force on a scale than an elephant on four legs. why

zlopas [31]

Answer:

no becaus force is mass multiplied by acceleration. the mass of the elephant does not change

7 0

2 years ago