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3 years ago

If a car was cruising down the highway at 80 km/hr, how fast would the people inside the car be traveling?

2 answers:

6 0

Because the people in the car are attached to the vehicle, the people inside the vehicle are going the same speed as the vehicle.

Hope this helps! :)

luda_lava [24]3 years ago

3 0

The people inside the car are at rest (speed = zero) relative to the car, moving 80 km/hr relative to the pavement, and slightly slightly less relative to a worm slithering on the road in the same direction as the car.

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Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to

jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

#SPJ4

7 0

2 years ago

(Will give brainliest)

Sveta_85 [38]

Answer:

b.) Length

Explanation:

The length of the string can be changed by removing it from the slotted bracket and placing it back in. You can change the mass by varying the number of washers on the mass hanger. The amplitude can be changed by varying the starting angle of the pendulum (low, medium, and high angle). sorry if wrong

8 0

3 years ago

An electron of mass 9.11×10-31kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to t

SOVA2 [1]

Answer:

(A) Acceleration will be $240.3846\times 10^{12}m/sec^2$

(b) Time taken will be $1.4\times 10^{-8}sec$

Explanation:

We have given that electron starts from rest so initial velocity u = 0 m/sec

Final velocity $v=2.50\times 10^6m/sec$

Mass of electron $m=9.11\times 10^{-31}kg$

Distance traveled by electron $s=1.30cm =0.013m$

From third equation of motion we know that $v^2=u^2+2as$

(a) So $(2.5\times 10^6)^2=0^2+2\times a\times 0.013$

$a=240.3846\times 10^{12}m/sec^2$

(b) From first equation of motion we know that v = u+at

So $2.50\times 10^6=0+240.3846\times 10^{12}t$

$t=0.014\times 10^{-6}=1.4\times 10^{-8}sec$

$F=ma=9.11\times 10^{-31}\times 240.3846\times 10^{12}=2189.9\times 10^{-19}N$

7 0

3 years ago

What us the difference in the ways objects move at a speed of a car and an object mkvinf close to the speed of light?

Charra [1.4K]

Answer:

The difference is in who or what is observing the speed.

Explanation:

Giving that speed is relative between the objects and the reference point from which it is being observed.

It is concluded that speed alone has no direct effect on a moving object, hence it is just a determining unit for the difference in distance between two objects.

Therefore, in this case, the difference is in who or what is observing the speed.

5 0

2 years ago

A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels in the negative x direction with a speed of 30.0 m

lesya [120]

Answer:

a) A =0.021525m

b) $\phi=0.37869rad$

c) $v_{max}=5.4098\frac{m}{s}$

d) $y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

Explanation:

1) Notation

A= Amplitude

v= velocity

$\lambda$ = wavelength

k= wave number

$\omega$ = angular frequency

f= frequency

2) Part a and b

The equation of movement for a transverse sinusoidal wave is gyben by (1)

$y(t)=Acos(kx+ \omega t +\phi)$ (1)

At x=0 ,t=0 we have that:

$0.02=Acos(\phi)$

The velocity would be the derivate of the position, so taking the derivate of (1) respect to t we got (2)

$v(t)=-\omega Asin(kx+ \omega t+\phi)$ (2)

And replacing the conditions at x=0, t=0 we got

$-2\frac{m}{s}=-\omega Asin(\phi)$

Now we can find the angular frequency with equation (3)

$\omega =\frac{2\pi}{T}$ (3)

Replacing the values obtained we got:

$\omega =\frac{2\pi}{0.025s}=80\pi \frac{rad}{s}$

From equation (1) we have:

$Acos(\phi)=0.02$ (a)

$-2=-80\pi Asin(\phi)$ (b)

So from condition (b) we have:

$Asin(\phi)=\frac{1}{40\pi}$ (c)

If we divide condition (c) by condition (a) we got:

$\frac{Asin(\phi)}{Acos(\phi)}=tan(\phi)=\frac{1}{0.02x40\pi}=\frac{1}{0.8\pi}=0.39789$

If we solve for $\phi$ we got:

$\phi =tan^{-1}(0.39789)=0.37869$

And now since we have $\phi$ we can find A from equation (a)

$Acos(0.37869)=0.02$

So then Solving for A we got $A=\frac{0.02}{cos(0.37869)}=0.021525$

3) Part c

From equation (2) we can see that the maximum speed occurs when $sin(\omega t+\phi)=1$ , so on this case we have:

$v_{max}=\omega A=80\pi \frac{rad}{s}x0.021525m=5.4098\frac{m}{s}$

4) Part d

On this case we need an equation like (1), and we have everything except the wave number, and we can obtain this from the following expression:

$v=\lambda f=\frac{2\pi}{k}\frac{\omega}{2\pi}=\frac{\omega}{k}$ (4)

And solving for k from equation (4) we got

$k=\frac{\omega}{v}=\frac{80\pi \frac{rad}{s}}{30\frac{m}{s}}=\frac{8\pi}{3}m^{-1}}$

And with the k number we have everythin in order to create the wave function, given by:

$y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

7 0

3 years ago