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koban [17]
2 years ago
7

John and Caroline go out for a walk one day. This graph represents their distance from home.Which statement accurately describes

their walk?
They walked uphill quickly, then they were still, then they walked downhill slowly.
They walked faster and faster until they walked at a constant rate, then they walked slower and slower.
They walked quickly, then they were still, then they walked slowly.
They walked in one direction, then they were still, then they walked in the opposite direction.

Physics
2 answers:
kotegsom [21]2 years ago
7 0
I get where they need to go
Zepler [3.9K]2 years ago
3 0

Answer:

C) They walked in one direction, then they were still, then they walked in the opposite direction.  

Explanation:

yw.

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When I wave a charged golf tube at the front of the classroom with a frequency of two oscillations per second, I produce an elec
borishaifa [10]

To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.

From the definition we know that the wavelength is described under the equation,

\lambda = \frac{c}{f}

Where,

c = Speed of light (vacuum)

f = frequency

Our values are,

f = 2Hz

c = 3*10^8km/s

Replacing we have,

\lambda = \frac{c}{f}

\lambda = \frac{3*10^8km/s}{2Hz}

\lambda = 1.5*10^8m

<em>Therefore the wavelength of this wave is 1.5*10^{8}m</em>

8 0
2 years ago
Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ram
Sonbull [250]
<h2><u>Full Question:</u></h2>

Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ramps.Which statement is TRUE?

answer choices

Marble 1 has a faster speed than Marble 2.

Marble 2 has a faster speed than Marble 1.

Both the marbles travel at the same speed.

There is not enough data to compare the speeds of marbles.

<h2><u>Answer:</u></h2>

Marble 2 has a faster speed than Marble 1.

Option B.

<h3><u>Explanation:</u></h3>

The speed is defined as the distance covered per unit time. Here in the question, 2 balls cover equal distances in same time.

Time taken by the ball = 10 seconds.

Distance covered by 1st ball = 20 cm.

Distance covered by 2nd ball = 3cm.

So speed of the 1st ball = 2cm/sec.

Speed of the 2nd ball = 3 cm /sec.

So,it's very much evident that speed of 2nd Marble is much higher than the speed of the 1st marble.

5 0
3 years ago
1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms.
lidiya [134]

Answer:

The answer would be 0.04ohms.

Explanation:

Hopefully this helps

4 0
2 years ago
Read 2 more answers
A force of 960 newtons stretches a spring 4 meters. A mass of 60 kilograms is attached to the end of the spring and is initially
Drupady [299]

Answer:

x(t) = - 6 cos 2t

Explanation:

Force of spring = - kx

k= spring constant

x= distance traveled by compressing

But force = mass × acceleration

==> Force = m × d²x/dt²

===> md²x/dt² = -kx

==> md²x/dt² + kx=0   ------------------------(1)

Now Again, by Hook's law

Force = -kx

==> 960=-k × 400

==> -k =960 /4 =240 N/m

ignoring -ve sign k= 240 N/m

Put given data in eq (1)

We get

60d²x/dt² + 240x=0

==> d²x/dt² + 4x=0

General solution for this differential eq is;

x(t) = A cos 2t + B sin 2t   ------------------------(2)

Now initially

position of mass spring

at time = 0 sec

x (0) = 0 m

initial velocity v= = dx/dt=  6m/s

from (2) we have;

dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)

put t =0 and dx/dt = v(0) = -6 we get;

-2A sin 2(0)+2Bcos(0) =-6

==> 2B = -6

B= -3

Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get

x(t) = - 6 cos 2t

==>  

4 0
3 years ago
two objects are connected by a light string that passes over a frictionless pulley as in the figure below, where m1 A) a1=g<br>
Shtirlitz [24]
You didn't attach the figure. Your text is incomplete. And you never got around to asking a question. Other than that, we're on it.
5 0
3 years ago
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