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Nataly_w [17]
3 years ago
13

Katie designed an experiment to show a water cycle process. The steps of the experiment are shown below.

Physics
1 answer:
GuDViN [60]3 years ago
6 0
Precipitation makes the most sense here. In basic terms, infiltration refers to the process whereby water enters the soil, transpiration refers to the movement of water out of a plant into the atmosphere and runoff is the flow of water over the earth's surface
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4. If you swing wide to the left before turning right, another driver may try to pass you on the right. True or False
olchik [2.2K]

Answer:

True.

Explanation:

Don't turn wide to the left as you start the turn. A driver behind may think you are turning left and try to pass you on the right. You may crash into the other vehicle as you complete your turn.

Instead, slowly give yourself and others more time to avoid problems, keep the rear of the vehicle close to the curb. This will stop other drivers from passing you on the right. This is called (button Hook)

If you are driving a truck or bus that cannot make the right turn without swinging into the other lane, turn wide as you complete the turn.

8 0
3 years ago
Question 9 (28 points)
ladessa [460]
Im sure the answer is letter B
5 0
3 years ago
Read 2 more answers
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
Help pleasessssssssssssss
irinina [24]

Sorry I had the answer but it wont let me type numbers :(.:

5 0
2 years ago
Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If
jeka57 [31]

Let F_{1}=F_{2}=F.

Normal force equals (using Newton's third law) N=mg+F\sin30^{o}-F\sin37^{o}.

F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o}), but F_{f}\leq F(\cos30^o+\cos37^o) for all F_{f} (in order to start moving the break). Therefore F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o}), solving for F: F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}

7 0
3 years ago
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