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Artist 52 [7]
3 years ago
13

You are driving home from school steadily at 91 km/h for 160 km . It then begins to rain and you slow to 64 km/h instantly. You

arrive home after driving 4.0 hours.How far is your hometown from school?
Physics
1 answer:
Korolek [52]3 years ago
3 0

Total distance = (average speed) x (total time)

What do we know, and what else can we figure out ?

-- Total time = 4 hours.

-- First piece of the trip:

. . . distance . . . 160 km

. . . speed . . . 91 km/h.  

. . . time . . . (160 km) / (91 km/hr) = 1.758 hours

-- Second piece of the trip:

. . . speed . . . 64 km/h

. . . time . . . (4 - 1.758) = 2.242 hours

. . . distance (64 km) x (2.242 hr) = 143.47 km

What we're trying to find is the total distance.  Looking back over our work for the two pieces of the trip, we see that we have the distance for each piece now, so all we have to do is addummup.

Total distance = (160 km) + (143.47 km)

<em>Total distance = 303.47 km</em>

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Answer: Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source. You can find total resistance in a Parallel circuit with the following formula: 1/Rt = 1/R1 + 1/R2 + 1/R3 +.

Hope this helps!

3 0
3 years ago
An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t
Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0

4 0
4 years ago
What's the difference mass and weight
yanalaym [24]

Answer:

Mass doesn't change.

Weight is measured based on gravitational pull.

Explanation:

4 0
2 years ago
An airport has runways only 198 m long. a small plane must reach a ground speed of 39 m/s before it can become airborne. what av
Anarel [89]
S: 198 m 
v=39 m/s 
u=0
t=? 
a=?

v²=u²+2as
(39)²=(0)²+2(a)(198) 
1521=396a
1521/396=a
3.84 m/s^2 = a 

Hope I helped :) 
8 0
3 years ago
A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement do
Lubov Fominskaja [6]

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

\theta=90-40=50^o

From the triangle OAB

cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}

1340.57=35^2+25^2-x^2

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

4 0
3 years ago
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