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yawa3891 [41]
4 years ago
7

Two straight roads diverge at an angle of 45°. Two cars leave the intersection at 2:00 P.M., one traveling at 32 mi/h and the ot

her at 54 mi/h. How far apart are the cars at 2:30 P.M.? (Round your answer to two decimal places.)
Physics
1 answer:
svp [43]4 years ago
3 0

Answer:19.34 miles

Explanation:

Given

First car travels with 32 mi/h

second travels with 54 mi/h

they leave at 2:00 Pm at 45

distance traveled by them is 0.5 hr

suppose First is traveling towards x axis and another travels 45 to it

Distance traveled by first car is 32\times 0.5=16 miles

Position vector of first car after 0.5 hr

\vec{r_1}=32\times 0.5(\cos (45)\hat{i})

Position vector of second car after 0.5 hr

\vec{r_2}=54\times 0.5(\cos (45)\hat{i}+\sin (45)\hat{j})

distance between them

\vec{r_{21}}=27(\cos (45)\hat{i}+\sin (45)\hat{j})-16(\cos (45)\hat{i})

\vec{r_{21}}=3.09\hat{i}+27\hat{j}\sin (45)

distance between them|\vec{r_{21}}|=19.34

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a water-balloon launcher with mass 4 kg fires a 0.5 kg balloon with a velocity of 3 m/s to the east. what is the recoil velocity
kotykmax [81]
I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 =  4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s

Therefore; 4 V1 = 0.5 × 3
                   4V1= 1.5
                     V1= 1.5/4
                          = 0.375 m/s










5 0
3 years ago
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A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin
Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0
3 years ago
Convertir 1200 ms a cs<br> Convertir 0,3 mm a cm.
Vaselesa [24]

Answer:

You can do the reverse unit conversion from cm/s to m/s, or enter any two units below: Metre per second (U.S. spelling: meter per second) is an SI derived unit of both speed (scalar) and velocity (vector quantity which specifies both magnitude and a specific direction), defined by distance in metres divided by time in seconds.

Explanation:

3 0
3 years ago
Hey help me plzzzzz i will mark brainliest​
vodomira [7]

Answer:

The answer to your question is given below.

Explanation:

Mechanical advantage (MA) = Load (L)/Effort (E)

MA = L/E

Velocity ratio (VR) = Distance moved by load (l) / Distance moved by effort (e)

VR = l/e

Efficiency = work done by machine (Wd) /work put into the machine (Wp) x 100

Efficiency = Wd/Wp x100

Recall:

Work = Force x distance

Therefore,

Work done by machine (wd) = load (L) x distance (l)

Wd = L x l

Work put into the machine (Wp) = effort (E) x distance (e)

Wp = E x e

Note: the load and effort are measured in Newton (N), while the distance is measured in metre (m)

Efficiency = Wd/Wp x100

Efficiency = (L x l) / (E x e) x 100

Rearrange

Efficiency = L/E ÷ l/e x 100

But:

MA = L/E

VR = l/e

Therefore,

Efficiency = L/E ÷ l/e x 100

Efficiency = MA ÷ VR x 100

Efficiency = MA / VR x 100

7 0
3 years ago
The time constant for RC circuit with the values of R1 and C1 is 5ms. What will be the time constant for a new RC circuit with t
lions [1.4K]

Answer:

d. 25 ms

Explanation:

  • In a RC circuit we call time constant to the product of the resistance times the capacitance, which represents the time when the charge reaches to the 63% of the final value, as follows:

       \tau_{1} = R_{1} *C_{1}  = 5 ms (1)

  • If we have a new circuit with new values for R and C, the time constant will be defined in the same way, as follows:

       \tau_{2} =10* R_{1} *0.5*C_{1}  = 5*(R_{1}* C_{1}) = 5* \tau_{1} = 5* 5 ms = 25 ms (2)

6 0
3 years ago
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