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zepelin [54]
2 years ago
10

Evan is just leaving his house to visit his grandmother. Normally, the trip takes him 35 minutes on the freeway, going 70 mph. B

ut tonight he's running 5 minutes late. How fast will he need to drive on the freeway to make up the 5 minutes?
Physics
1 answer:
sdas [7]2 years ago
7 0

Answer:

Evan drive at speed = 81.67 m/hr

Explanation:

Normal speed of Evan = 70 m/hr

Time taken by Evan to visit his grandmother = 35 min = \frac{35}{60} hr

Distance covered by Evan = {speed}\times{time taken} = 70\times\frac{35}{60}

                                            = \frac{245}{6} miles

According to the problem, one day Evan was running 5 minutes late. So, the time Evan has to visit his grandmother is 30 mins.

To reach on time, Evan should be drive at the speed :

Speed = \frac{Distance}{Time}

           = \frac{\frac{245}{6} }{\frac{30}{60} }

          = 81.67 m/hr

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You are wallpapering two walls of a room. One wall measures 15 ft by 12 ft and the other measures 9 ft by 12 ft. The wall paper
enyata [817]

Answer:

5.76 round off to 6

Explanation:

wall 1 = 15 × 12 = 180

wall 2 = 9 × 12 = 108

now 1 roll covers 50 square feet

formula = wall 1 + wall 2 / 50

= 180 + 108 / 50

= 288÷ 50

= 5.76

4 0
2 years ago
If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
Butoxors [25]
(a) 3.56 m/s 
(b) 11 - 3.72a 
(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56 
So the velocity is 3.56 m/s  
(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
 y' = 11 - 22
 y' = -11 
 So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
3 0
2 years ago
A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga
andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

Cp-Cv= R

Cp=7/2 R

Now by putting the values

\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}

\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

Process is adiabatic

\Delta S_{surr}=0

\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}

\Delta S_{univ}= 20.79 +0

\Delta S_{univ}= 20.79

3 0
2 years ago
Find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α. express your answer in terms of ω1
VMariaS [17]
T1 = ω1/α.............
3 0
2 years ago
Read 2 more answers
1. A student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.
Snezhnost [94]

Explanation:

(a) Displacement of an object is the shortest path covered by it.

In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.  She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.

0.4 miles = 0.64 km

displacement = 0.7-0.3+0.64 = 1.04 km

(b) Average velocity = total displacement/total time

t = 15 min = 0.25 hour

v=\dfrac{1.04\ km}{0.25\ h}\\\\v=4.16\ km/h

Hence, this is the required solution.

8 0
2 years ago
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