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zhuklara [117]
3 years ago
6

Jack, a student, travels to school 10 miles every morning. The speed limit is 25 miles/hour. How long it takes Jack to get to hi

s school?
Physics
2 answers:
taurus [48]3 years ago
8 0

Answer:

24 minutes

Explanation:

• If the speed limit is 25miles/hour, it takes 2.4 minutes to travel 1 mile.

60mins/25miles

= 2.4 minutes

• Therefore to travel 10 miles;

2.4 minutes × 10

= 24 minutes

Stells [14]3 years ago
6 0

Answer:

0.4 hour or 24 minutes

Explanation:

Speed =displacement /time

25miles/hour=10/time

10/25=time

0.4hour=time

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An increase in average speed of 1 km/h typi- cally results in a 3% higher risk of a crash involving injury, with a 4–5% increase for crashes that result in fatalities. — Speed also contributes to the severity of the impact when a collision does occur. Hope this helps!
4 0
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What is the nineth plant closet to the sun
marusya05 [52]
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4 0
3 years ago
Read 2 more answers
What pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c?
Ronch [10]

The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

Therefore, option A is correct option.

Given,

Mass m = 14g

Volume= 3.5L

Temperature T= 75+273 = 348 K

Molar mass of CO = 28g/mol

Universal gas constant R= 0.082057L

Number of moles in 14 g of CO is

n= mass/ molar mass

= 14/28

= 0.5 mol

As we know that

PV= nRT

P × 3.5 = 0.5 × 0.082057 × 348

P × 3.5 = 14.277

P = 14.277/3.5

P = 4.0794 atm

P = 4.1 atm.

Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

learn more about pressure:

brainly.com/question/22613963

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5 0
1 year ago
What is the most important safety rule in science? A. Never work with chemicals. B. Always use unbreakable glassware. C. Always
sasho [114]
Well based on the three I would think it's C
4 0
3 years ago
A rookie quarterback throws a football with an initial upward velocity component of 17.0 m/s and a horizontal velocity component
vitfil [10]

Answer:

(a) 1.73 s

(b) 14.75 m

(c) 3.36 s

(d) double

(e) 63.32 m

Explanation:

Vertical component of initial velocity, uy = 17 m/s

Horizontal component of initial velocity, ux = 18.3 m/s

(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.

Use first equation of motion in vertical direction

vy = uy - gt

0 = 17 - 9.8 t

t = 1.73 seconds

(B) Let the highest point is at height h.

Use III equation of motion in vertical direction

v^{2}=u^{2}-2gh

0 = 17 x 17 - 2 x 9.8 x h

h = 14.75 m

(C) The time taken by the ball to return to original level is T.

Use second equation of motion i vertical direction.

h = ut + 0.5at^2

h = 0 , u = 17 m/s

0 = 17 t - 0.5 x 9.8 t^2

t = 3.46 second

(D) It is the double of time calculated in part A

(E) Horizontal distance = horizontal velocity x total time

d = 18.3 x 3.46 = 63.32 m

8 0
3 years ago
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