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charle [14.2K]
3 years ago
12

A car travels at a speed of 64 km/h for 45 minutes how far does it travel

Physics
1 answer:
Likurg_2 [28]3 years ago
7 0
<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Use the formula:

distance = speed x time

Convert 45 minutes into hours:

45 minutes ==> 0.75 hours

Substitute the values in:

distance = 64 x 0.75

Solve:

distance = 48km

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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Assume a reaction takes 7.5 moles of anhydrous calcium chloride and energy transfer occurs, which we record as 21.2 joules of en
OLga [1]

Answer: 2.83 J/mol

Explanation:

Heat of solution, sometimes interchangeably called enthalpy of solution, is said to be the energy released or absorbed when the solute dissolves in the solvent. A solute is that which can dissolve in a solvent, to form a solution

Given

No of moles of CaCl = 7.5 mol

Total energy used = 21.2 J

Heat of solution = q/n where

q = total energy

n = number of moles

Heat of solution = 21.2 / 7.5

Heat of solution = 2.83 J/mol

8 0
3 years ago
Read 2 more answers
Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the
joja [24]

As it is given that the air bag deploy in time

t = 10 ms = 0.010 s

total distance moved by the front face of the bag

d = 20 cm = 0.20 m

Now we will use kinematics to find the acceleration

d = v_i*t + \frac{1}{2}at^2

0.20 = 0 + \frac{1}{2}a*0.010^2

0.20 = 5 * 10^{-5}* a

a = 4000 m/s^2

now as we know that

g = 10 m/s^2

so we have

a = 400g

so the acceleration is 400g for the front surface of balloon

3 0
3 years ago
A 350-N child is in a swing that is attached to a pair of ropes 2.10 m long. Find the gravitational potential energy of the chil
o-na [289]

Answer:

a)  U = 735 J , b) U = 125.7 J , c)   U = 0 J

Explanation:

The gravitational power energy is

      U = mg y - mg y₀

The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part

a) Rope is horizontal

The height in this case is the same length of the rope

     y = 2.10 m

    w = mg = 350 N

    U = 350 2.10

    U = 735 J

b) when the angle is 34º

     y = L - L cos 34

    y = L (1- cos34)

    y = 2.10 (1- cos 34)

    y = 0.359 m

    U = 350 0.359

    U = 125.7 J

c) in this case this point coincides with the reference system

     y = 0

     U = 0 J

4 0
3 years ago
Please give answer with solution​
UNO [17]
250 m. for a longer explanation or solution look at this article, i’m sorry.
https://www.quora.com/A-projectile-is-thrown-so-it-travels-a-maximum-range-of-1000m-How-high-will-it-rise
3 0
2 years ago
You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving
Alecsey [184]
<h2>The balloon is moving when it is halfway down the building at 20.78 m/s.</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s  

Acceleration, a = 9.81 m/s²  

Displacement, s = 0.5 x 44 = 22 m

Substituting  

v² = u² + 2as

v² = 0² + 2 x 9.81 x 22

v² = 431.64

v = 20.78 m/s

Velocity at 22 m = 20.78 m/s

The balloon is moving when it is halfway down the building at 20.78 m/s.

7 0
3 years ago
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