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velikii [3]
3 years ago
5

A cart is pushed at a supermarket for a distance of 30 m South, then 15 m West and finally 5 m West. This entire motion lasts fo

r 30 minutes. What was the average speed of the cart? What was the average velocity of the cart?
(how would i find the average speed and velocity?)
Physics
2 answers:
babunello [35]3 years ago
5 0

Let the displacement vector S be 30 m South and W 20 m West

S^2 + W^2 = D^2   where D is the magnitude of the total displacement vector

D = (30^2 + 20^2)^1/2 = 36.1 m

So the magnitude of the velocity is D / 30 = 1.20 m/min

The direction of the displacement is given by

tan theta = 20 /30 = 2/3    and theta = 33.7 deg

So the average velocity is 1.2 meter/min at 33.7 deg West of South

The average speed is 50 m / 30 min = 1.67 meter/min

PSYCHO15rus [73]3 years ago
3 0

the speed would have to be somewhere between 1.667 or 1.666 miles per minute.

add 30, 15, 5 together and you will get 50

50 is the total miles traveled

then you multiply 30 by 1.666 and you will get 49.98 but it's still not quite 50 miles

if you multiply 30 by 1.667 and you will get 50.01 but its a little over to miles

i got the 30 from the minutes traveled

anyway that's the best i got for speed

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A balloon filled with helium occupies 20.0 l at 1.50 atm and 25.0◦
bija089 [108]
At stp (standard temperature and pressure), the temperature is T=0 C=273 K and the pressure is p=1.00 atm. So we can use the ideal gas law to find the number of moles of helium:
pV=nRT
where p is the pressure (1.00 atm), V the volume (20.0 L), n the number of moles, T the temperature (273 K) and R=0.082 atm L K^{-1} mol^{-1} the gas constant. Using the numbers and re-arranging the formula, we can calculate n:
n= \frac{pV}{RT}= \frac{(1.00atm)(20.0L)}{(0.082 LatmK^{-1}mol^{-1})(273 K)}=0.89 mol
5 0
3 years ago
Please help I will mark you brainliest
Radda [10]

I believe the answer is a

7 0
3 years ago
Read 2 more answers
2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or
krek1111 [17]

a) The work done by the tree is -1.44\cdot 10^5 J

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

W=Fd

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

W=-1.44\cdot 10^5 J is the work done

d=50.0 cm = 0.50 m is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0
4 years ago
4. A student measures a temperature several times. The readings lie between 29.6 and 30.2 K. This
Alona [7]

Answer:

(C) apparently written incorrectly - it should be 29.9 +- .3 K

(read 29.9 plus or minus .3 K)

8 0
3 years ago
I can’t figure it out! <br> Please help I have a test tomorrow
nignag [31]

it's 20m/s i.e. no. (2)


7 0
3 years ago
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