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Montano1993 [528]
2 years ago
8

Can somebody help meeeee...?​

Physics
1 answer:
Luba_88 [7]2 years ago
7 0

Answer:

{ \boxed{ \rm{pressure =   \frac{weight}{area} }}} \\  \\ { \rm{pressure =  \frac{200}{2} }} \\  \\ { \underline{ \underline{ \rm{ \:  \: pressure = 100 \: pascals \:  \: }}}}

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A shopper walks westward 2.8 meters in the first 5 minutes and then eastward 9.2 meters in the next 10 minutes. What is the shop
Alecsey [184]

Answer:

12m

Explanation:

Given parameters:

Distance walked westward  = 2.8m

Time of travel  = 5min

Distance walked eastward = 9.2m

Time of travel  = 10min

Unknown:

The total shopper's travel distance  = ?

Solution:

Total distance traveled is the sum of the length of path covered by a body. It is a scalar quantity.

 Total distance = distance walked westward + distance walked eastward

  Total distance  = 2.8m + 9.2m  = 12m

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2 years ago
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The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole huma
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Explanation:

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6 0
3 years ago
Name the type of component that has a greater resistance as the current through it increases​
Gnesinka [82]

Answer:

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Explanation:

4 0
3 years ago
A 500 kg sack of coal falls vertically onto a 2000 kg railroad flatcar which was initially moving horizontally at 3 m/s. no exte
Zinaida [17]

Since there are no external forces, including friction, act on the flatcar. after the sack rests on the flatcar, we would assume that momentum is conserved. This means that

total momentum of car before collision = total momentum of car after collision.

Recall,

momentum = mass x velocity

From the information given,

mass of car before collision = 2000

velocity of car before collision = 3

Thus,

total momentum of car before collision = 2000 x 3 = 6000

Also,

mass of sack = 500

mass of car and sack after collision = 500 + 2000 = 2500

velocity after collision = v

momentum after collision = 2500 x v = 2500v

Since momentum is conserved, then

6000 = 2500v

v = 6000/2500

v = 2.4

the speed of the flatcar is 2.4 m/s

6 0
10 months ago
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