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AnnyKZ [126]
3 years ago
8

Finish this equation useing the term "average speed" and "time interval ": total distance =

Physics
2 answers:
Natasha2012 [34]3 years ago
8 0

Total distance during the time interval  = 

         (average speed during the time interval)
times
         (length of the time interval) . 
enyata [817]3 years ago
8 0
It's speed equals ( distance Divided by time ) I hope this helps
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Can anyone explain how tides work
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The position of the sun and the moon affect how high the tide is 
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3 years ago
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The maximum force that a grocery bag can withstand without ripping is 250 n. Suppose that the bag is filled with 20 kg of grocer
zhannawk [14.2K]

Answer:

Groceries stay in the bag.

Explanation:

Given:

Maximum force = 250 N

Bag filled with = 20 kg

Lifted acceleration = 5.0\ m/s^2

Solution:

We need to calculate the exerted force on the grocery bag by using Newton's second law.

F = ma

Where:

F = Exerted force on the object.

m = Mass of the object in kg

a = Acceleration of the object in m/s^2

Now, we substitute m = 20 kg and a = 5.0\ m/s^2 in Newton's second law,

F = 20\times 5.0

F = 100\ m/s^2

Since, the exerted force on the bag is less than 250 N, the groceries will stay in the bag.

3 0
3 years ago
A bird with a mass of 0.88 kg has a potential energy of 96 J. How high off the ground is the bird?
Ray Of Light [21]

Answer:

Data:-m=0.88kg ,g=9.8m/sec² ,P.E=96J ,h=?

Explanation:

solution ,P.E=mgh here we have to find h so h=P.E/mg ,h=96/0.88×9.8 ,h=96/8.624=11.131m and if you want to verify so just put the value of h in same formula, likewise :-P.E=mgh ,P.E=0.88×9.8×11.131=96J so we got the same value of P.E as it is given the question (verified).

5 0
3 years ago
A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire
oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v).  From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

6 0
3 years ago
Two groups of students were tested to compare their speed working math problems,Each group was given the same problems.One group
Zanzabum
Your hypothesis can be that the group with calculators would finish faster than the group without calculators.
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