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alina1380 [7]
3 years ago
8

In track, there is a 1500 meter event. Is this distance more or less than a mile?

Physics
1 answer:
fenix001 [56]3 years ago
5 0
Less than because a mile is 1600 meters
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Physics question, answer completely with work
Alenkasestr [34]

The force is 2.0 N east

Explanation:

The impulse exerted by a force is defined as the product between the force itself and the time interval during which the force is applied. Mathematically, it is equal to the change in momentum experienced by the object on which the force is acting:

I=F\Delta t = \Delta p

Where

I is the impulse

F is the force

Delta t is the time interval during which the force is applied

\Delta p is the change in momentum

In this problem,

\Delta t = 3.0 s is the time interval

I=6.0 N\cdot s (east) is the impulse

Therefore, the magnitude of the force is

F=\frac{I}{\Delta t}=\frac{6.0}{3.0}=2.0 N

And the direction is the same as the impulse (east).

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

7 0
2 years ago
How far does a boat travel in 5 hours at 32 miles per hour? 162 mi 160 mi 210 mi
sleet_krkn [62]
We know that:
d=vt
d=32mph*5h
d=160mi
4 0
3 years ago
Read 2 more answers
What will most likely happen ifa light wave moves through a solid?
levacccp [35]
It will decrease speed
7 0
2 years ago
Apakah yang dimaksud dengan gaya
tangare [24]
Nah gaya seperti titik fashion seperti apa yang Anda mana dan hal-hal seperti itu. (tell me if you cant understand)
8 0
3 years ago
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A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0
3 years ago
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