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murzikaleks [220]
3 years ago
11

If the position of a particle on the x-axis at time t is −5t2, then the average velocity of the particle for 0 ≤ t ≤ 3 is

Physics
1 answer:
Drupady [299]3 years ago
3 0

Answer:

v = 15 m / s

Explanation:

In this exercise we are given the position function

          x = 5 t²

and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval

          v= \frac{v_{f} - v_{o} }{t_{f} - t_{o} }

let's look for the displacements

        t = 0     x₀ = 0 m

        t = 3     x_{f} = 5 3 2

                     x_{f} = 45 m

 

we substitute

           v = \frac{45 -0}{3 - 0}

           v = 15 m / s

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In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the ancho
melisa1 [442]

Answer:

largest lead = 3 m

Explanation:

Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.

So, first we need to calculate the velocities of both the anchorman  

given data:

Distance = d = 100 m

Time arrival for A = 9.8 s

Time arrival for B = 10.1 s

Velocity of anchorman A = D / Time arrival for A

=100/ 9.8 = 10.2 m/s

Velocity of anchorman B = D / Time arrival for B

=100/10.1 = 9.9 m/s

As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use

d = vt

= 9.9 x 9.8 = 97 m  

So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.

largest lead = 100 - 97 = 3 m

So if his lead no more than 3 m anchorman A win the race.

5 0
3 years ago
Physics {deceleration}
seropon [69]
Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)
7 0
3 years ago
A man does 500 j work pushing a car a distance of 2m how much force does he apply
olasank [31]

Answer: 250 N

Explanation:

Use equation for work

W=F*d

d=2m

W=500J

F=?

-----------------------

W=Fd

F=W/d

F=500J/2m

F=250N

6 0
3 years ago
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According to Einstein's theory of relativity, what happens to the mass of an object as the object approaches the
Anuta_ua [19.1K]

Answer:

B

Explanation:

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8 0
3 years ago
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Pressure is about 1000 hPa at sea level and about 500 hPa at an altitude of 5.5 km. Why doesn’t this vertical pressure gradient
saul85 [17]

Answer:

A. The upward pressure gradient force is balanced by gravity.

Explanation:

A. is correct because the pressure difference is actually generated by gravity. As in the following formula for the pressure at different points:

p = p_0 + \rho g h

where p, p_0 are the pressure at 2 points, ρ is the density of the fluid, g is the gravitational constant, and h is the height difference.

B is incorrect because friction in air is too small to make an effect.

C is incorrect because the Coriolis force is horizontal, not vertical.

D is incorrect because a difference of 500 hPa = 50000 Pa, this is half of the atmospheric pressure.

E is incorrect because temperature cannot generate force.

6 0
3 years ago
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