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3 years ago

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Two drivers make a 100-km trip. 1 completes the trip in 2 hours. driver 2 takes 3 hours but stops for an hour halfway. which dri

ver had a greater average

1 answer:

expeople1 [14]3 years ago

7 0

Neither because if driver 2 hadn't stopped, they both would've finished in the same amount of time.

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a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r

olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0

3 years ago

A 73.0 kg firefighter climbs a flight of stairs 9.0 m high. how much work is required? j

pshichka [43]

The strength of the fireman in vertical direction will be given by F = m * g. Then, the work done will be given by definition by W = F * d. Substituting the expression of the Force in that of the work, we have that the work will be W = m * g * d. Substituting the given values and assuming that g = 10m / s ^ 2, we have a total work of W = (73) * (10) * (9) = 6570 J

7 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u> $m_{1}$ <u>:</u>

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

<u>For </u> $m_{2}$ <u>:</u>

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

<u>For </u> $m_{3}$ <u>:</u>

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

In order to increase the amount of work completed it is necessary to

navik [9.2K]

Do all you can in one big day that you have time off or work on one thing then work on the other at the same time

3 0

3 years ago

Read 2 more answers

(a) An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kin

Iteru [2.4K]

Answer:

A) The smaller object; B) The larger object

Explanation:

A) Lets say the small object is 2 kg and the large one is 6 kg. Lets say they also have 30 kg*m/s of momentum each. The small object would have 15 m/s velocity and the large would have 5 m/s.

Now for kinetic energy(.5*m*v²), the small object is .5*2*15², which is 225 J

The large object is .5*6*5², which is 75 J, so the smaller object has more Kinetic energy. Since velocity is squared, it is more important than how large mass is.

B) Same masses as part A. Lets say the kinetic energy is 45 J for both of them. For the small object, 45=.5*2*v²

.5*2 is 1, so 45/1 is 45. Take the square root and we get v= 6.71 m/s

For the large object, 45=.5*6*v²

.5*6 is 3, so 45/3 is 15. Take the square root and we get v=3.87 m/s

Now we plug the velocities into p=mv

For the small object, p=2*6.71, which gives us p=13.42 kg*m/s

For the large object, p=6*3.87, which gives us p=23.22 kg*m/s

The larger object has the larger momentum.

Hope this helps

5 0

3 years ago