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astraxan [27]
3 years ago
8

A car (mass = 1200 kg) is traveling at 31.1 m/s when it collides head-on with a sport utility vehicle (mass = 2830 kg) traveling

in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?
Physics
1 answer:
IrinaVladis [17]3 years ago
4 0

Answer:

13.18 m/s

Explanation:

Let the velocity of sports utility car is

-u as it is moving in opposite direction.

mc = 1200 kg, uc = 31.1 m/s

ms = 2830 kg, us = - u = ?

Using conservation of momentum

mc × uc + ms × us = 0

1200 × 31.1 - 2830 × u = 0

u = 13.18 m/s

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You have been training and can run a mile in 6.5 minutes. How long would it take to run a lap around the outside track, which is
Paha777 [63]

Answer:

1 minute 36.85 seconds

Explanation:

First we need to convert the miles into meters, as the demanded result should be in meters.

1 mile = 1,609.34 meters

Also, 6.5 minutes should be converted into seconds.

1 minute = 60 seconds

6.5 x 60 = 390 seconds

Now we need to divide the miles with the seconds to see how much meters have been run in a second.

1,609.34 / 390 = 4.13 meters

The suggested meters now should be divided with the distance run in one second.

400 / 4.13 = 96.85 seconds

So we get a result of 96.85 seconds, or 1 minute 36.85 seconds.

5 0
3 years ago
Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always threethree times its height. Supp
hammer [34]

Answer:

159241.048 cm³/s

Explanation:

r = Radius = 3×height = 3h

h = height = 16 cm

Height of the pile increases at a rate = \frac{dh}{dt}=22\ cm/s

\text{Volume of cone}=\frac{1}{3}\pi r^2h\\\Rightarrow V=\frac{1}{3}\pi (3h)^2h\\\Rightarrow V=3\pi h^3

Differentiating with respect to time

\frac{dv}{dt}=9\pi h^2\frac{dh}{dt}\\\Rightarrow \frac{dv}{dt}=9\pi 16^2\times 22\\\Rightarrow \frac{dv}{dt}=159241.048\ cm^3/s

∴ Rate is the sand leaving the bin at that​ instant is 159241.048 cm³/s

5 0
3 years ago
A 5cm object is located 12 cm from a convex mirror with a focal length of 14cm. Calculate the
Tema [17]

Answer:

-86.415485655

Explanation:

3 0
3 years ago
A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin
tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       v_{oy} = v₀ sin 40

       v_{oy} = 50.0 sin40

       v_{oy} = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and v_{y} = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    p_{oy} = 0

After the break

X axis

   p_{fx} = m₂ v₂

Axis y

     p_{fy} = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       v_{fy}² = v_{oy}² - 2 gy

       0 =  v_{oy}²-2gy

       y =  v_{oy}² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      v_{f}² =  v_{y}² - 2 g y₂

      0 =  v_{y}² - 2 g y₂

     y₂ =  v_{y}² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

8 0
3 years ago
Question is in the picture
Svetradugi [14.3K]

the more pressure put on the string, the more frequency and higher pitch.

4 0
3 years ago
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