Ask question

Ask question

All categories

2 years ago

15

An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=514 kg,m1=514 kg, m2=792 kg,m2=792 kg, and m

3=322 kgm3=322 kg have blocked a busy road. The rocks are lined up from left to right in the order m1,m1, m2,m2, and m3.m3. The city calls a local contractor to use a bulldozer to clear the road. The bulldozer applies a constant force to m1m1 in order to slide the rocks off the road. Assuming the road is a flat, frictionless surface and the rocks are all in contact, what force, FA,FA, must be applied to m1m1 to slowly accelerate the group of rocks from the road at 0.250 m/s2?

1 answer:

hjlf2 years ago

3 0

Answer:

278.5 N

Explanation:

First, let us find the total mass to be moved. We add up all the given masses together.

The total mass to be moved, M

M = 514 + 792 + 322 = 1628 kg.

Next, we find the force that is applied to the blocks. Remember that F = ma, so

Force applied to all the blocks, F =

F = 1628 * 0.25 = 407 N

Now we find the force that was applied to only the first rock

F = 514 * 0.25 = 128.5 N

From this, we can get the remaining force that was applied to the last two rocks, by subtracting them from each other

F' = 407 - 128.5 = 278.5 N

You might be interested in

A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

Black_prince [1.1K]

<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y$ is the height of the stone at 6s (the value we want to find)

$y_{o}=925ft$ is the initial height of the stone

$V_{o}=20ft/s$ is the initial velocity of the stone

$t=6s$ is the time at which we need to find the height

$g=32ft/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $y$ from (1):

$y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2}$ (2)

Finally:

$y=469ft$ This is the height of the stone at t=6s

4 0

2 years ago

The graph below plots four main sequence stars according to their luminosity and temperature.

saul85 [17]

There is a positive correlation between luminosity and mass of stars, meaning the more luminous a star is, the more massive it is likely to be as well. Given this, the masses of the stars should be in descending order of brightness.

Star 1 is the most luminous, so it should be heaviest, and the luminosity descends to Star 4.
Option B is the only chart that conforms to this, so it is the answer.

Answer is B

7 0

2 years ago

Read 2 more answers

A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released

GREYUIT [131]

Answer:

$6.75\mu C/m^2$

Explanation:

We are given that

Diameter,d= $1\mu m=1\time 10^{-6} m$

$1\mu m=10^{-6} m$

Radius,r= $\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m$

Density, $\rho=900kg/m^3$

Total number of electrons,n=39

Charge on electron = $1.6\times 10^{-19} C$

Total charge= $q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C$

Distance,s=2mm= $2\times 10^{-3} m$

Mass = $density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg$

Initial velocity,u=0

Final speed,v=4.5 m/s

$v^2-u^2=2as$

$(4.5)^2-0=2a(2\times 10^{-3})$

$20.25=4a\times 10^{-3}$

$a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2$

Force,F=ma

$qE=ma$

$q(\frac{\sigma}{2\epsilon_0})=ma$

$\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}$

$\epsilon_0=8.85\times 10^{-12}$

$\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2$

6 0

2 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.33 times a second. A tack is stuck in the tire a

Troyanec [42]

Answer: 6.47m/s

Explanation:

The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.

The time taken for 1 revolution is, 1/3.33 = 0.30s

velocity of the wheel = d/t

Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t

V = 2π * 0.309 / 0.3

V = 1.94/0.3

V = 6.47m/s

The tangential speed of the tack is 6.47m/s

7 0

2 years ago

Read 2 more answers

Смоктати півень мій друг

kaheart [24]

What uhhhhhhhhhhhhhhhhhhhhh

3 0

2 years ago