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3 years ago

8

If you exert a horizontal force of 200 N to slide a crate across a factory floor at constant velocity, how much friction is exer

ted by the floor on the crate?

1 answer:

kati45 [8]3 years ago

3 0

Answer:

200 N

Explanation:

Applied force, F = 200 N

Let the friction force is f.

By use of Newton's second law

Net force = mass × acceleration

F - f = m × a

Here acceleration a is zero as crate moves with constant velocity.

So, F - f = 0

f = F

Friction force = 200 N

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Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. You have determined t

luda_lava [24]

The horizontal distance traveled by the ball (range of motion) is given by the following equation:

$x= v_{0x} t$

In which $x$ is the range of motion, $v_{0x}$ is the horizontal component of the initial velocity, and $t$ is the time of motion.

First, lets calculate the horizontal component of the initial velocity:

$v_{0x}= v_{0}cos( \alpha)=13.9cos(35)=11.39$

Now, we calculate the time of motion from the equation that described the motion of the ball in the vertical axis:

$y= \frac{1}{2}at^2+ v_{0y}t$

In which $y$ is the position of the ball vertically, $v_{0y}$ is the vertical component of the initial velocity, $a$ is the acceleration in the vertical axis (which is gravity), and $t$ is time of motion.

We want to find the time when the ball lands, hence, when $y=0$ ; so the equation becomes:

$0= \frac{1}{2}at^2+ v_{0y}t=\frac{1}{2}at^2+ v_{0}sin( \alpha )t=\frac{1}{2}(-9.8)t^2+ 13.9sin(35 )t$

We rewrite it a bit more:

$-4.9t^2+7.97t=0$

This is a quadratic equation, so we use the quadratic equation formula to solve for time (we'll get two answers):

$t= -\frac{1}{9.8} [{-7.97}+-\sqrt{(7.97)^2}]$

Clearly, one of the answers is $t=0$ , this is before you kick the ball (it is on the ground), we want the nonzero answer (when it lands) so:

$t= -\frac{1}{9.8} ({-7.97}-7.97})=1.63$

Now, we plug-in the time value to the equation of the motion's range:

$x= v_{0x} t=(11.39)(1.63)=18.57$

The ball will travel 18.57 meters.

6 0

3 years ago

Read 2 more answers

In a certain city, electricity costs $0.13 per kW·h.

Alex_Xolod [135]

The question is asking to compute the annual cost of electricity to power a lamp post for 7.5 hrs per day with the following power, and base on my calculation, the answers would be the following:
#1. 7,580,769.23
#2. 1,895,195.31
I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications.

4 0

3 years ago

Read 2 more answers

M<br> An object has a mass of 5.52 g and a volume of 12 mL. What is the density of the object

shutvik [7]

Answer:

0.46 g/mL

Explanation:

Density can be found using the following formula.

$d=\frac{m}{v}$

where <em>d</em> is density, <em>m</em> is mass, and <em>v</em> is volume.

We know that the object has a mass of 5.52 grams and a volume of 12 milliliters.

m=5.52 g

v=12 mL

Substitute these values into the formula.

$d=\frac{5.52 g}{12 mL}$

Divide 5.52 by 12

5.52/12=0.46

d=0.46 g/mL

The density of the object is 0.46 g/mL, and option C is correct.

4 0

3 years ago

A block of mass 3 kg slides down a frictionless inclined plane of length 6 m and height 4 m. If the block is released from rest

Ludmilka [50]

(if it's frictionless the length doesn't even matter :) )
It would have the same kinetic energy down as the potential energy up. That is, $mgh=\frac{mv^2}{2}$ or $2gh=v^2$ (the mass doesn't even matter). The result is $\sqrt{2gh}$ , so only the height matters really. It is almost 9 (it is $\sqrt{80}=4\sqrt{5}$ ).

6 0

3 years ago

A water storage tank has the shape of a cylinder with diameter 14 ft. It is mounted so that the circular cross-sections are vert

Allisa [31]

Answer:

$\%A_F=77.335\%$

Explanation:

Given:

diameter of tank, $d=14\ ft$

level of the tank filled in its horizontal position, $h=13\ ft$

Now refer the schematic that show water with blue colour.

The triangle ORQ is symmetric about OS as it comes from center O on the cord QR at S.

$RO=QO=7\ ft$ (∵ radius of the cylinder)

$RS=\sqrt{RO^2-OS^2}$

$RS=3.6055\ ft$

Now the area of triangle ORQ:

$A_t=\frac{1}{2}\times QR\times OS$

$A_t=RS\times OS$

$A_t=3.6055\times 6$

$A_t=21.6333\ ft^2$

Now the angle ROS:

$\cos\theta=\frac{OS}{OR}$

$\theta=\cos^{-1}(\frac{6}{7} )$

$\theta=31.0027^{\circ}$

<u>Therefore the reflex angle ROQ:</u>

$\rangle ROQ=360^{\circ}-\theta^{\circ}$

$\rangle ROQ=360^{\circ}-31.0027^{\circ}$

$\rangle ROQ=328.9973^{\circ}$

Now the area of sector ROQPR:

We have the area of full circle, $A=\pi.r^2$

where:

r = radius of the circle

hence for sector:

$A_S=\pi\times 7^2\times \frac{328.9973}{360}$

$A_S=140.6811\ ft^2$

Now the cross sectional area filled with water:

$A_F=A_S+A_t$

$A_F=140.6811-21.6333$

$A_F=119.0478\ ft^2$

Total cross sectional area of tank:

$A=\pi.r^2$

$A=\pi\times 7^2$

$A=153.9380\ ft^2$

Now the percentage of total capacity used:

$\%A_F=\frac{A_F}{A}\times 100\%$

$\%A_F=\frac{119.0478}{153.9380} \times 100$

$\%A_F=77.335\%$

5 0

2 years ago