Ask question

Ask question

All categories

2 years ago

7

A person weighing 55 kg walks by applying 160 N of force on the ground, while pushing a 10-kg object. If the person accelerates

at 2 m/s2, what is the force of friction experienced by the system consisting of the person and the object?

1 answer:

lozanna [386]2 years ago

6 0

Answer:

idk

Explanation:

since i didn't answer it right, keep up with my questions and u can answer it with idk for 100 brainly points.

You might be interested in

The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati

Vedmedyk [2.9K]

Answer:

The tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, $m=33.2\ kg$

Coefficient of static friction is, $\mu = 0.214$

Angle of inclination is, $\theta = 31.5$ °

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are $mg\cos \theta$ and normal $N$ . Now, there is no motion in the direction perpendicular to the inclined plane. So,

$N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N$

Consider the direction along the inclined plane.

The forces acting along the plane are $mg\sin \theta$ and frictional force, $f$ , down the plane and tension, $T$ , up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

$T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N$

Therefore, the tension in the rope is 229.37 N.

3 0

2 years ago

A boat's capacity plate gives the maximum weight and/or number of people the boat can carry safely in certain weather conditions

erastova [34]

•wind
•snow
•high tide/low tide
•thunder/lightning storms

5 0

2 years ago

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

7nadin3 [17]

Answer:

A). σ = 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B). $\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C). $U=10.322$ J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

$\sigma =k.\varepsilon _{0}.E$

where, E is resultant electric field = 1.2 x $10^{6}$ V/m

$\varepsilon _{0}$ is permittivity of free space = 8.85 x $10^{-12}$ $C^{2}$ /N- $m^{2}$

k is dielectric constant = 3.6

∴ $\sigma =k.\varepsilon _{0}.E$

= 3.6 x 8.85 x $10^{-12}$ x 1.2 x $10^{6}$

= 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

$\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )$

$\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )$

$\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C).

Area of the plate, A = 2.5 $cm^{2}$

= 2.5 x $10^{-4}$ $m^{2}$

diameter of the plate, d = 1.8 mm

= 1800 m

∴ Total energy stored in the capacitor

$U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad$

$U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800$

$U=10.322$ J

4 0

2 years ago

What is the formula that can be used to find velocity if kinetic energy and mass are known?

viva [34]

The formula for kinetic energy is $\frac{1}{2}m\Delta v^2$ . Thus, the equation for velocity is $v= \sqrt{ \frac{2TotalKineticEnergy}{m} }$ .

6 0

2 years ago

Read 2 more answers

Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o

IRINA_888 [86]

Answer:

a) $W=-0.0103125\ J$

b) $W=0.0059375\ J$

c) Compressing is easier

Explanation:

Given:

Expression of force:

$F=kx-bx^2+cx^3$

where:

$k=100\ N.m^{-1}$

$b=700\ N.m^{-2}$

$c=12000\ N.m^{-3}$

$x$ when the spring is stretched

$x$ when the spring is compressed

hence,

$F=100x-700x^2+12000x^3$

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, $x=-0.05\ m$

Now,

The spring constant at this instant:

$j=\frac{F}{x}$

$j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}$

$j=-8.25\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times -8.25\times (-0.05)^2$

$W=-0.0103125\ J$

b)

When compressing the spring by 0.05 m

we have, $x=0.05\ m$

<u>The spring constant at this instant:</u>

$j=\frac{F}{x}$

$j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}$

$j=4.75\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times 4.75\times (0.05)^2$

$W=0.0059375\ J$

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0

2 years ago