All categories

2 years ago

A force of 20 newtons is exerted on a chair from its right and a force of 12 newtons is applied from

Physics

1 answer:

tankabanditka [31]2 years ago

5 0

Answer:

The net force acting on the chair is 8 newtons, and the chair moves toward the left.

Explanation:

We can better understand this problem if we analyze it through a free body diagram. This free body diagram is attached as an image.

If we make a sum of forces acting over the chair we can find -20 + 12 = -8 [N]

The minus sign indicates that the chair is moving to the left.

You might be interested in

You perform the Hooke's Law experiment and create a plot of Displacement vs. Force. You add a linear fit and find the following

balandron [24]

Answer: 0.192 N/m

Explanation:

Well, generally when a Hooke's Law experiment is performed the plot is in fact Force vs Displacement, being the Force (in units of Newtons) in the Y-axis and the Displacement (in units of meters) in the X-axis.

In addition, if we add a linear fit the resultant equation will be the Line equation of the form:

$Y=mX+b$

Where $m=\frac{Y_{2}-Y_{1}}{X_{2}-X_{1}}$ is the slope and $b$ is the point where the line intersects the Y-axis.

So, if the equation is:

$Y=0.192X+0.011$

The slope of this line is $0.192 N/m$ which is also the spring constant $k$ .

7 0

3 years ago

Help with this please!

DochEvi [55]

Answer:

I have no clue.

Explanation:

3 0

3 years ago

Describe the general relationship between the weight of adult men and the average number of Calories needed daily to maintain a

Alexeev081 [22]

Answer:

The average number of calories needed daily represents the average quantity of calories eliminated by human body due to metabolism and must be compensated by eating and drinking.

The amount of calories contained in the food we eat every day must represent the amount of calories eliminated by the body in that time to have a steady weight.

Explanation:

The average number of calories needed daily represents the average quantity of calories eliminated by human body due to metabolism and must be compensated by eating and drinking. If total quantity of calories in the food we consume every day is higher that the average number of calories needed daily, then weight increases by fat accumulation.

3 0

2 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago

A ball is thrown into the air with a vertical velocity of 50 m/s and a horizontal

daser333 [38]

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Projectile Motion.

Since, here given that, vertical velocity= 50m/s

we know that u*sin(theta) = vertical velocity

so the time taken to reach the maximum height or the time of Ascent is equal to

T = Usin(theta) ÷ g, here g = 9.8 m/s^2

so we get as,

T = 50/9.8

T = 5.10 seconds

thus the time taken to reach max height is 5.10 seconds.

5 0

2 years ago