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3 years ago

Which of the following statements best explains why a book resting on a table is in equilibrium?

1 answer:

3 0

I'm pretty sure the answer is d.The weight of the book and the table's upward force on the book are equal in magnitude but opposite in direction.

You might be interested in

Quanto vale a resultante de duas forças de mesmo módulo mesma direção e sentidos opostos atuando sobre um corpo de massa igual a

Crank

A resultante das duas forças será zero, já q os sentidos são opostos e sãos iguais em módulo, elas se anulam. Logo, se a força resultante é zero, e F=ma, aceleração também será igual a zero.

6 0

3 years ago

Read 2 more answers

A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a rad

sergey [27]

Explanation:

Formula for angle subtended at the center of the circular arc is as follows.

$\theta = \frac{S}{r}$

where, S = length of the rod

r = radius

Putting the given values into the above formula as follows.

$\theta = \frac{S}{r}$

= $\frac{4}{2}$

= $2 radians (\frac{180^{o}}{\pi})$

= $114.64^{o}$

Now, we will calculate the charge density as follows.

$\lambda = \frac{Q}{L}$

= $\frac{18 \times 10^{-9} C}{4 m}$

= $4.5 \times 10^{-9} C/m$

Now, at the center of arc we will calculate the electric field as follows.

E = $\frac{2k \lambda Sin (\frac{\theta}{2})}{r}$

= $\frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}$

= 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.

5 0

3 years ago

I need help with this question how to solve it for Brass and Cooper

Ksenya-84 [330]

Take into account that density and relative density are given by:

$\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}$

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

$\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}$

Cooper:

$\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}$

3 0

1 year ago

A 25 newton force applied on an object moves it 50 meters. The angle between the force and displacement is 40.0 degrees. What is

zvonat [6]

Work done is when a force is exerted to cause a displacement in a certain object.
the equation for work done ;
work done = force applied * displacement of the object
when the force applied is not in the same direction as that of the displacement of the object then the effect of the force is not its whole value. The force is then applied at an angle to that of the displacement of the object, then the resultant force is the force exerted* cos of the angle between force and displacement, in this instance the angle is 40 °.
the new equation is then;
work done = force cos 40° * displacement
after substitution,
work = 25 N * 0.76 * 50 m
= 957.55 J
round it off
= 9.6 *10² J
the correct answer is B

5 0

3 years ago

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While water skiing behind her father’s boat, Letty is pulled at constant speed by a force of 164 N from the tow rope that makes

kap26 [50]

Answer:

0.265

Explanation:

Draw a free body diagram. There are four forces:

Normal force Fn pushing up.

Weight force mg pulling down.

Tension force T at an angle θ.

Friction force Fn μ pushing left.

Sum the forces in the y direction:

∑F = ma

Fn + T sin θ − mg = 0

Fn = mg − T sin θ

Sum the forces in the x direction:

∑F = ma

T cos θ − Fn μ = 0

Fn μ = T cos θ

μ = T cos θ / Fn

μ = T cos θ / (mg − T sin θ)

Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:

μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)

μ = 0.265

7 0

3 years ago