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2 years ago

If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be?

Physics

1 answer:

kicyunya [14]2 years ago

6 0

For this case we have that by definition:

$F = ma$

Where,

m: mass of the object
a: acceleration of the object
F: summation of forces

We have then:

$F = 10 + 5\\F = 15 N$

Then, by clearing the acceleration we have:

$a = \frac {F} {m}$

Substituting values we have:

$a = \frac {15} {5}\\a = 3 \frac {m} {s ^ 2}$

Answer:

The acceleration of the box is equal to:

$a = 3 \frac {m} {s ^ 2}$

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A train runs from New Delhi to Hyderabad it covers first of 420 km in 7 hours and the next distance of 360 km in 6 hours

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Explanation:

the answer is 16.7 miles or 60kmph

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2 years ago

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Visualize five horizontal sedimentary strata exposed in a cliff or canyon wall identified by consecutive numbers, 1 being the lo

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Answer:

Only the 4 th statement is true that is bed 1 and 3 are older than 4.

Explanation:

The 5 beds are numbered from 1 to 5 , 1 being the lowest and 5 being the topmost bed.

We are given 4 statements and we have to find out which all are true.

(a)Bed 4 is older than bed 2

This is wrong because the lower beds are older than beds that are higher.

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2 years ago

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A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and if $A_1=\frac{\pi D_{1}^{2}}{4}$ , then the expression would be: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ ,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
But, $D_2=2D_1$ and $L_2=2L_1$ , substituting in the expression for $Q_2$ : $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the last declaration of $Q_2$ , it could be noted that the expressión inside the parenthesis is actually $Q_1$ , then: $Q_2=\frac{2^2}{2}(25W)=50W$ .
It should be noted, that the temperatures in the hot and cold reservoirs never change.

7 0

2 years ago

A slender rod is 80.0 cm long and has mass 0.370 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05

nataly862011 [7]

Answer:

1.10 m/s

Explanation:

Linear speed is given by

$v=r\omega$

Kinetic energy is given by

$KE=0.5I\omega^{2}$

Potential energy

PE= mgh

From the law of conservation of energy, KE=PE hence

$0.5I\omega^{2}=mgh$ where m is mass, I is moment of inertia, $\omega$ is angular velocity, g is acceleration due to gravity and h is height

Substituting m2-m1 for m and 0.5l for h, $\frac {2v}{L}$ for $\omega$ we obtain

$0.5I(\frac {2v}{L})^{2}=0.5Lg(m2-m1)$

$(\frac {2v}{L})^{2}=\frac {gl(m2-m1)}{I}$ and making v the subject

$v^{2}=\frac {gl^{3}(m2-m1)}{4I}$

$v=\sqrt {\frac {gl^{3}(m2-m1)}{4I}}$

For the rod, moment of inertia $I=\frac {ML^{2}}{12}$ and for sphere $I=MR^{2}$ hence substituting 0.5L for R then $I=M(0.5L)^{2}$

For the sphere on the left hand side, moment of inertia I

$I=m1(0.5L)^{2}$ while for the sphere on right hand side, $I=m2(0.5L)^{2}$

The total moment of inertia is therefore given by adding

$I=\frac {ML^{2}}{12}+ m1(0.5L)^{2}+ m2(0.5L)^{2}=\frac {L^{2}(M+3m1+3m2)}{12}$

Substituting $\frac {L^{2}(M+3m1+3m2)}{12}$ for I in the equation $v=\sqrt {\frac {gL^{3}(m2-m1)}{4I}}$

Then we obtain

$v=\sqrt {\frac {gL^{3}(m2-m1)}{4(\frac {L^{2}(M+3m1+3m2)}{12})}}=\sqrt {\frac {3gL^{3}(m2-m1)}{L^{2}(M+3m1+3m2)}}$

This is the expression of linear speed. Substituting values given we get

$v=\sqrt {\frac {3*9.81*0.8^{3}(0.05-0.02)}{0.8^{2}(0.39+3(0.02)+3(0.05))}} \approx 1.08 m/s$

8 0

2 years ago

Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment

riadik2000 [5.3K]

Answer:

291.598 N-m

291.6 N-m

Explanation:

Let's first take a look at the free bodily diagrammatic representation.

The first diagram will aid us in answering question (a), so as the second diagram will facilitate effective understanding when solving for question (b).

Let's first determine our angle θ from the diagram

To find angle θ ; we have :

tan θ = $\frac{360+240}{450}$

tan θ = $\frac{600}{450}$

tan θ = 1.333

θ = tan⁻¹ (1.333)

θ = 53.13°

Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.

We have:

$M__B}=(Fcos \theta *240)-(Fsin \theta *450)$

where Force(F) = Force in the cord AC = 1350 N and θ = 53.13° ; we have:

$M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)$

$M__B}= 194400.463-485999.348$

$M__B}=-291598.885 N-mm\\$

$M__B}=-291.598 N-m$

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m

b) From the second diagram, taking the moment at point B $(M__B})$ ,