All categories

2 years ago

Which pair of graphs represent the same motion of an object

Physics

2 answers:

posledela2 years ago

7 0

Answer:

Explanation:

In this pair of graph we have the displacement versus time graph as concave up and if we consider it to be parabolic function then as we know that the velocity is given by the slope of the displacement-time graph.

$v=\frac{ds}{dt}$

where:

s= displacement

t= time

We get the linear inclined slope graph of the velocity versus time graph which stands true if we differentiate a parabolic function.

Also observe that at time t=0 we have slope of displacement graph as zero so corresponding to it we get velocity as zero in the velocity graph.

Viktor [21]2 years ago

3 0

The answer would be C

You might be interested in

A person weighing 0.70 kn rides in an elevator that has an upward acceleration of 1.5 m/s2. what is the magnitude of the normal

creativ13 [48]

First of all, we can find the mass of the person, since we know his weight W:
$W=mg=0.70 kN=700 N$
And so
$m= \frac{W}{g}= \frac{700 N}{9.81 m/s^2}=71.4 kg$

We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
$\sum F = ma$
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
$R_v -W = ma$
We know the acceleration of the system, $a=1.5 m/s^2$ (upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
$R_v = ma+W=(71.4 kg)(1.5 m/s^2)+700 N =807N$

8 0

2 years ago

3. In 1989, Michel Menin of France walked on a tightrope suspended under a

Tamiku [17]

Answer: 80m

Explanation:

Distance of balloon to the ground is 3150m

Let the distance of Menin's pocket to the ground be x

Let the distance between Menin's pocket to the balloon be y

Hence, x=3150-y------1

Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

3070m

Hence, recall equation 1

x = 3150 - 3070

80m

I hope this solve the problem.

6 0

2 years ago

A 1.25 in. by 3 in. rectangular steel bar is used as a diagonal tension member in a bridge truss. the diagonal member is 20 ft l

pentagon [3]

Answer:

axial stress in the diagonal bar =36,000 psi

Explanation:

Assuming we have to find axial stress

Given:

width of steel bar: 1.25 in.

height of the steel bar: 3 in

Length of the diagonal member = 20ft

modulus of elasticity E= 30,000,000 psi

strain in the diagonal member ε = 0.001200 in/in

Therefore, axial stress in the diagonal bar σ = E×ε

= 30,000,000 psi× 0.001200 in/in =36,000 psi

5 0

2 years ago

To learn to apply the concept of current density and microscopic Ohm's law. A "gauge 8" jumper cable has a diameter d of 0.326 c

alukav5142 [94]

Answer:

Resistivity = (1.726 × 10⁻⁸) Ωm

Comparing this answer with the resistivity of the listed materials in the question, this resistivity matches that of Copper the most.

Hence, the material is Copper.

Option A is correct. Copper.

Explanation:

Given,

I = 30.0 A

d = 0.326 cm = 0.00326 m

E = 0.062 V/m

The current density for the wire is given as

J = (I/A)

where J = current density = ?

I = current = 30.0 A

A = Cross sectional Area of the wire = (πd²/4) = [π×(0.00326²)/4] = 0.0000083503 m²

J = (30 ÷ 0.0000083503) = 3,592,685.3 A/m²

On a microscopic scale, Ohm's law can be stated as

E = Jρ

where E = Electric field = 0.062 V

J = Current density = 3,592,685.3 A/m²

ρ = Resistivity of the material = ?

ρ = (E/J)

ρ = 0.062 ÷ 3,592,685.3 = (1.726 × 10⁻⁸) Ωm

Comparing this answer with the resistivity of the listed materials in the question, this resistivity matches that of Copper the most.

Hence, the material is Copper.

Hope this Helps!!!

8 0

2 years ago

Read 2 more answers

A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against

valkas [14]

Answer:

(A) Work done will be 87.992 KJ

(B) Work done will be 167.4 KJ

Explanation:

We have given mass of methane m = 4.5 gram = 0.0045 kg

Volume occupies $V_1=12.7dm^3=12.7liters$

And volume is increased by $3.3dm^3$ so $V_2=12.7+3.3=16liters$

Temperature T = 310 K

Pressure is given as 200 Torr = 26664.5 Pa

(a) At constant pressure work done is given by

$W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj$

(b) At reversible process work done is given by $W=nRTln\frac{V_2}{V_1}$

We have given mass = 4.5 gram

Molar mass of methane = 16

So number of moles $n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125$

So work done $W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J$

7 0

2 years ago