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3 years ago

How do you find distance on a velocity time graph?

Physics

1 answer:

vekshin13 years ago

8 0

The region under the speed time chart gives the relocation.

In simple to-recall, straightforward terms, remove levels with the result of speed and time. So when you discover the region under a speed time diagram, you are really increasing speed and time. This gives the separation.

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Two 1-kg objects, C and D, increase in temperature by the same amount, but the

Ede4ka [16]

The object D is made up of material Lead. The correct option is D.

<h3>What is specific heat?</h3>

The specific heat is the amount of heat required to change the temperature by 1°C. It is denoted by C.

Two 1-kg objects, C and D, increase in temperature by the same amount, but the thermal energy transfer of object C is greater than the thermal energy transfer of object D. The object C has a specific heat of 235 J/kg-K.

Q = m C ΔT

Qc > Qd

The energy transfer is proportional to specific heat.

Specific heat of D must be less. The possible material with specific heat less than the given value is for Lead material.

Thus, the correct option is D.

Learn more about specific heat,

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4 0

1 year ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

horrorfan [7]

Answer:

B. 6HgO → 6Hg + 3O $_{2}$

Explanation:

A decomposition reaction is a reaction in which a single reactant is broken down into 2 or more products.

6 0

3 years ago

When a 58g tennis ball is served, it accelerates from rest to a constant speed of 36 m/s. The impact with the racket gives the b

inysia [295]

We first calculate the acceleration on the ball using:
2as = v² - u²; u = 0 because ball is initially at rest
a = (36)²/(2 x 0.35)
a = 1850 m/s²
F = ma
F = 0.058 x 1850
= 107.3 Newtons

5 0

3 years ago

Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass

kirza4 [7]

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

5 0

3 years ago