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3 years ago

15

An object in free fall has a velocity of 5m/s in the upward direction. What is the instantaneous velocity of the object one seco

nd later?

1 answer:

aleksandrvk [35]3 years ago

7 0

Answer:

Explanation:

Near the earth's surface where gravity is approximately 10 m/s² downward

v = u + at

v = 5 + (-10)(1) = -5 m/s

so it has the same speed but in the opposite (downward) direction.

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Can an object have both kinetic energy and gravitational potential energy? Explain.

Wewaii [24]

Answer:

Yes

Explanation:

An object can be moving (have kinetic energy) and be elevated above the ground at the same time (and also have potential energy).

3 0

2 years ago

Before digital filmmaking, what tool was used to control the speed of movement on the screen after filming?

nikitadnepr [17]

Answer:

The appropriate response is "Optical printer ".

Explanation:

A photographic printer used mostly for optical aberrations, comprised simply of either a camera that captures the frame to expand, minimize, deform, respectively. through magnifying lenses.
A projector that always, as distinct from some kind of touch printer, transferred the image to something like the printing supply.

4 0

4 years ago

the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp

Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0

3 years ago

CAN SOME ONE TALK TO ME PLZ ASAP IM GOING THROUGH A BAD TIME

ss7ja [257]

Hey! How's it going? If you need anything, feel free to send me a friend request and message me.

Don't worry if things get wrong, they will surely get better, if not, I'm here to talk to you. :)

7 0

3 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago