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3 years ago

10

If a planet's orbital speed is 20 km/s when it's at its average distance from the sun which is most likely orbital speed when it

is nearest the sun?

1 answer:

Alona [7]3 years ago

8 0

Answer:25km/s

Explanation:

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The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re

mihalych1998 [28]

Answer:

Explanation:

Given

$\omega =13-\frac{1}{2}\cdot t^2$

Motor reverse its direction when \omega =0

$13-0.5t^2=0$

$26=t^2$

$t=\sqrt{26}=5.099\approx 5.1 s$

(b)

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

$\int d\theta =\int_{0}^{5.1}\omega dt$

$\int d\theta =\int_{0}^{t}(13-.05t^2)dt$

$\theta =(13t-0.1667\times t^3)_0^{5.1}$

$\theta =44.192^{\circ}$

4 0

3 years ago

Do you think that solids can undergo convection

Alex_Xolod [135]

<span>Convection means the movement happens inside a fluid by the cause of too much heat and lesser density to rise, and also the colder cause denser materials to sink.
Thus, Solid cannot undergo convection because the molecules inside the solid are closer and can hardly move unlike liquid.
So, when the temperature got hotter or colder, the molecules on solid will only vibrate and not convert.
Therefore, this concludes that the solid do not undergo convection rather conduction of heat.</span>

7 0

4 years ago

Imagine you are holding a box in your hand, the force required for you to hold the box with mass 4.54kg and you want to accelera

svetlana [45]

Answer:

(a) Fₓ = 0 N

$F_y$ = 9.08 N

(b)

(a) Fₓ = 0 N

<u></u> $F_y$ = 9.08 N

(c) $F_y$ = 0 N

Fₓ = 9.08 N

Explanation:

The magnitude of the force will remain the same in each case, which is given as follows:

F = ma (Newton's Second Law)

where,

F = force = ?

m = mass = 4.54 kg

a = acceleration = 2 m/s²

Therefore,

F = (4.54 kg)(2 m/s²)

F = 9.08 N

Now, we come to each scenario:

(a)

Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:

<u>Fₓ = 0 N</u>

For upward direction the force will be positive:

<u></u> $F_y$ <u> = 9.08 N</u>

<u></u>

(b)

Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:

<u>Fₓ = 0 N</u>

For upward direction the force will be negative:

<u></u> $F_y$ <u> = - 9.08 N</u>

<u></u>

(c)

Since the motion is in the horizontal direction. Therefore the magnitude of the force in y-direction will be zero:

<u></u> $F_y$ <u> = 0 N</u>

<u>Fₓ = 9.08 N</u>

3 0

3 years ago

A wire with a circular cross section and a resistance R is lengthened to 9.66 times its original length by pulling it through a

damaskus [11]

Answer:

The resistance of the wire after it is stretched is 93.31R.

Explanation:

Resistance is the property of the material to oppose the current flow through it. It is given by the relation :

R = (ρl)/A

Here ρ is resistivity, l is length of wire and A is the area of the wire.

Let l₀, and A₀ are the original length and original circular cross section area of the wire. while l₁ and A₁ are the new length and new circular cross section area of the wire.

Volume of the original wire, V₀ = A₀ x l₀

Volume of the new wire, V₁ = A₁ x l₁

According to the problem. volume remain same. So,

V₀ = V₁

A₀ x l₀ = A₁ x l₁

It is given that l₁ = 9.66 x l₀. Substitute this value in the above equation;

A₀ x l₀ = A₁ x 9.66 x l₀

A₁ = A₀/9.66

Resistance of the original wire, R = (ρl₀)/A₀

Resistance of the new wire, R₁ = (ρl₁)/A₁

Substitute the value of l₁ and A₁ in the above equation.

R₁ = (ρ x l₀ x 9.66)/(A₀/9.66) = 93.31 x (ρl₀)/A₀

But (ρl₀)/A₀ = R. hence,

R₁ = 93.31 R

8 0

3 years ago

A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a

Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial Velocity of Each object

$V_f$ = Final Velocity

Our values are given as

$m_1 = 2190Kg$

$v_1 =10.5m/s$

$m_2 = 3220kg$

$V_f = 4.74m/s$

Replacing we have that

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

$(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)$

$v_2 = 0.8224m/s$

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0

4 years ago