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2 years ago

At 12 seconds what is the direction of the velocity of the rocket what is the direction of the acceleration rocket

Physics

1 answer:

algol [13]2 years ago

8 0

Answer:

rocket at the moment of clearance, the velocity and acceleration have the same direction, vertical upwards.

Explanation:

In a rocket at the moment of clearance, the velocity and acceleration have the same direction, vertical upwards.

In the case of landing we have a downward acceleration due to the force of gravity and upward velocity.

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Which element of variation would not be affected by adding the data value 15 to the data set {3, 5, 6, 8, 9, 10, 13, 14}?

Anna11 [10]

The answer to this would be 15

4 0

3 years ago

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Burka [1]

Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .

At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .

At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .

7 0

2 years ago

Read 2 more answers

Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27

densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

$\rho=n\times\dfrac{m}{V_{c}\times N_{A}}$

$n=\dfrac{\rho\timesV_{c}\times N}{m}$ ....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

$\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100$

$\rho=5.98$

We calculate the mass of the alloy

$m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100$

$m=111.15$

Put the value into the equation (I)

$n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}$

$n=1.99\approx 2\ atoms/cell$

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0

3 years ago

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago

Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both fil

MatroZZZ [7]

Answer:

A₁/A₂ = 0.44

Explanation:

The emissive power of the bulb is given by the formula:

P = σεAT⁴

where,

P = Emissive Power

σ = Stefan-Boltzman constant

ε = Emissivity

A = Surface Area

T = Absolute Temperature of Surface

FOR BULB 1:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₁T₁⁴ ----------- equation 1

where,

A₁ = Surface Area of Bulb 1

T₁ = Temperature of Bulb 1 = 3000 k

FOR BULB 2:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₂T₂⁴ ----------- equation 2

where,

A₂ = Surface Area of Bulb 2

T₂ = Temperature of Bulb 1 = 2000 k

Dividing equation 1 by equation 2, we get:

P/P = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁(3000)²/A₂(2000)²

A₁/A₂ = (2000)²/(3000)²

A₁/A₂ = 0.44

6 0

3 years ago