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Jobisdone [24]
2 years ago
7

Julia jumps straight upward on mars, where the acceleration due to gravity is 3.7\,\dfrac{\text{m}}{\text{s}^2}3.7 s 2 m ​ 3, po

int, 7, space, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction downward. After 3\,\text{s}3s3, space, s, julia is falling downward with a velocity of 3.1\,\dfrac{\text{m}}{\text{s}}3.1 s m ​ 3, point, 1, space, start fraction, m, divided by, s, end fraction.
Physics
1 answer:
liberstina [14]2 years ago
6 0

We know the equation of motion v = u+ at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

In this case Final velocity = -3.1 m/s, negative sign indicates it is pointing downward.

acceleration = - 3.7m/s^2 . Negative means acceleration is towards center of planet Mars.

Time taken = 3 seconds

v = u + at\\ \\-3.1 = u-3.7*3\\ \\u = 8 m/s

So jumping velocity of Julia = 8 m/s

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One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit
nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

\lambda = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

\theta=\dfrac{1.22\lambda}{D}

The width is given by

d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m

The required width is 0.000003782 m

Minimum resolvable line separation is given by

\dfrac{0.000003782}{2}=0.000001891\ m

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when \lambda=157\ nm

d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m

The new minimum resolvable line separation between adjacent lines is

\dfrac{0.00000239425}{2}=0.000001197125\ m

6 0
2 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
2 years ago
Derive an expression for the acceleration of the car. Express your answer in terms of D and vt Determine the time at which the s
Bezzdna [24]

Answer:

V(car) = V(truck) at t = Dt/2

acceleration = v(car) = D/t^2

Explanation:

acceleration = v(car) = D/t^2

Since the average velocities must be the same, the car's final velocity must be twice the trunk velocity assuming the car start with zero velocity, since acceleration remain the same throughout the journey velocities at half-time point must be equal.

3 0
2 years ago
Look at the two question marks between zinc (Zn) and arsenic (As). At the time, no elements were known
marissa [1.9K]

Answer:

Mendeleev predicted the atomic mass of each element along with compounds they each should form.

Explanation:

Based on other elements in the same group he predicted the existence of eka-aluminum and eka-silicon, later to be named gallium (Ga) and germanium (Ge).

6 0
2 years ago
Maria throws an apple vertically upward from a height of 1.3 m with an initial velocity of +2.8 m/s. Will the apple reach a frie
evablogger [386]

Answer:

No, the apple will reach 4.20041 m below the tree house.

Explanation:

t = Time taken

u = Initial velocity = 2.8 m/s

v = Final velocity = 0

s = Displacement

g = Acceleration due to gravity = -9.81 m/s² = a (negative as it is going up)

Equation of motion

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-2.8^2}{2\times -9.81}\\\Rightarrow s=0.39959\ m

The height to which the apple above the point of release will reach is 0.39959 m

From the ground the distance will be 1.3+0.39959 = 1.69959 m

Distance from the tree house = 5.9-1.69959 = 4.20041 m

No, the apple will reach 4.20041 m below the tree house.

The values in the option do not reflect the answer.

5 0
2 years ago
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