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3 years ago

Please help me with this question :(

Physics

1 answer:

laila [671]3 years ago

3 0

Answer:

$\boxed{\text{\sf \Large 50 N}}$

Explanation:

Sum of the vertical forces is zero

F₁ + F₂ - Z = 0

Z is negative since Z acts in the opposite direction from F₁ and F₂

30 + 20 - Z = 0

Z = 50

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The diameter of an atom is 1.1×10−10m and the diameter of its nucleus is 1.0×10−14m. Part A What percent of the atom's volume is

77julia77 [94]

To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,

$\% Percent = \frac{V_{nucleus}}{V_{atom}}*100$

$\% Percent = \frac{4/3 \pi (d_{nucleus}/2)^3}{4/3 \pi (d_{atom}/2)^3}*100$

$\% Percent = \frac{(d_{nucleus}/2)^3}{ (d_{atom}/2)^3}*100$

$\% Percent =\frac{(1.0*10^{-14}/2)^3}{ (1.1*10^{-10}/2)^3}*100$

$\% Percent = 7.51*10^{-13}*100$

$\% Percent = 7.51*10^{-11}\%$

Therefore the percent of the atom's volume is occupied by mass is $7.51*10^{-11}\%$

3 0

2 years ago

In odd nuclei, what determines the final spin of the nucleus?

Katen [24]

Answer:

In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.

Explanation:

The meaning of odd nuclei is atomic mass is odd.

A=odd number.

A=Z+n

Here, Z is proton either it will odd or n will odd which is neutron.

Now according to the shell model the left out proton or neutron will contribute to the spin and parity.

For example,

Take the case of isotope of nitrogen-15.

Here Z is 7, and n is 8 will not contribute in spin.

Now, for Z=7.

$1S^{2} _{\frac{1}{2} }, 1P^{4} _{\frac{3}{2} }, 1P^{1} _{\frac{1}{2} }$

Here,

$j=\frac{1}{2}$

and, L=1.

Fort parity,

$(-1)^{L}$

Put the value of L.

Parity will be -1.

Now, spin will be

$S=(\frac{1}{2} )^{-1}$ .

7 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

Um avião, após deslocar-se 120 km para nordeste (NE), desloca-se 160 km para sudeste (SE). Sendo um quarto de hora o tempo total

Svetach [21]

Answer:

E) 800 km/h

Explanation:

The computation of the average vector velocity module of the plane, at that time is shown below:

The displacement vector is

d^2 = d1^2 + d2^2

where,

d1 = northeast displacement

d2 = southeast displacement

Now

d^2 = (120)^2 + (160)^2

= 14400 + 2560

= 40000

= √40000

d = 200 km

Now the average velocity is

V = ΔS ÷ Δt

= 200 ÷ 1 ÷ 4

= 200 × 4

= 800 km/h

4 0

2 years ago

Proposed Exercise - Circular Movement

notka56 [123]

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

5 0

2 years ago