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ololo11 [35]
3 years ago
15

Your friend has slipped and fallen. To help her up, you pull with a force F, as the drawing shows. The vertical component of thi

s force is 110 newtons (N), while the horizontal component is 122 N. Find (a) the magnitude of F and (b) the angle0.
Physics
1 answer:
sineoko [7]3 years ago
3 0

This question can be solved using the concept of the composition of the vector.

(a) The magnitude of F is b "164.27 N".

(b) The angle of the vector from horizontal is " ".

(a)

From the composition of the rectangular components of a vector, we know that the magnitude of the resultant vector is given by the following formula:

F = \sqrt{F_x^2+F_y^2}

where,

F_x = x-component of the force = 122 N

F_y = y-component of the force = 110 N

Therefore,

F = \sqrt{(122\ N)^2+(110\ N)^2}

<u>F = 164.27 N</u>

(b)

The angle of the vector is given by the following formula:

\theta = tan^{-1}(\frac{F_y}{F_x})\\\\\theta = tan^{-1}(\frac{110\ N}{122\ N})\\\\

<u>θ = 42.04°</u>

Learn more about the composition of vector here:

brainly.com/question/24313520?referrer=searchResults

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JulijaS [17]
I mean if he flies 5g that means that's his average speed too
5 0
4 years ago
Bright and dark fringes are seen on a screen when light from a single source reaches two narrow slits a short distance apart. Th
Dima020 [189]

Answer:

Explanation:

Let the thickness of the film is t and the refractive index of the material of film is n.

When light travels through a sheet of thickness t, the optical path traveled is nt.

When the path of one of slit is covered by a sheet of thickness t, the optical path becomes

x = ( n - 1) t

As the one fringe is shift, so the optical path changed by one wavelength.

i.e., x = λ

So, λ = ( n - 1) t

t=\frac{\lambda }{n-1}

7 0
3 years ago
If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the
kodGreya [7K]

Answer:

a)

v = 4.048 *10^6 m/s

b)  

Angular frequency =  1.92 * 10^7

Explanation:

As we know

v =  \frac{qBr}{m}

q is the charge on the electron = 3.2 * 10^{-19} C

B is the magnetic field in Tesla = 0.4 T

r is the radius of the circle = 0.21 m

mass of the electrons = 6.64 * 10^{-27} Kg

a)

Substituting the given values in above equation, we get -

v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6m/s

b)  

Angular frequency =

\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7

8 0
3 years ago
If the radius of a blood vessel drops to 84.0% of its original radius because of the buildup of plaque, and the body responds by
erma4kov [3.2K]

To develop this problem it is necessary to apply the equations concerning Bernoulli's law of conservation of flow.

From Bernoulli it is possible to express the change in pressure as

\Delta P = \frac{1}{2}\rho (v_1^2-v_2^2)+ \rho g (h_1h_2)

Where,

v_i =Velocity

\rho = Density

g = Gravitational acceleration

h = Height

From the given values the change of flow is given as

R = r^4P

Therefore between the two states we have to

\frac{R_2}{R_1} = \frac{r_2^4 P_2}{r_1^4 P_1} *100\%

\frac{R_2}{R_1} = \frac{84^4 (110)}{100^4*(100)} *100\%

\frac{R_2}{R_1} = 54.77\%

The flow rate will have changed to 54.77 % of its original value.

4 0
3 years ago
A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33 °C. Given Kf = 1.86 °C/m for wa
Fantom [35]

Answer:

62.06 g/mol

Explanation:

We are given that a solution containing 10 g of an unknown liquid and 90 g

Given mass of solute =W_B=10 g

Given mass of solvent=W_A=90 g

k_f=1.86^{\circ}C/m

Freezing point of solution =-3.33^{\circ}C

Freezing point of solvent =0^{\circ}C

Change in freezing point =Depression in freezing point

=Freezing point of solvent - freezing point of solution=0+3.33=3.33^{\circ}

\Delta T_f=\frac{W_B\times K_f\times 1000}{W_A\times M_B}

M_B=\frac{10\times 1.86\times 1000}{3.33\times 90}

M_B=62.06 g/mol

Hence, molar mass of unknown liquid is 62.06g/mol.

6 0
3 years ago
Read 2 more answers
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