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3 years ago

Plz help me out in need u to help me

Physics

2 answers:

Feliz [49]3 years ago

8 0

Answer:

First and third one

Explanation:

Took test, and it can be the other ones because they dont make any sense

Westkost [7]3 years ago

6 0

1 and 3...... I have done this before

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Two ropes are attached to a 35 kg object. The first rope applies a force of 20 N and the second applies a force of 55 N. If the

Goshia [24]

Answer:

$a=1.672\ m.s^{-2}$

Explanation:

Given:

mass of the object, $m=35\ kg$

forces by two mutually perpendicular ropes of the attached to the object:

$F_x=20\ N$

$F_y=55\ N$

<u>Now we find the resultant force effect due to the two given forces:</u>

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

$F=\sqrt{(20)^2+(55)^2}$

$F=58.52\ N$

<u>Now the acceleration will be due to this resultant force:</u>

$a=\frac{F}{m}$

$a=\frac{58.52}{35}$

$a=1.672\ m.s^{-2}$

3 0

3 years ago

Read 2 more answers

Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vert

chubhunter [2.5K]

Answer:

362.41 km/h

Explanation:

F = Force

m = Mass = 84 kg

g = Acceleration due to gravity = 9.81 m/s²

C = Drag coefficient = 0.8

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.04 m²

v = Terminal velocity

F = ma

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow mg=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{mg}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{20\times 9.81}{1.21\times 0.8\times 0.04}}\\\Rightarrow v=100.66924\ m/s$

Converting to km/h

$100.66924\times 3.6=362.41\ km/h$

The terminal velocity of the stone is 362.41 km/h

5 0

3 years ago

Read 2 more answers

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box= $u=13m/s$

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

$v^2-u^2=2as$

Substitute the values

$(11.5)^2-(13)^2=2a(8.5)$

$132.25-169=17a$

$-36.75=17a$

$a=\frac{-36.75}{17}=-2.2m/s^2$

We know that

Acceleration= $a=\frac{v-u}{t}$

Substitute the values

$-2.2=\frac{11.5-13}{t}$

$-2.2=\frac{-1.5}{t}$

$t=\frac{1.5}{2.2}=0.68 s$

Hence, the time taken by box to slow by this amount=0.68 s

8 0

3 years ago

________ is a measure of how many waves pass by in one second.

Sergio [31]

C. Frequency is a measure of how many waves pass by in one second.

4 0

3 years ago

HELP ASAP

amid [387]

Answer:

3200 has the fewest number of sig figs

Explanation:

sorry i was late

7 0

3 years ago