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Dvinal [7]
3 years ago
12

Science please help!​

Physics
1 answer:
IceJOKER [234]3 years ago
7 0

Explanation:

spherical lenses which are curved outward are CONVEX lenses

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An observer standing near a window 5 m high observe that an object falling downwards is passing across the window in 0.5 s. Find
Valentin [98]

Answer:

Explanation:

An object in free fall, NOT experiencing parabolic motion, has an equation of

h(t)=\frac{1}{2} gt^2+h_0 which says:

The height of an object with respect to time in seconds is equal to the pull of gravity times time-squared plus the height from which it was dropped. Normally we use -9.8 for gravity but you said to use 10, so be it.

For us, h(t) is 5 because we are looking for the height of the window when the object is 5 m off the ground at .5 seconds;

g = 10 m/s/s, and

t = .5sec

5=\frac{1}{2}(-10)(.5)^2+h and

5 = -5(.5)² + h and

5 = -5(.25) + h and

5 = -1.25 + h so

h = 6.25

That's how high the window is above the ground.

8 0
2 years ago
Grant sprints 50m to the right with an average velocity of 3.0 m/s
zlopas [31]

I need help with my math

6 0
3 years ago
A 72-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.
Shalnov [3]

Answer:

Explanation:

During the first .8 s , the elevator is under acceleration . It starts from initial velocity u = 0 , final velocity v = 1.2 m /s , time = .8 s

v = u + at

1.2 = 0 +  .8 a

a = 1.2 / .8

= 1.5 m /s²

During the acceleration in upward direction , let reaction force of ground on man be R .

Net force on man = R - mg

Applying Newton's 2 nd law

R - mg = ma

R = m ( g + a )

= 72 ( 9.8 + 1.5 )

= 813.6 N .

This reaction force will be measured by spring scale , so reading of spring scale will be 813.6 N .

3 0
2 years ago
A 20 cm-radius ball is uniformly charged to 71 nC.
artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

q_1 = 1.11 nC

q_2 = 8.88 nC

q_3 = 71 nC

Part c)

E_1 = 3996 N/C

E_2 = 7992 N/C

E_3 = 15975 N/C

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

V = \frac{4}{3}\pi R^3

V = \frac{4}{3}\pi(0.20)^3

V = 0.0335 m^3

now the charge density of the ball is given as

\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3

Part b)

Now the charge enclosed by the surface is given as

q = \rho V

at radius of 5 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3

q = 1.11 nC

at radius of 10 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3

q = 8.88 nC

at radius of 20 cm

q = 71 nC

Part c)

As we know that electric field is given as

E = \frac{kq}{r^2}

so we have electric field at r = 5 cm

E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}

E_1 = 3996 N/C

electric field at r = 10 cm

E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}

E_2 = 7992 N/C

electric field at r = 20 cm

E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}

E_3 = 15975 N/C

3 0
3 years ago
Mesopotamians also were the first to use ___________.
rusak2 [61]
I think it may be c i learned about this last year
7 0
3 years ago
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