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Maurinko [17]
3 years ago
8

What is the net force on this object?​

Physics
1 answer:
Mariulka [41]3 years ago
5 0
Answer=8N

Explantion:
20-20=0
22-14=8N
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Please help/ show work !!! Also don’t answer by putting a link!!
Zepler [3.9K]
I hope this helps you out!

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2 years ago
The left face of a biconvex lens has a radius of curvature of magnitude 11.7 cm, and the right face has a radius of curvature of
SIZIF [17.4K]

Answer:

Explanation:

R1 = + 11.7 cm

R2 = - 18.7 cm

n = 1.52

(a) Use lens maker formula

\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )

\frac{1}{f}=\left ( 1.52-1 \right )\left ( \frac{1}{11.7}}+\frac{1}{18.7} \right )

\frac{1}{f}=\frac{0.52\times 30.4}{218.79}

f = 13.84 cm

(b)

R1 = + 18.7 cm

R2 = - 11.7 cm

n = 1.52

(a) Use lens maker formula

\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )

\frac{1}{f}=\left ( 1.52-1 \right )\left ( \frac{1}{18.7}}+\frac{1}{11.7} \right )

\frac{1}{f}=\frac{0.52\times 30.4}{218.79}

f = 13.84 cm

5 0
4 years ago
HELP ME PLEASE ASAP!!!!! WILL GIVE BRAINLIEST, FIVE STARS, AND HEART!!!!!<br>(picture included)
harina [27]

Answer:

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3 0
3 years ago
Read 2 more answers
A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0
3 years ago
Constants Part A wo charged dust particles exert a force of 7.5x102 N n each other What will be the force if they are moved so t
alexandr1967 [171]

Answer:

New force, F'=48\times 10^3\ N

Explanation:

It is given that,

Force acting between two charged particles, F=7.5\times 10^2\ N

We need to find the force if they are moved so they are only one-eighth as far apart.

The force between two charged particles separated at a distance of r is given by :

F=k\dfrac{q_1q_2}{r^2}............(1)

If the charges are one-eighth as far apart then, r' =(1/8)r and new force is given by :

F'=k\dfrac{q_1q_2}{(\dfrac{r}{8})^2}..........(2)

Dividing equation (1) and (2) :

\dfrac{F}{F'}=\dfrac{1}{64}

F'=7.5\times 10^2\ N\times 64

F' = 48000 N

or

F'=48\times 10^3\ N

Hence, this is the required solution.

8 0
3 years ago
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