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sweet-ann [11.9K]
2 years ago
5

Franny drew a diagram to compare images produced by concave and convex lenses.

Physics
2 answers:
Zina [86]2 years ago
8 0

Answer:

B. Virtual

Good Luck!

Tasya [4]2 years ago
6 0

Answer:

Virtual

Explanation:

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Differences between Pressure and upthrust​
Angelina_Jolie [31]

Answer:

Pressure is equal to the ratio of thrust to the area in contact. Upthrust is a force exerted by the fluids on an object placed in the fluid . Upthrust acts in upward direction.

4 0
3 years ago
An object starts at the 100 m mark and ends up at the -100 m mark, 50
horsena [70]
It traveled 200 m in 50 seconds. 200/50 can be simplified to 4 m/s!
The velocity is -4 m/s (negative because it travelled from 100 to -100 or backwards)
8 0
2 years ago
If opposite poles repel each other, why does the north end of a compass point to the north pole?
lina2011 [118]
Ratios are fractions
6 0
3 years ago
Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football
faltersainse [42]

Answer:

a=159.32\ m/s^2

Explanation:

It is given that,

Angular speed of the football spiral, \omega=6.9\ rev/s=43.35\ rad/s

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

v=r\omega

v=0.085\times 43.35

v = 3.68 m/s

The centripetal acceleration is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(3.68)^2}{0.085}

a=159.32\ m/s^2

So, the centripetal acceleration of the laces on the football is 159.32\ m/s^2. Hence, this is the required solution.

6 0
3 years ago
Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

    a = 2M / (1 + 2) M g

    a = 2/3 g

We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
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