All categories

3 years ago

In American football how many points.... look at the pic

Physics

2 answers:

guajiro [1.7K]3 years ago

3 0

two points im pretty sure have a nice day

tino4ka555 [31]3 years ago

3 0

Answer:

A safety is worth 2 positive points for the team on defense and 2 negative for the team on offense.

Explanation:

You might be interested in

A uniformly charged conducting sphere of 1.3 m diameter has a surface charge density of 8.1 µc/m2. (a) find the net charge on th

8_murik_8 [283]

<span>The surface charge density = q/A So q = surface charge density x Area The surface area of a sphere of radius R is 4*Pi*R^2. R = d/2 where d is diameter. This leaves us with 1.3/2 = 0.65. Area = 4 * pie * (0.65)^2 = 5.30998. So the net charge q = 8.1 * 10^(-6) * 5.30998 = 42.47998 * 10^(-6) The Total electric flux = Q/e_0 where , 8.854 Ă— 10â’12, e_0 is permitivity of free space. So Flux = 42.47998 * 10^(-6) / 8.854 * 10(â’12) = 4.833 * 10^(-6 - (-12)) = 4.833 * 10^(6)</span>

8 0

3 years ago

Two convex lenses are placed 15 cm apart. The left lens has a focal length of 10 cm, and the right lens a focal length of 5 cm.

Minchanka [31]

Answer:

The image distance from right lens is 2.86 cm and image is real.

Explanation:

Given that,

Focal length of left lens = 10 cm

Focal length of right lens = 5 cm

Distance between the lenses d= 15 cm

Object distance = 50 cm

We need to calculate the image distance from left lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{-50}$

$\dfrac{1}{v}=\dfrac{3}{25}$

$v=8.33\ cm$

We need to calculate the image distance from right lens

The object distance will be

$u = 15-8.33 = 6.67\ cm$

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{5}-\dfrac{1}{-6.67}$

$\dfrac{1}{v}=\dfrac{1167}{3335}$

$v=2.86\ cm$

The image is real.

Hence, The image distance from right lens is 2.86 cm and image is real.

4 0

3 years ago

The batteries in the figure above are connected in

Elena-2011 [213]

The answer is a, series circuit.

5 0

3 years ago

Read 2 more answers

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batt

Natali [406]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?