PLS DO THIS FOR ME I NEED DONE ASAP

Mariana [72]

I think it would be 77.9 grams

5 0

2 years ago

Consider a collection of charges in a given region and suppose all other charges are distant and have a negligible effect. Furth

AlladinOne [14]

Answer:

E. Some charges in the region are positive, and some are negative.

Explanation:

Electric potential is given as;

$V = \frac{W}{Q}$

where;

W is the work done in moving a charge between two points which have a difference in potential

Q is quantity of charge in the given region

If the electric potential at a given point in the region is zero, then sum of the charges in the given region must be equal to zero. For the charges to sum to zero, some will be positive while some will be negative,.

Therefore, the correct statement in the given options is "E"

E. Some charges in the region are positive, and some are negative.

3 0

3 years ago

A swimming pool has the shape of a box with a base that measures 30 m by 10 m and a uniform depth of 2.5 m. How much work is req

pav-90 [236]

Answer:

18,375 kJ

Explanation:

Parameters given:

Density of water: 1000 kg/m³

Acceleration due to gravity = 9.8 m/s²

Dimensions of pool = 30m × 10m × 2.5m

Depth of pool = 2.5 m

First we need to find the mass of the water in the pool:

Density = mass/volume

Mass = density * volume

Since the pool is full of water, the volume of the water is equal to the volume of the pool:

Volume = 30 * 10 * 2.5 = 750 m³

Mass = 1000 * 750

Mass = 750,000 kg

We can now find the force required to pump the water out of the pool:

F = m * g

Where m = mass of water

g = acceleration due to gravity

F = 750,000 * 9.8 = 7,350,000 N

The work needed to be done is the product of the force required by the distance the water would need to move to be pumped out (depth of pool):

Work = F * d

Work = 7,350,000 * 2.5

Work = 18,375,000 J = 18,375 kJ

4 0

3 years ago

Read 2 more answers

A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.

Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

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

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

$a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2$

And now we can calculate the force exerted on the running back, by using Newton's second law:

$F=ma=(100 kg)(-10 m/s^2)=-1000 N$

so, the magnitude of the force is 1000 N.

6 0

3 years ago

Read 2 more answers

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

1 year ago