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3 years ago

Please give answer with solution

Physics

1 answer:

UNO [17]3 years ago

3 0

250 m. for a longer explanation or solution look at this article, i’m sorry.
https://www.quora.com/A-projectile-is-thrown-so-it-travels-a-maximum-range-of-1000m-How-high-will-it-rise

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A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?

Olenka [21]

Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
$F = k*x$

Solving:
$F = k*x$
$F = 50*0.15$
$\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark$

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
$E = \frac{k*x^2}{2}$

Solving:(Energy associated with this stretching)
$E = \frac{k*x^2}{2}$
$E = \frac{50*0.15^2}{2}$
$E = \frac{50*0.0225}{2}$
$E = \frac{1.125}{2}$
$\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark$

7 0

3 years ago

A test charge of -3x10^-7 C is located 7 cm to the right of a charge of -9x10^-6 C and 20 cm to the left of a charge of +10x10^-

musickatia [10]

Answer:

5.634 N rightwards

Explanation:

qo = - 3 x 10^-7 C

q1 = - 9 x 10^-6 C

q2 = 10 x 10^-6 C

r1 = 7 cm = 0.07 m

r2 = 20 cm = 0.2 m

The force on test charge due to q1 is F1 which is acting towards right

According to the Coulomb's law

$F_{1}=\frac{Kq_{1}q_{0}}{r_{1}^{2}}$

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)

F1 = 4.959 N rightwards

The force on test charge due to q2 is F1 which is acting towards right

According to the Coulomb's law

$F_{2}=\frac{Kq_{2}q_{0}}{r_{2}^{2}}$

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)

F2 = 0.675 N rightwards

Net force on the test charge

F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards

3 0

3 years ago

What is the function of a capacitor in an electric circuit

topjm [15]

Capacitors function is to store up electricity and release it when the current is flowing. When there is no current, Capacitors still have electricity stored in them and can be discharged accidentally if something conductive touches them.

7 0

3 years ago

What are the three types of mechanical waves?

Black_prince [1.1K]

Answer:

I'm sure its transverse, longitudinal and surface

Explanation:

sorry if I'm wrong

4 0

2 years ago

Approximating the eye as a single thin lens 2. 75 cmcm from the retina, find the eye's near-point distance if the smallest focal

RSB [31]

The near-point distance of the eye is 11 cm.

<h3>What is the eye's near-point distance?</h3>

The near-point distance of the eye is the closest possible distance an object can be from the eye in order for its image to be formed on the retina. It can also be termed the closest distance of accommodation.

The near-point distance of the eye in the given scenario can be calculated using the lens formula given below:

1/f = 1/v + 1/u

where;

f = focal length

v = image distance

u = object distance

From the data provided;

f = 2.20 cm

v = 2.75 cm

u = ?

Solving for u:

1/u = 1/f - 1/v

1/u = 1/2.20 - 1/2.75

1/u = 0.91

u = 11 cm

In conclusion, the lens formula is used to determine the eye's near-point distance.

Learn more about eye's near-point distance at: brainly.com/question/16391605

#SPJ1

5 0

2 years ago

Please give answer with solution​

Please give answer with solution