All categories

2 years ago

A spring has a spring constant of 5N/cm. What is its extension when loaded with 15N?

Physics

2 answers:

cluponka [151]2 years ago

7 0

Answer:

As per given Information

Spring Constant ,K = 5N/cm
Force , F = 15N

we have to find the extension when loaded with given magnitude of force.

Using Formulae

$\bf \: F = Kx \:$

where, x is extension

Putting the value we get

$\sf \twoheadrightarrow \: 15 \: = 5 \times x \\ \\ \sf \twoheadrightarrow \: \frac{15}{5} = x \\ \\ \sf \twoheadrightarrow \cancel \frac{15}{5} = x \\ \\ \sf \twoheadrightarrow \: 3cm \: = x$

Hence, the extension is 3 cm

So, required option is (d) 3 cm

kakasveta [241]2 years ago

3 0

Pretty sure a it’s d since every cm increases by 5N meaning 3 cm would increase it by 15N

You might be interested in

Someone please answer this, ill give you brainliest and your earning 50 points.

Tema [17]

All of the following are non-renewable resources except

O natural gas

O oil

O minerals

O water ✓ 

Water is a renewable source because evaporation and condensation takes place everytime on our planet

3 0

1 year ago

Read 2 more answers

A student environmental group is creating a campaign for locally sourced energy resources.What would be best for them to feature

Ivenika [448]

Answer:

A farmer with a field of solar panels.

Explanation:

The closest to a locally sources energy would have been

A coal mine located in their county.

But coal as an energy source is not environmentally friendly due to carbon emission, and should not be what the group should advocate for.

The best bet for them is

A farmer with a field of solar panels.

As solar panels are a source of green energy and green energy is what the environmental group should often and always advocate for

3 0

3 years ago

What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force i

CaHeK987 [17]

Answer:

The friction coefficient's minimum value will be "0.173".

Explanation:

The given query seems to be incomplete. Below is the attached file of the complete question.

According to the question,

(a)

The net friction force's magnitude will be:

⇒ $F_{net}=ma$

$=5\times 1.7$

$=8.5 \ N$

(b)

For m₃,

⇒ $ma=\mu m_3 g$

Or,

⇒ $\mu=\frac{a}{g}$

$=\frac{1.7}{9.8}$

$=0.173$

5 0

2 years ago

A 72 kg skydiver can be modeled as a rectangular "box" with dimensions 21 cm × 41 cm × 170 cm . if he falls feet first, his drag

LuckyWell [14K]

Formula for terminal velocity is:

Vt = √(2mg/ρACd)
Vt = terminal velocity = ?
m = mass of the falling object = 72 kg
g = gravitational acceleration = 9.81 m/s^2
Cd = drag coefficient = 0.80
ρ = density of the fluid/gas = 1.2 kg/m^3
A = projected area of the object (feet first) = 0.21 m * 0.41 m = 0.0861 m^2

Therefore:

Vt = √(2 * 72 * 9.81 / 1.2 * 0.0861 * 0.80)

Vt = 130.73 m/s

4 0

3 years ago

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

Pepsi [2]

Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

$a_r=\omega^2R$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2$

Now, tangential acceleration is given by the formula:

$a_t=R\alpha$

Plug in the given values and solve for $a_t$ . This gives,

$a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2$

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

$a_{Total}=\sqrt{(a_r)^2+(a_t)^2}$

Plug in the given values and solve for total acceleration, $a_{Total}$ . This gives,

$a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2$

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

$\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°$

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

4 0

3 years ago