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2 years ago

URGENT PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST !!!

Physics

2 answers:

Salsk061 [2.6K]2 years ago

8 0

Sound waves because sound requires a medium such as air or water

user100 [1]2 years ago

6 0

Answer:

sound waves

Explanation:

have good day

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Work of 2 joules is done in stretching a spring from its natural length to 11 cm beyond its natural length. what is the force (i

Sidana [21]

The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.

If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is

W = k[{x₂-x₁)² - (x₁-xn)²],

where
xn is the natural length
x₁ is the stretched length
x₂ is also the stretched length when stretched even further than x₁

In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.

Then, substituting the values,

2 J = k (0.11² -0²)
k = 165.29 N/m

Finally, we use the value of k to the Hooke's Law to determine the Force.

F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons

5 0

3 years ago

Find the net force and acceleration. 15 points. Will give brainliest!

gladu [14]

Answer:

$\boxed{F_{net} = 28.7 \ N}$

$\boxed{a = 2.1 \ m/s^2}$

Explanation:

Finding the net force:

Firstly , we'll find force of Friction:

$F_{k} = (micro)_{k}mg$

Where $(micro)_{k}$ is the coefficient of friction and m = 13.6 kg

$F_{k} = (0.16)(13.6)(9.8)\\$

$F_{k} = 21.32 \ N$

Now, Finding the net force:

$F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\$

$F_{net} = 28.7 \ N$

Finding Acceleration:

$a = \frac{F_{net}}{m}$

$a = \frac{28.7}{13.6}$

$a = 2.1 \ m/s^2$

8 0

3 years ago

Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi

Orlov [11]

The number converted is $0.0467 \frac{(kg)(s)}{m^3}$

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

$1 kg = 1000 g = 10^6 mg$

$1 min = 60 s$

$1 m^3 = 1000 L$

The original unit that we have is

$\frac{mg\cdot min}{L}$

Therefore, it can be rewritten as:

$=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot 60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}$

Therefore, since the initial number was 0.779, the final value is

$0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}$

#LearnwithBrainly

5 0

3 years ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

u₁ = √(2 × 9.8 × 5)

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

x = 2.86 m

8 0

3 years ago

What substance is used by plants as a source of food

harkovskaia [24]

Answer:

dart

Explanation:

dart and sun and water so that the plant be okay

7 0

2 years ago