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Setler79 [48]
3 years ago
5

your neighbour is throttling his recent bought motorbike to show off. the sound intensity measured at your window 16m away is 0.

25W/m^2. what is the sound intensity level in dB at your friend's house, a distance of 28m away from the noisy bike?
Physics
1 answer:
aivan3 [116]3 years ago
8 0

Answer:

hi your pinterest I'd or Twitter I'd pls.

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Describe briefly how you would check that the rule is horizontal in this experiment. You  may draw a diagram.​
aliina [53]

Answer:

We want to describe how to graph a linear equation.

Explanation:

The given equation is:

y = -4x - 1

a) To graph it, we need to find two points that belong to the line, then we graph the points, and then we connect them with a line.

To get the points, we just need to evaluate the function in two different values of x.

for x = 0

y = -4*0 - 1 = -1

So we have the point (0, -1)

for x = 1

y = -4*1 - 1 = -5

So we have the point (1, - 5)

Now we just need to find these two points and connect them with a line, the graph can be seen below.

b) To check if the graph is correct we can see two things:

in y = -4*x - 1

The y-intercept is -1, this means that the graph should intersect the y-axis at  y = -1

The slope is -4, this means that for each unit increase on x, we should see that the y-value decreases by 4.

Checking those two things we can see if our graph is correct or not.

pls give brainliest!

3 0
2 years ago
A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar
m_a_m_a [10]

Answer:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

#Where u is in meters and t in seconds.

Explanation:

Given that :m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s

From \omega=kL we have:

k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2

From F_d(t)=-\gamma u\prime(t) we have that:

\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m

Now,given that the initial value problem is given by:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

Hence,the position of u at time t is given by

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\, u in meters,t in seconds.

4 0
3 years ago
Answer choices
marshall27 [118]

Answer:

Option-C (Lipoprotein profile)

4 0
2 years ago
Longitudinal waves cannot travel through a vacuum. True False
JulsSmile [24]
FALSE <span>Only electromagnetic </span>waves<span> can </span>travel through a vacuum<span>; mechanical </span>waves<span> such as sound </span>waves<span> require a particle-interaction to transport their energy. There are no particles in a </span>vacuum<span>. </span>Waves<span> are either </span>longitudinal<span> or transverse.
</span>Hoped i helped   :):)
4 0
3 years ago
On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1
Ad libitum [116K]

Answer:

D) The worker will accelerate at 2.17  m/s²  and the box will accelerate at 1.08  m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m =  W/g

Where:

W : Weight (N)

g : acceleration due to gravity  (m/s²)

Data

W₁ =1.8 kN   : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁    m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}

a₂=  -2.17 m/s² : acceleration of the worker

5 0
3 years ago
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