Whats the measure of how much energy a sound wave carries, the loudness of a sound?

A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in

katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= $10+1.50$

= $11.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v = \frac{d}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{11.5}$

$=26.08 \ s$

Throughout the opposite direction, when the boat seems to be travelling then,

= $10-1.50$

= $8.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v=\frac{v}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{8.5}$

$=35.29 \ s$

hence,

The time taken by the boat will be:

= $26.08+35.29$

= $61.37 \ s$

8 0

2 years ago

Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani

Dahasolnce [82]

Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

$n = \frac{t}{t_{1/2}}$

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

$t_{1/2}$ = half-life = 700 million years

Therefore,

$n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3$

Now, we will calculate the number of uranium nuclei left ( $n_u$ ):

$n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u = 800\ million$

and the rest of the uranium nuclei will become thorium nuclei ( $u_{th}$ )

$n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million$

dividing both:

$\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u$

<u>Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.</u>

6 0

2 years ago

A table-tennis ball is thrown at a stationary bowling ball. The table-tennis ball makes a one-dimensional elastic collision and

Alecsey [184]

<u>Option b. </u>A smaller magnitude of momentum and more kinetic energy.

<h3>What is a momentum?</h3>

In Newtonian physics, an object's linear momentum, translational momentum, or simply momentum is defined as the product of its mass and velocity.
It has both a magnitude and a direction, making it a vector quantity. The object's momentum, p, is defined as: p=mv if m is the object's mass and v is its velocity (also a vector quantity).
The kilogram metre per second (kg m/s), or newton-second in the International System of Units (SI), is the unit used to measure momentum.
The rate of change of a body's momentum is equal to the net force exerted on it, according to Newton's second law of motion.

To know more about momentum, refer:

brainly.com/question/1042017

#SPJ4

7 0

1 year ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

2 years ago

A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec

miss Akunina [59]

The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
$W=F d cos \theta$
where the force is F=qE, d=0.556 and $\theta=55.2^{\circ}$ . Using the value of q and E given by the problem, we find
$W=qEdcos\theta = 6.39\cdot10^{-5}J$

3 0

3 years ago