PHYSICS HIGH SCHOOL

An explosion occurs 340 km away. Given that sound travels at 340 m/s, the time the sound takes to reach you is a. 1 s.
b. 10 s.
c. 100 s.
d. 200 s.
e. more than 200 s.

Answers

Answer 1
Answer:
(340,000 meters) / (340 meters/sec)  =  1,000 seconds

That's more than 200 sec.
Answer 2
Answer: first convert 340km into 340,000m, and 340000/340=1000. So the answer is e.

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Maria’s father started a fire in the fireplace. He crumpled some paper, lit a match, and soon the logs in the fireplace were burning. In this case, the stored chemical energy in the logs was changed intoA) electrical energy.
B) mechanical motion.
C) heat and light energy.
D) electrical and heat energy.

Answers

As per the question the crumpled paper was burnt .

When the paper was burnt,it produces light.One also feels the sensation of heat.

The electrical energy is due to the flow of electrons. Here no electronic motion takes place.Hence no electrical energy will be produced.

Mechanical energy is the sum total of kinetic and potential energy.The kinetic energy is the energy gained due to the motion of the body and potential energy is the energy gained due to the position or configuration of the body.

Hence from above,we can get that the correct answer to the question is-C] Heat and light energy.
It was changed into heat and light energy.

A disk-shaped space station 175 m in diameter spins uniformly about an axis perpendicular to the plane of the disk through its center. How many rpm (rev/min) must this disk make so that the acceleration of all points on its rim is g/2?

Answers

The angular velocity of the disk must be 2.25 rpm

Explanation:

The centripetal acceleration of an object in circular motion is given by

a=\omega^2 r

where

\omega is the angular velocity

r is the distance of the object from the axis of rotation

For the space station in this problem, we have

a=(g)/(2)=(9.8)/(2)=4.9 m/s^2 is the centripetal acceleration

The diameter of the disk is

d = 175 m

So the radius is

r=(175)/(2)=87.5 m

So, a point on the rim has a distance of 87.5 m from the axis of rotation. Therefore, we can re-arrange the previous equation to find the angular velocity:

\omega = \sqrt{(a)/(r)}=\sqrt{(4.9)/(87.5)}=0.237 rad/s

And this is the angular velocity of any point along the disk. Converting into rpm,

\omega=0.236 (rad)/(s)\cdot (60 s/min)/(2\pi rad/rev)=2.25 rpm

Learn more about circular motion:

brainly.com/question/2562955

#LearnwithBrainly

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be balanced? The answer is 13.4 In the mobile what is the value for m3 to the nearest hundredth of a kilogram?

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm = m_2 * xcm

Making x the subject of the formula  

                x = (m_1 * 15)/(m_2)

                    = (0.42 * 15)/(0.47)

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1 \ and \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20 = m_3 * 30

          (0.42 + 0.47) * 20 = 30 * m_3

 Making m_3 subject

          m_3 = ((0.42 + 0.47) * 20 )/(30 )

                m_3 = 0.59 kg

Endothermic reactions give off heat as they proceed.
a. True
b. False

Answers

b. False
Endothermic reactions absorb heat.

Two objects attract each other gravitationally with a force of 3.0×10−10N when they are 0.45m apart. Their total mass is 4.0kg . find the individual masses

Answers

If everything is in meters and kilograms, then the gravitational force
between 2 masses is

F = G M₁ M₂ / R²

G = 6.67 x 10⁻¹¹
M₁ and M₂ =  the masses of the two masses
R = the distance between their centers.

When we plug in 3 x 10⁻¹⁰ for the force, and 0.45 for the distance,
we get the product of the 2 objects  M₁M₂ = about 0.911 , and
that's as far as we can go without some more information.

The question says that the two masses add up to 4.0 kg.
So now we know that M₁M₂ = about 0.911 and (M₁ + M₂) = 4.0 kg.
With this, we immediately step into a big squishy quadratic equation,
whose solutions are:

one mass = 3.758 kg
the other mass = 0.242 kg .

A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at rest, how fast is it moving after it travels 3 meters?

Answers

The box has 3 forces acting on it:

• its own weight (magnitude w, pointing downward)

• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)

• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ F = -f + w sin(35°) = m a

• net perpendicular force:

∑ F = n - w cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

n = w cos(35°)

n = (15 kg) (9.8 m/s²) cos(35°)

n ≈ 120 N

Solve for the mag. of friction:

f = µn

f = 0.25 (120 N)

f ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a

a ≈ (54.3157 N) / (15 kg)

a ≈ 3.6 m/s²

Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:

v² - v₀² = 2 a ∆x

v² = 2 (3.6 m/s²) (3 m)

v = √(21.7263 m²/s²)

v ≈ 4.7 m/s
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