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Kitty [74]
3 years ago
15

While filming an intense action sequence for the next James Bond movie, a controlled explosion detonates 1.3 km away from the ac

tors. If the speed of sound through solid rock is 3000 m/s on average, the actors will feel the explosion before they hear it. How much time will pass between when they feel the explosion and when they hear it?
Physics
1 answer:
brilliants [131]3 years ago
8 0

To solve this problem we must basically resort to the kinematic equations of movement. For which speed is defined as the distance traveled in a given time. Mathematically this can be expressed as

v = \frac{d}{t}

Where

d = Distance

t = time

For which clearing the time we will have the expression

t = \frac{d}{v}

Since we have two 'fluids' in which the sound travels at different speeds we will have that for the rock the time elapsed to feel the explosion will be:

t = \frac{1300m}{3000m/s}

t = 0.433s

In the case of the atmosphere -composite of air- the average speed of sound is 343m / s, therefore it will take

t = \frac{1300m}{343m/s}

t = 3.79s

The total difference between the two times would be

\Delta t = 3.79s-0.433s

\Delta t = 3.357s

Therefore 3.357s will pass between when they feel the explosion and when they hear it

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What is the gravitational force between Mars and the sun? 7.43 × 1030 N 1.79 × 1026 N 1.65 × 1021 N 3.76 × 1032 N
VMariaS [17]

The gravitational force between Mars and the Sun is 1.65\cdot 10^{21} N

Explanation:

The magnitude of the gravitational force between two objects is given by  the equation:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we have:

m_1 = 1.99\cdot 10^{30} kg is the mass of the Sun

m_2 = 6.39\cdot 10^{23} kg is the mass of Mars

r=229\cdot 10^6 km = 229\cdot 10^9 m is the average distance Mars-Sun

Substituting into the equation, we find the gravitational force:

F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(6.39\cdot 10^{23})}{(229\cdot 10^9)^2}=1.62\cdot 10^{21} N

So, the closest answer is

1.65\cdot 10^{21} N

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0
3 years ago
Which has more thermal energy: lake or a cup of hot chocolate?
lawyer [7]
Though the hot cocoa would have a higher temperature, the lake would have more thermal energy because it has more molecules with a greater total internal energy.
4 0
3 years ago
Help for brainlist easy science plz
zalisa [80]

Answer:

the pics upside down fam

Explanation:

6 0
2 years ago
Read 2 more answers
1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g
Delvig [45]

1) 6.67\cdot 10^{-4} N

The force of gravitation between the two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2} is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N


2)  2.7\cdot 10^{-3} N

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

F \sim \frac{1}{r^2}

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F

So, the gravitational force increases by a factor 4. Therefore, the new force will be

F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N


3)  12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

\tau=Fd

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

\tau=(25 N)(0.5 m)=12.5 Nm


4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity (\omega) is:

L=I\omega

In this problem, we have

I=0.007875 kgm^2

\omega=6.28 rad/s

So, the angular momentum is

L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s

6 0
3 years ago
A 5.00 L air sample at a temperature of -50 °C has a pressure of 107 kPa. What will be the new pressure if the temperature is ra
Serjik [45]
Its simple use formuila ,
PV=nRT
n,R is constant as the both have same moles.
so,
(p1v1)/T1 = (p2v2)/T2
so, 128.53338kpa
4 0
2 years ago
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