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Inessa [10]
2 years ago
6

In a thunderstorm, the thunder occurs at the same time as lightning. Explain how does a person hears the thunder 5s after he see

s the lightning
Physics
1 answer:
sergey [27]2 years ago
4 0

Answer:

If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close. Keep in mind that you should be in a safe place while counting.

Explanation:

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A ball is dropped off a cliff and falls for 24 seconds. How far did the ball fall from the top down
DerKrebs [107]

Answer:

9266 feet

Explanation:

with Earth's gravity and long it fell that's as good as it gets if there was no other factors like wind mass weight but your welcome

4 0
2 years ago
If the radius of an atom is 60 pm and the radius of the Earth is 6000 km, by how many orders of
kherson [118]

Answer:

1 x 10¹⁷

               

Explanation:

Given data:

        Radius of the earth  = 6000km

        Radius of an atom  = 60pm

Now, how many orders is the radius of the earth larger than an atom

Solution:

To solve this problem, let us express both quantity as the same unit;

         1000m  = 1km

    6000km  = 6000 x 10³m   =  6 x 10⁶m

    60pm;

            1 x 10⁻¹²m  = 1pm

    60pm  = 60 x  1 x 10⁻¹²m  = 6 x 10⁻¹¹m

Now;

The order:   \frac{6 x 10^{6} }{6 x 10^{-11} }   = 1 x 10¹⁷

               

6 0
2 years ago
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0
3 years ago
Estimate the moment of inertia of a bicycle wheel 66 cm in diameter. The rim and tire have a combined mass of 1.2 Kg
Fed [463]
<span>For a point mass the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr^2. That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection of point masses. So the I = (1.2 kg)(0.66m/2)^2 = 0.1307 kg m2</span>
3 0
3 years ago
An ideal heat engine absorbs 97.2 kJ of heat and exhausts 83.8 kJ of heat in each cycle. What is the efficiency of the engine?
r-ruslan [8.4K]

Answer:

13.7%

Explanation:

Given that,

Heat absorbed by the engine = 97.2 kJ

Heat exhausted by the engine in each cycle = 83.8 kJ

We need to find the efficiency of the engine. It is calculated by the formula.

\eta=1-\dfrac{Q_e}{Q_a}\\\\=1-\dfrac{83.8}{97.2}\\\\=0.137\\\\=13.7\%

so, the efficiency of heat engine is 13.7%.

5 0
3 years ago
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