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DerKrebs [107]
3 years ago
12

Nerve impulses in the human body travel at a speed of about 100 m/s. A 1.6 m tall man accidentally drops a hammer on his toe. Ho

w long does it take for the nerve impulse to travel from his toe to his brain?
Physics
1 answer:
EastWind [94]3 years ago
4 0

time=distance/speed

1.6/100 secs = 0.016secs=16millisecs

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Use the image to determine the volume of the rock.
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Answer:

12 ml

Explanation:

The initial volume in the cylinder is 20 ml

 adding the rock adds volume to the cylinder

       the new volume is 32 ml .....the increase in volume is the volume of the rock :   32 - 20 = 12 ml    volume of rock

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What is the frequency of this wave?
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Can you find the net force of one vertical force and one horizontal force, such as in the picture below? Yes or no, and why?
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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
If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?
leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

<span>[density * volume]_salt water = [density * volume]_fresh water                 ---> 1</span>

 

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

 

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

<span>h = 0.72 m        (ANSWER)</span>

8 0
3 years ago
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