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____ [38]
3 years ago
5

How much energy in calories does it take to melt 10 grams of ice?

Physics
2 answers:
Nikitich [7]3 years ago
3 0

Answer:

800 calories

Explanation:

gavmur [86]3 years ago
3 0
750-800 should be correct
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Less than or equal to the magnitude of the vector
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Which of the following is a problem that some people blame on technology?
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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
The ____ of an object is a measure of the amount of matter it contains. on the other hand ____ is a measure of the gravitational
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An object with a mass of 20 kg has a net force of 80 N acting on it. What is the acceleration of the object?
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=4m/s^2

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