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Novay_Z [31]
3 years ago
10

A rocket takes off from Earth. It travels 825km in 75 seconds. What is the

Physics
1 answer:
Naya [18.7K]3 years ago
4 0

Answer:

11 km/s

Explanation:

v=s/t

v=825km/75s

v=11km/s

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Explain how thermal energy (temperature) affects chemical changes.
Monica [59]
If bonds are broken, the energy is released, and if bonds are formed, energy is absorbed. During conversions from chemical energy to thermal energy, the energy stored in the chemical bonds are released and this energy causes surrounding molecules to move faster thus increasing the thermal energy of a substance.
8 0
3 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
How could you increase the kinetic energy of a wagon without increasing it's mass
Zarrin [17]

Answer:

send the wagon down a higher hill

3 0
3 years ago
Read 2 more answers
A motorist enters a freeway at 45 km/h and accelerates uniformly to 99 km/h. From the odometer in the car, the motorist knows th
Marat540 [252]

Answer:

a)  19440 km/h²

b) 10 sec

Explanation:

v₀ = initial velocity of the car = 45 km/h

v = final velocity achieved by the car = 99 km/h

d = distance traveled by the car while accelerating = 0.2 km

a = acceleration of the car

Using the kinematics equation

v² = v₀² + 2 a d

99² = 45² + 2 a (0.2)

a = 19440 km/h²

b)

t = time required to reach the final velocity

Using the kinematics equation

v = v₀ + a t

99 = 45 + (19440) t

t = 0.00278 h

t = 0.00278 x 3600 sec

t = 10 sec

5 0
3 years ago
Consider the rocket in the image. Which unbalanced force accounts for the direction of the net force of the rocket ?
Wittaler [7]

Answer:

unbalenced force

Explanation:

make Brainliest  plz

3 0
3 years ago
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