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jasenka [17]
3 years ago
14

The two sleds shown below are about to collide. If they stick together in the collision, what will happen after the collision? (

In this collision, there is no friction
between the sleds and the snow.)
A. The two sleds will move to the left.
O B. The two sleds will move to the right
O C. The two sleds will be motionless in the middle.
O D. There is not enough information to know which sled will be pushed backward,
om
Physics
1 answer:
zhannawk [14.2K]3 years ago
3 0

Answer:

B. The two sleds will move to the right.

Explanation:

I majored in physics

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How do you find the kinetic energy of an object? ​
Sergeeva-Olga [200]

Answer:

KE = 1/2 * m * v^{2}

Explanation:

use the formula:

KE = 1/2 * m * v^{2}

KE = kinetic energy in joules (J)

m = mass in kg

v = velocity in m/s

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"State the process that divides one nucleus into two genetically identical nuclei"
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Answer:

<em>Mitosis</em>

Explanation:

Mitosis: Mitosis is a cell or nuclear division following the duplicate of the chromosomes, whereby each daughter cell or nucleus has exactly the same chromosomes content as the parent. In other words, mitosis is a cell division in which daughter cells have the same number of chromosomes as the parent cell.

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2 years ago
What is the net force on this object?​
Mariulka [41]
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3 years ago
Energy transferred as heat occurs between two bodies in thermal contract when they differ in which of the following properties?
Hatshy [7]
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3 years ago
A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package hits the ground, how high
Lelu [443]

Answer:

The package was released at a height of 1015.296 meters.

Explanation:

The package is dropped at an initial velocity different of zero, decelerated and later accelerated by gravity. Let assume that final height is equal to zero, the final height is given by the following equation of motion:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}

Where:

v_{o} - Initial velocity, measured in meters per second.

y - Final height, measured in meters.

y_{o} - Initial height, measured in meters.

t - Time, measured in seconds.

g - Gravitational constant, measured in meters per square second.

(Positive sign - Package is moving upward, Negative sign - Package is moving downward)

The initial height is now cleared:

y_{o} = y - v_{o}\cdot t - \frac{1}{2}\cdot g \cdot t^{2}

Given that y = 0\,m, v_{o} = 15\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} and t = 16\,s, the final height of the package is:

y_{o} = 0\,m - \left(15\,\frac{m}{s} \right)\cdot (16\,s) - \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (16\,s)^{2}

y_{o} = 1015.296\,m

The package was released at a height of 1015.296 meters.

7 0
3 years ago
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