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hoa [83]
3 years ago
6

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

age. In one such accident, a 1850 kg car traveling to the right at 1.60 m/s collides with a 1400 kg car going to the left at 1.10 m/s . Measurements show that the heavier car's speed just after the collision was 0.270 m/s in its original direction. You can ignore any road friction during the collision.
Physics
1 answer:
iris [78.8K]3 years ago
5 0

Answer:

Explanation:

Given:

U1 = 1.6 m/s

U2 = -1.1 m/s

M1 = 1850 kg

M2 = 1400 kg

V1 = 0.27 m/s

Using momentum- collision equation,

M1U1 + M2U2 = M1V1 + M2V2

1850 × 1.6 - 1400 × 1.1 = 1850 × 0.27 + 1400 × V2

1420 = 499.5 + 1400V2

V2 = 0.6575 m/s

B.

KE = 1/2 × MV^2

KEa1 + KEa2 = KEb1 + KEb2

Delta KE = KE2 - KE1

KEa1 = 2368 J

KEb1 = 847 J

KEa2 = 67.433 J

KEb2 = 302.6 J

KE1 = KEa1 + KEb1

= 3215 J

KE2 = 370.033 J

Delta KE = -2845 J.

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PLEASE HELPPPPP!!Calculate the net force acting on obiect A. In your answer, be sure to include the number, unit, and
asambeis [7]

Answer:

4 N to the right

Explanation:

For the object in this problem, we have 2 forces acting on it:

- A force of 2 N, to the right

- A force of 2 N, to the right

The two forces are in line: this means that the two vectors can be simply added by calculating their algebraic sum.

If we take right as positive direction, the two forces can be written as

F1 = +2 N

F2 = +2 N

So, the net force will be

F=F_1+F_2=+2+2=+4 N

And the positive sign means the direction is to the right.

6 0
3 years ago
A ball with a mass of 275 g is dropped from rest, hits the floor, and re-bounds upward. If the ball hits the floor with a speed
disa [49]

Answer:

a) \Delta p = 1.350\,\frac{kg\cdot m}{s}, b) \Delta p' = -0.454\,\frac{kg\cdot m}{s}, c) D. The magnitud of the change in the ball's momentum.

Explanation:

a) The magnitude of the change in the ball's momentum is:

\Delta p = (0.275\,kg)\cdot \left[\left(1.63\,\frac{m}{s} \right)-\left(-3.28\,\frac{m}{s} \right)\right]

\Delta p = 1.350\,\frac{kg\cdot m}{s}

b) The change in the magnitude of the ball's momentum:

\Delta p' = (0.275\,kg)\cdot \left[(1.63\,\frac{m}{s} )-(3.28\,\frac{m}{s} ) \right]

\Delta p' = -0.454\,\frac{kg\cdot m}{s}

c) The magnitude of the change in the ball's momentum is more directly related to the net force acting on the ball, as it measures the effect of the force on change in ball's motion at measured time according to the Impact Theorem. So, the right answer is option D.

3 0
3 years ago
What are the products of linear electron flow during the light reactions of photosynthesis?
Katena32 [7]

Answer:

NADPH and ATP

Explanation:

In the clear stage the light that "hits" chlorophyll excites an electron to a higher energy level. In a series of reactions, energy is converted (throughout an electron transport process) into ATP and NADPH. Water breaks down in the process releasing oxygen as a secondary product of the reaction. ATP and NADPH are used to make the C-C bonds in the dark stage.

Photophosphorylation is the process of converting the energy of the electron excited by light into a pyrophosphate bond of an ADP molecule. This occurs when water electrons are excited by light in the presence of P680. The energy transfer is similar to the chemosmotic electron transport that occurs in the mitochondria.

Light energy causes the removal of an electron from a P680 molecule that is part of Photosystem II, the electron is transferred to an acceptor molecule (primary acceptor), and then passes downhill to Photosystem I through a conveyor chain of electrons The P680 requires an electron that is taken from the water by breaking it into H + ions and O-2 ions. These O-2 ions combine to form O2 that is released into the atmosphere.

The light acts on the P700 molecule of Photosystem I, causing an electron to be raised to a higher potential. This electron is accepted by a primary acceptor (different from the one associated with Photosystem II).

The electron goes through a series of redox reactions again, and finally combines with NADP + and H + to form NADPH, a carrier of H needed in the independent phase of light.

Electron of photosystem II replaces the excited electron of the P700 molecule.

There is therefore a continuous flow of electrons (non-cyclic) from water to NADPH, which is used for carbon fixation.

Cyclic electron flow occurs in some eukaryotes and in photosynthetic bacteria. NADPH does not occur, only ATP. This also occurs when the cell requires additional ATP, or when there is no NADP + to reduce it to NADPH.

In Photosystem II, the "pumping" of H ions into the thylakoids (from the stroma of the chloroplast) and the conversion of ADP + P to ATP is motorized by an electron gradient established in the thylakoid membrane.

7 0
3 years ago
A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108
tiny-mole [99]

Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

W = F*d

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m  

Hence, the work is:

W = F*d = 108.915 N*2.38 m = 259.22 J

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!                                                

6 0
3 years ago
Circuit diagram showing three resistors in series with an ammeter,a cell and a switch​
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Answer:

the mayflower

Explanation:

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