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drek231 [11]
3 years ago
13

A crane exerts a net force of 900 N upward on a 750-kilogram car as the crane starts to lift the car from the deck of a cargo sh

ip. What is the acceleration of the car during this time?
Physics
1 answer:
melisa1 [442]3 years ago
5 0

Answer: 1.2m/s^2

Explanation: the force exerted on the car is 900N upwards

The mass of the car is 750kg

According to Newton's third law acceleration is proportional to force

F = ma

900 = 750a

a = 900/750

a = 1.2m/s^2

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Snowcat [4.5K]

My best guess is c) Dark colors reflect less radiation making them warmer.

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3 years ago
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An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
just olya [345]

Answer:

2.78\times 10^{-35}\ \text{kg m/s}

6.178\times 10^{-34}\ \text{m/s}

0.31\times 10^{-4}\ \text{m/s}

Explanation:

\Delta x = Uncertainty in position = 1.9 m

\Delta p = Uncertainty in momentum

h = Planck's constant = 6.626\times 10^{-34}\ \text{Js}

m = Mass of object

From Heisenberg's uncertainty principle we know

\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

The minimum uncertainty in the momentum of the object is 2.78\times 10^{-35}\ \text{kg m/s}

Golf ball minimum uncertainty in the momentum of the object

m=0.045\ \text{kg}

Uncertainty in velocity is given by

\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}

The minimum uncertainty in the object's velocity is 6.178\times 10^{-34}\ \text{m/s}

Electron

m=9.11\times 10^{-31}\ \text{kg}

\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}

The minimum uncertainty in the object's velocity is 0.31\times 10^{-4}\ \text{m/s}.

6 0
2 years ago
Two metal spheres are suspended from strings. The
nikdorinn [45]

Answer:

B. A repulsive force of 8.0*10^3 N.

Explanation:

As we know by Coulomb's law that the electrostatic force between two charges is given as

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = -2 \times 10^2 C

q_2 = -4 \times 10^{-8} C

r = 3.0 m

now we have

F = \frac{(9 \times 10^9)(2 \times 10^2)(4\times 10^{-8})}{3^2}

F = 8000 N

since both charges are similar charges so they will repel each other by the force we calculated above so correct answer will be

B. A repulsive force of 8.0*10^3 N.

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3 years ago
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