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melamori03 [73]

3 years ago

13

A 2 kg object is

1 answer:

Mademuasel [1]3 years ago

8 0

NO net force is required to keep a moving object moving in a straight
line at a constant speed. In fact, if you apply ANY force to it, in ANY
direction, then its speed, its direction, or both must change, and its
velocity won't be uniform any more.

I know we never see this in our daily life. Whenever we see an object
moving, it always stops. That's because the net force on it is never zero ...
there's always some gravity or some friction acting on it. That's what you
have to put up with when you live on Earth.

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What are the inputs (reactants) of photosynthesis?

Katyanochek1 [597]

Answer:

Explanation:

Carbon dioxide and water.

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

4 0

3 years ago

Striking a fixed object at 60 MPH is the same as falling from a _____ story building:

yulyashka [42]

speed of the striking object is given as

$v = 60 mph$

now this speed in SI units given as

$v = 60 \times \frac{1609 m}{3600 s}$

$v = 26.82 m/s$

So here we can use kinematic now to find the height from where it is dropped

so here we will have

$v_f^2 - v_i^2 = 2 a d$

now we have

$v_f = 26.82 m/s$

$v_i = 0$

$a = 9.81 m/s^2$

now from above equation

$26.82^2 - 0 = 2(9.81)d$

$d = 36.67 m$

now we know that average height of 1 story building is 3 meter

so approx total number of stories in the building will be

$N = \frac{36.67}{3} = 12 story$

6 0

3 years ago

Read 2 more answers

Intense physical activity that requires little oxygen but uses short bursts of energy is called Anaerobic exercise? True or fals

Ghella [55]

I believe that is true. Anaerobic means without (much) Oxygen. This usually builds up lactic acid in muscles.

8 0

3 years ago

How much work does the charge escalator do to move 2.30 μC of charge from the negative terminal to the positive terminal of a 3.

loris [4]

Answer:

$6.9\times 10^{-6} J$

Explanation:

We are given that

Charge=q= $2.3\mu C=2.3\times 10^{-6} C$

$1\mu C=10^{-6} C$

Potential difference=V=3 V

We know that

Work done= $V\times q$

Using the formula

Work done by charge to move from the negative terminal to the positive terminal of battery= $2.3\times 10^{-6}\times 3$ J

Work done by charge to move from the negative terminal to the positive terminal of battery= $6.9\times 10^{-6} J$

6 0

3 years ago

kap26 [50]

<h2>A is the correct answer!</h2><h3></h3><h3>I'm too lazy to explain :(</h3><h3></h3><h3><em>Please let me know if I am wrong.</em></h3>

6 0

3 years ago