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mario62 [17]
2 years ago
10

A car traveling initially at 7.35 m/s acceler-

Physics
1 answer:
Flura [38]2 years ago
4 0

Answer:

v_f=9,07~m/s

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes uniformly in time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

v_f=v_o+at

The car initially travels at vo=7.35 m/s and accelerates at a rate of a=0.824~m/s^2 during t=2.09 s.

The final velocity is:

v_f=7.35+0.824*2.09

\mathbf{v_f=9,07~m/s}

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4. A little boy pushes a wagon with his dog in it. The mass of the dog and
Artyom0805 [142]

Answer:

<h2>59.5 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 70 × 0.85

We have the final answer as

<h3>59.5 N</h3>

Hope this helps you

6 0
3 years ago
Which statements describe density? Check all that apply.
HACTEHA [7]

Density is a chemical property of an object.

The density of an object is constant.

Density is a derived unit of measure

7 0
2 years ago
Read 2 more answers
What is the name of the covalent compound P4S2?
Roman55 [17]

Tetraphosphorus disulfide


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6 0
3 years ago
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Find the deBroglie wavelength of an electron with 4.0 eV.
pantera1 [17]

Explanation:

It is given that,

Voltage, V=4 eV=4\times 1.6\times 10^{-19}\ V= 6.4\times 10^{-19}\ V

De broglie wavelength in terms of voltage is given by :

\lambda=\dfrac{h}{\sqrt{2meV} }

m and e are the mass and charge on electron. So,

\lambda=\dfrac{12.27}{\sqrt{V} }\ A

\lambda=\dfrac{12.27}{\sqrt{6.4\times 10^{-19}} }\ A  

\lambda=1.53\times 10^{10}\ A

\lambda=1.53\ m

So, the De broglie wavelength of an electron is 1.53 meters. Hence, this is the required solution.      

3 0
3 years ago
How many 100W light bulbs could be powered for one year by the direct conversion of 1g of matter into energy?
stellarik [79]

Answer:

No. of 100 W bulbs, n = 28,539 bulbs

Given:

Power of a single bulb = 100 W

Time period, T = 1 yr = 365\times 24\times 60\times 60 = 31,536,000 s

mass of matter, m = 1 g = 1\times 10^{-3}]

Solution:

According to Eintein's mass-energy equivalence:

E = mc^{2}

E = 1\times 10^{-3}\times (3\times 10^{8})^{2}

E =  9\times 10^{13} J

Power of a single bulb, P =  \frac{E}{T}

                                       P =  \frac{9\times 10^{13} }{31,536,000}

                                       P = 28,53,881 W

No. of 100 W bulbs, n = \frac{P}{power of one bulb}

                                 n = \frac{28,53,881}{100}

                                 n = 28,539

5 0
3 years ago
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