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3 years ago

Can someone pls help.

Physics

1 answer:

strojnjashka [21]3 years ago

4 0

Answer:take an apple out of the box (I THINK)

Explanation:

i what i think

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What the charges of the nucleus

saw5 [17]

Protons are positive, electrons are negative, and neutrons are neutral. In the nucleus, there are protons and neutrons, so the charge of a nucleus is positive.

4 0

3 years ago

If a train travels 500 kilometers from Stockholm to

o-na [289]

Hello,

Average speed is total distance divided by total time. From the problem, our total distance is given as 500 kilometers and given time is 5 hours. Therefore, the average speed is:

$\displaystyle{v_\text{average}=\sum_{i=1}^n \dfrac{s_i}{t_i}}\\\\\displaystyle{v=\dfrac{500\ \text{km}}{5 \ \text{h}}}\\\\\displaystyle{v=100 \ \text{km/h}}$

Therefore, the average speed is 100 km/h. Please let me know if you have any questions!

4 0

1 year ago

A roller coaster weighing 2000 kg is lifted to a height of 28 m within 30 seconds with a kind of elevator.Calculate how many kil

Vladimir [108]

Answer:

Explanation:

The equation for Power is

P = Work/time to do work and the equation for work is

Work = FΔx

We first need to find the amount of work done, then we can find the power it took to do that work.

W = 2000(9.8)(28) so

W = 550,000 N*m

Now we fill that into the power equation:

$P=\frac{550000}{30}$ gives us

P = 18000 Watts. But we need kW, so we divide by 1000 to get

P = 18 kW of power.

7 0

2 years ago

Read 2 more answers

Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0

Inessa05 [86]

Let $a_1$ be the average acceleration over the first 2.46 seconds, and $a_2$ the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let $v$ be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

$a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$

$a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}$

and we're also told that

$\dfrac{a_1}{a_2}=1.66$

(or possibly the other way around; I'll consider that case later). We can solve for $a_1$ in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

$1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}$