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3 years ago

The energy transferred by a force to a moving object?

Physics

2 answers:

mixer [17]3 years ago

8 0

Answer:

mechanical energy

Vikentia [17]3 years ago

3 0

Answer: mechanical energy

Explanation: I hope this helps! Also, please mark as brainliest, thanks!

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Please help me with this 29 points

Irina18 [472]

Answer:

)Give the definition of poverty line as defined by the World Bank.

7 0

3 years ago

Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4

kati45 [8]

Answer:

velocity = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = $\frac{\pi }{4}$ × (4.50× $10^{-2}$ )² × velocity

solve it we get

velocity = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = $\frac{\pi }{4}$ × 3 × (4.50× $10^{-2}$ )² × velocity

velocity = 52.4 m/s

5 0

3 years ago

I need help with egg drop assignments any ideas?

Jet001 [13]

Answer:

make a parachute out of the bag connecting to a bowl made out of paper filled with cotton balls. then put the egg in the bowl.

Explanation:

7 0

3 years ago

In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it

bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

$=\sqrt{2 \times 9.8 \times 0.35}$ (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}$

$=\frac{0.175}{0.525}+0$

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

3 0

3 years ago

Find the total electric charge of 1.7 kg of electrons. me=9.11×10−31kg, e=1.60×10−19C.

Gelneren [198K]

Answer:

$2.99\cdot 10^{11}C$

Explanation:

The mass of one electron is

$m_e = 9.11\cdot 10^{-31}kg$

So the number of electrons contained in M=1.7 kg of mass is

$N=\frac{M}{m_e}=\frac{1.7 kg}{9.11\cdot 10^{-31}kg}=1.87\cdot 10^{30}$

The charge of one electron is

$e=1.60\cdot 10^{-19} C$

So, the total charge of these electrons is equal to the charge of one electron times the number of electrons:

$Q=Ne=(1.87\cdot 10^{30})(1.6\cdot 10^{-19}C)=2.99\cdot 10^{11}C$

8 0

3 years ago