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3 years ago

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For horizontally-launched projectiles, which of the following describes acceleration in both directions with a = 0 and a = -9.8m

/s^2
1. Acceleration is constant in both directions
2. Acceleration is -9.8m/s^2 in both directions
3. Acceleration is zero in both directions
4. Acceleration is equal in both directions

1 answer:

natulia [17]3 years ago

5 0

Answer:

4 is the best option for it as there are some stuff in grade to ye

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How much would you have to eat for your appendix to explode ? I ate so much on thanksgiving and i need to know i’m very scared

dimulka [17.4K]

Answer:

I don't think your appendix can explode because you ate too much honestly. It's not even possible to eat so much that your appendix explodes, and if you're feeling any pain it definitely isn't because your appendix is about to explode, believe me. Also you could just type it into the internet, that'd be a much faster solution.

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3 years ago

A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the

vitfil [10]

Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

The average speed of the rider bicycle is

average speed = total distance/ total time

Mathematically, it can be computed as:

$v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}$

$v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}$

$v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}$

$v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}$

$\mathbf{v_{avg} =12.31 \ m/s}$

8 0

2 years ago

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Marianna [84]

Answer:

The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

When two point are at same height so ,

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$ ....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

$Q=A_{1}v_{1}$

$v_{1}=\dfrac{Q}{A_{1}}$

$v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}$

$v_{1}=56.58\ m/s$

For output velocity,

$v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}$

$v_{2}=6.28\ m/s$

Put the value into the formula

$P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)$

$\Delta P=1.58\times10^{6}\ N/m^2$

(b). We need to calculate the maximum height

Using formula of height

$\Delta P=\rho g h$

Put the value into the formula

$1.58\times10^{6}=1000\times9.8\times h$

$h=\dfrac{1.58\times10^{6}}{1000\times9.8}$

$h=161.22\ m$

Hence, The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

8 0

3 years ago

A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N1 turns. A second toroidal solenoid w

lilavasa [31]

Answer:

Mutual inductance, $M=2.28\times 10^{-5}\ H$

Explanation:

(a) A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N₁ turns. A second thyroidal solenoid with N₂ turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius.

Mutual inductance is given by :

$M=\dfrac{\mu_oN_1N_2A}{2\pi r}$

(b) It is given that,

$N_1=550$

$N_2=290$

Radius, r = 10.6 cm = 0.106 m

Area of toroid, $A=0.76\ cm^2=7.6\times 10^{-5}\ m^2$

Mutual inductance, $M=\dfrac{4\pi \times 10^{-7}\times 550\times 290\times 7.6\times 10^{-5}}{2\pi \times 0.106}$

$M=0.0000228\ H$

or

$M=2.28\times 10^{-5}\ H$

So, the value of mutual inductance of the toroidal solenoid is $2.28\times 10^{-5}\ H$ . Hence, this is the required solution.

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3 years ago

Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{Mm}{r^{2}}$ (1)

Where:

$F$ is the gravitational force between Earth and Moon

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$m=7.349(10)^{22} kg$ is the mass of the Moon

$r=3.9(10)^{8} m$ is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to $F$ :

$F=m.a_{C}$ (2)

Where $a_{C}$ is the centripetal acceleration given by:

$a_{C}=\frac{V^{2}}{r}$ (3)

Being $V$ the orbital velocity of the moon

Making (1)=(2):

$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

Finally:

$V=1010.92 m/s$

5 0

2 years ago