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4 years ago

Which of the following is group 1? a. alkali metals b. alkaline earth metals c. halogens d. transition metals

Physics

2 answers:

scoundrel [369]4 years ago

6 0

I might be D i think try D

Elanso [62]4 years ago

4 0

The answer is A. Alkali metals

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I need help ASAP plzzzz

Fiesta28 [93]

Answer:

a) 1.75s b) 17.2 m/s (down)

Explanation:

d1= 15m d2= 0m (because it hits ground)

a= -9.81 m/s^2 t=???

Equation

the triangle means change in so d2-d1

Δd= v1 * t + 1/2 * a * t^2

0m-15m= v1*t + 1/2 a t^2

-15 m= 0m/s*t (goes away) + 1/2* a *t^2

-15mx2= t^2

-15mx2/a= t^2

Square root (-30/-9.81m/s^2)

t=1.75 s

b) now v2!!

Im going to use v2= v1 + a*t

v2= 0m/s + -9.81 x 1.75s

v2 = -17.2 m/s or you can say 17.2 m/s down!!!

7 0

3 years ago

The ideal mechanical advantage of an inclined plane is 3.5, and its efficiency is 0.6. What is the mechanical advantage of the i

Nadya [2.5K]

Mechanical advantage = ideal mechanical advantage x efficiency = 3.5 x 0.6 = 2.1
The mechanical advantage of the inclined plane is 2.1

7 0

3 years ago

Suppose the coefficient of static friction between the road and the tires on a car is 0.638 and the car has no negative lift. Wh

larisa [96]

Answer:

12.6332454263 m/s

Explanation:

m = Mass of car

v = Velocity of the car

$\mu$ = Coefficient of static friction = 0.638

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of turn = 25.5 m

When the car is on the verge of sliding we have the force equation

$\dfrac{mv^2}{r}=\mu mg\\\Rightarrow v=\sqrt{\mu gr}\\\Rightarrow v=\sqrt{0.638\times 9.81\times 25.5}\\\Rightarrow v=12.6332454263\ m/s$

The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s

4 0

3 years ago

How to use kinetic energy to calculate velocity

BartSMP [9]

As we know the formula of kinetic energy is

$KE = \frac{1}{2} mv^2$

here given that

KE = 150,000 J

mass = 120 kg

we can use this to find speed

$150,000 = \frac{1}{2} * 120 * v^2$

$v^2 = 2500$

$v = 50 m/s$

So speed of above object is 50 m/s

7 0

3 years ago

series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s

andre [41]

Answer:

$d=1.84\ mm$

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

$\displaystyle C=\frac{\epsilon_o A}{d}$

where

$\epsilon_o=8.85\cdot 10^{-12}\ F/m$

A = area of the plates = $\pi r^2$

d = separation of the plates

$\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}$

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

$\displaystyle X_c=\frac{1}{wC}$

where w is the angular frequency

$w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s$

Solving for C

$\displaystyle C=\frac{1}{wX_c}$

The reactance can be found knowing the total impedance of the circuit:

$Z^2=R^2+X_c^2$

Where R is the resistance, $R=15 K\Omega=15000\Omega$ . Solving for Xc

$X_c^2=Z^2-R^2$

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

$\displaystyle Z=\frac{V}{I}$

The rms current is the peak current Ip divided by $\sqrt{2}$ , thus

$\displaystyle Z=\frac{\sqrt{2}V}{I_p}$

$I_p=0.65\ mA/1000=0.00065\ A$

Now collect formulas

$\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2$

Or, equivalently

$\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}$

$\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}$

$X_c=36176.34\ \Omega$

The capacitance is now

$\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F$

The radius of the plates is

$r=18\ cm/2=9 \ cm = 0.09 \ m$

The separation between the plates is

$\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}$

$d=0.00184\ m$

$\boxed{d=1.84\ mm}$

8 0

3 years ago