What is the chemical formula for magnesium sulfide?

Which was the first object made by humans to orbit earth?

The first human made object made to orbit earth was the Soviet Union's Sputnik 1, it launched the 4th of October 1957.

6 0

2 years ago

You want the current amplitude through a inductor with an inductance of 4.70 mH (part of the circuitry for a radio receiver) to

goldenfox [79]

Answer:

f = 1.69*10^5 Hz

Explanation:

In order to calculate the frequency of the sinusoidal voltage, you use the following formula:

$V_L=\omega iL=2\pi f i L$ (1)

V_L: voltage = 12.0V

i: current = 2.40mA = 2.40*10^-3 A

L: inductance = 4.70mH = 4.70*10^-3 H

f: frequency = ?

you solve the equation (1) for f and replace the values of the other parameters:

$f=\frac{V_L}{2\pi iL}=\frac{12.0V}{2\pi (2.4*10^{-3}A)(4.70*10^{-3}H)}=1.69*10^5Hz$

The frequency of the sinusoidal voltage is f

3 0

3 years ago

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.7 m tall window in 0.129 sec. From how high above th

vladimir1956 [14]

Answer:

h = 22.35 m

Explanation:

given,

initial speed of the rock,u = 0 m/s

length of the window,l = 2.7 m

time taken to cross the window,t = 0.129 s

Speed of the rock when it crosses the window

$v = \dfrac{l}{t}$

$v = \dfrac{2.7}{0.129}$

v = 20.93 m/s

height of the building above the window

using equation of motion

v² = u² + 2 g h

20.93² = 0² + 2 x 9.8 x h

h = 22.35 m

Hence, the height of the building above the top of window is equal to h = 22.35 m

8 0

2 years ago

Joseph studied whether different materials can block certain electromagnetic waves by testing television reception in different

Sedbober [7]

Joseph's experiment could be improved by using the same antenna at each part of the house during each trial instead of using different antenna. By doing so, he can obtain accurate results how is the signal in different part of the house under the same conditions (despite the location). So, he will see the dependence of the signal on the location. If he uses different antenna, than this antenna can also have influence of the signal.

5 0

3 years ago

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

3 years ago