How does increasing the prong angle affect the launch angle and initial speed of the projectile?

What is the charge of an atom that has lost one electron?

Luden [163]

Answer:

positive

An atom has a even amount of protons, electrons, and a little less neutrons. If the atom lost one electron, then it now has a positive charge.

Hope it helps!

5 0

4 years ago

A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler

Pepsi [2]

Answer:

$331665750000\ m/s^2$

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = $1.989\times 10^{30}\ kg$

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2$

The gravitational acceleration at the surface of such a star is $331665750000\ m/s^2$

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s$

The velocity of the object would be 3257806.62409 m/s

8 0

4 years ago

I need help the option for D is the golfball isnt accelerating

algol13

Answer:

es b

Explanation:

no se

8 0

3 years ago

How many grams of sugar are in 0.5mol C6H12O6?

Anastasy [175]

Answer:

90.07794 grams

Explanation:

5 0

3 years ago

Read 2 more answers

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

RideAnS [48]

Answer:

a) 2.33 m/s

b) 5.21 m/s

c) 882 m³

Explanation:

Using the concept of continuity equation

for flow through pipes

$A_{1}\times V_{1} = A_{2}\times V_{2}$

Where,

A = Area of cross-section

V = Velocity of fluid at the particular cross-section

given:

$A_{1} = 0.070 m^{2}$

$V_{1} = 3.50 m/s$

a) $A_{2} = 0.105 m^{2}$

substituting the values in the continuity equation, we get

$0.070\times 3.50 = 0.105\times V_{2}$

or

$V_{2} = \frac{0.070\times 3.5}{0.150}m/s$

or

$V_{2} = 2.33m/s$

b) $A_{2} = 0.047 m^{2}$

substituting the values in the continuity equation, we get

$0.070\times 3.50 = 0.047\times V_{2}$

or

$V_{2} = \frac{0.070\times 3.5}{0.047}m/s$

or

$V_{2} = 5.21m/s$

c) we have,

Discharge $Q = Area (A)\times Velocity(V)$

thus from the given value, we get

$Q = 0.070m^{2}\times 3.5m/s\$

$Q = 0.245 m^{3}/s$

Also,

Discharge $Q = \frac{volume}{time}$

given time = 1 hour = 1 ×3600 seconds

substituting the value of discharge and time in the above equation, we get

$0.245m^{3}/s = \frac{volume}{3600s}$

or

$0.245m^{3}/s\times 3600 = Volume$

volume of flow = $882 m^{3}$

8 0

4 years ago